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Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes

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Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes

 

Savvas Learning Co Geometry Student Edition Chapter 1 Exercise 1.2 Points Lines And Planes Solutions Page 16  Exercise 1  Problem 1

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 1 Problem 1 Geometrical 1

To find what are two other names for \(\overleftarrow{X Y}\).

Using the method geometrical representation.

From the given
Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 1 Problem 1 Geometrical 2

A line is named by any two points (in any order) on the line such as AB or by a single lowercase such as line l.

\(\overleftarrow{X Y}\) can also be named as \(\overleftarrow{Y X}\) or \(\overleftarrow{X R}\) or \(\overleftarrow{R X}\) or\(\overleftarrow{R Y}\) or\(\overleftarrow{Y R}\)

The other names for \(\overleftarrow{X Y}\) are\(\overleftarrow{Y X}\) or \(\overleftarrow{X R}\) or \(\overleftarrow{R X}\) or\(\overleftarrow{R Y}\) or\(\overleftarrow{Y R}\)

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Exercise 1.2 Points Lines And Planes Savvas Geometry Answers Page 16  Exercise 2  Problem 2

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes

The intersection of the two planes is \(\overleftarrow{R S}\) or \(\overleftarrow{S R}\)

 

Points Lines And Planes Solutions Chapter 1 Exercise 1.2 Savvas Geometry Page 16  Exercise 3  Problem 3

Given: Ray, \(\overline{AB}\)and \(\overline{AB}\)

To find the given rays are same.Using the method Functions of Ray.

A ray is a part of a line that has one endpoint and goes on infinitely in only one direction. Cannot measure the length of a ray.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 3 Problem 3 Direction of ray

A ray is named using its endpoint first, and then any other point on the ray (for example \(\overline{B A}\)).

These two are not the same ray because their directions are different. One starts at A then moves towards B and continues in the direction of B.

The other starts at B and goes toward A continuing in A’s direction.

The Rays \(\overline{A B}\) and \(\overline{A B}\) are not same

 

Page 16  Exercise 4  Problem 4

Given: \(\overleftarrow{E F}\)

To explain about two arrowheads.Using the method Functions of Line.

Can say that a line is an infinitely thin, infinitely long collection of points extending in two opposite directions.

When draw lines in geometry, use an arrow at each end to show that it extends infinitely

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 4 Problem 4 Directions of rays

A line can be named either using two points on the line (for example,\(\overleftarrow{A B}\)).

Arrowheads are used to imply that the lines extend infinitely in both directions.

Arrowheads are used to imply that the lines extend infinitely in both directions.

 

Chapter 1 Exercise 1.2 Points Lines And Planes Savvas Learning Co Geometry Explanation Page 16  Exercise 5  Problem 5

Given: Ray and Line.

To Differentiate.

Using the method Functions of Ray and Line.

The naming of a ray and a line is similar in the sense that use two points named by letters to name each.

However, with a line the order of the letters does not matter because the line continues in both directions beyond these points.

A ray on the other hands has a specific direction, so the order of the letters determines which ray have named because the first letter is always the endpoint with the second letter being the direction in which the ray continues.

The naming of a ray and a line is similar in the sense that use two points named by letters to name each. The order of the letters determines which ray have named because the first letter is always the endpoint with the second letter being the direction in which the ray continues.

 

Solutions For Points Lines And Planes Exercise 1.2 In Savvas Geometry Chapter 1 Student Edition Page 16  Exercise 6  Problem 6

Given:
Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 6 Problem 6 Non collinear Point plane 1

To find other ways to name plane C.

Using the method Functions of Plane.

Given

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 6 Problem 6 Non collinear Point plane 2

Can name a plane either by a letter or by a sequence of 3 noncollinear points which belong to the plane (because3 noncollinear points uniquely determine a plane):GEB, GEF, GEF.

The other way to name plane C is,GEB,GBF,GEF.

 

Solutions For Points Lines And Planes Exercise 1.2 In Savvas Geometry Chapter 1 Student Edition Page 16  Exercise 7  Problem 7

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 7 Problem 7 Collinear Point plane 1

To name three collinear points.

Using the method Functions of collinear.

Three or more points that lie on the same line are collinear points.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 7 Problem 7 Collinear Point Line s

 

The points A,B and C lie on the line m they are collinear.

The points D,B and E lie on the line n they are collinear.

There is no line that goes through all three points A,B and D.

So, they are not collinear.

In this given Figure

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 7 Problem 7 Collinear Point plane 2
Collinear points are,E,B and F because that are all on line n.

Collinear points are E,B and F.

 

Exercise 1.2 Points Lines And Planes Savvas Learning Co Geometry Detailed Answers Page 16  Exercise 8  Problem 8

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 8 Problem 8 Coplanar plane 1

To name coplanar points.

Using the method Functions of Coplanar.

Points or lines are said to be coplanar if they lie in the same plane.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 8 Problem 8 Coplanar plane 2

The points P,Q and R lie in the same plane A.

They are coplanar.

In the given plane
Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 8 Problem 8 Coplanar plane 3

G,E,B and F are all coplanar because they are on plane C .

In the given plane G,E,B and F are all coplanar because they are on plane C.

 

Geometry Chapter 1 Points Lines And Planes Savvas Learning Co Explanation Guide Page 16  Exercise 9  Problem 9

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 9 Problem 9 Line sigment 1

To name the segments.

Using the method Functions of Line segment.

A line segment has two endpoints.

It contains these endpoints and all the points of the line between them.

Can measure the length of a segment, but not of a line.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 9 Problem 9 Line sigment 2

A segment is named by its two endpoints, for example \(\overline{A B}\)).

A segment is any two points that form part of a line. They go from one point to the other, with each point acting as an end point.

Therefore have to name all of the segments that are unique, meaning that they are between different points, and in doing so get all of the ones.

The segments of the figure are, \(\overline{R S}, \overline{S T}, \overline{T W}, \overline{R W}, \overline{W S}, \overline{T R}\)

The segments of the figure are, \(\overline{R S}, \overline{S T}, \overline{T W}, \overline{R W}, \overline{W S}, \overline{T R}\)

 

Page 16  Exercise 10  Problem 10

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 10 Problem 10 Ray of direction line 1

To name the rays in the figure.

Using the method Functions of Ray.

Example for Ray

A ray is a part of a line that has one endpoint and goes on infinitely in only one direction. Cannot measure the length of a ray.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 10 Problem 10 Ray of direction line 2

A ray is named using its endpoint first, and then any other point on the ray (for example \(\overline{B A}\)

To ensure that name every ray have to treat each point as an endpoint for a ray and then follow it in both directions or the line, finding a second point (if there is one) to name it by. The reason don’t name both \(\overline{R S}\) and \(\overline{R T}\) as rays is because they have the same endpoint and then continue in the same direction, making them the same ray.

Therefore these rays would also be acceptable answers in place of another that describes the same ray.

The rays in the given figure are \(\overline{R S}, \overline{S R}, \overline{S T}, \overline{T S}, \overline{T W}, \overline{W T}\)

The rays in the given figure are \(\overline{R S}, \overline{S R}, \overline{S T}, \overline{T S}, \overline{T W}, \overline{W T}\)

 

Page 16  Exercise 11  Problem 11

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 11 Problem 11 End point

 

To name the pair of opposite rays with endpoint T.

Using the method of endpoint function.

Let

Opposite rays are rays that share the same endpoint, but continue along the line in different directions.

Instead of \(\overline{T S}\) we could have also named this ray as \(\overline{T R}\) , because they both describe the same ray that starts at T and continues in the direction of R and S .

The pair of opposite rays with endpoint \(\overline{T R}\)and \(\overline{T R}\).

 

Page 16  Exercise 11  Problem 12

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 11 Problem 12 Opposite Rays End point

To name another pair of opposite rays.

Using the method of endpoint function.

Let

Opposite rays are rays that share the same endpoint, but continue along the line in different directions.

Instead of \(\overline{T S}\) we could have also named this ray as \(\overline{T R}\) because they both describe the same ray that starts at T and continues in the direction of R and S.

The another pair of opposite rays is TW.

 

Page 16   Exercise 12  Problem 13

Given planes UXV and WVS.

To name the intersection of each pair of planes.

Using methods of geometry.

If two distinct planes intersect, then they intersect in exactly one line.

Here UXV and WVS intersect at \(\overrightarrow{V W}\).

In the figure the plane UXV and WVS intersect at \(\overrightarrow{V W}\).

 

Page 16  Exercise 13  Problem 14

Given planes XWV and UVR.

To name the intersection of each pair of planes.

Using methods of geometry.

If two distinct planes intersect, then they intersect in exactly one line.

Here, planes XWV and UVR intersect at \(\overrightarrow{U V}\).

In the figure the plane XWV and−URV intersect at \(\overrightarrow{U V}\)

 

Page 16  Exercise 14  Problem 15

Given plane TXW and TQU.

To name the intersection of each pair of planes.

Using methods of geometry.

If two distinct planes intersect, then they intersect in exactly one line.

The planes TWX and TQU intersect at \(\overrightarrow{X T}\).

In the figure the plane TWX and TQU intersect at \(\overrightarrow{X T}\).

 

Page 16  Exercise 15  Problem 16

Given line TU.

To name two planes that intersect in the given line.

Using methods of geometry.

If two distinct planes intersect, then they intersect in exactly one line.

Here, \(\overrightarrow{T S}\) is the intersection to the back and bottom face of the figure so a possible name pair for them are XTS and QTS.

In the figure the plane XTS, and QTS intersect at line \(\overrightarrow{T S}\)

 

Page 16  Exercise 16  Problem 17

Given Points R,V,W

To Shade the plane that contains the given points.

Using methods of geometry.

The possible plane is RSWV which contains the given points.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 16 Problem 17 RSWV plane

In the figure the shaded plane contains R,V,W points.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 16 Problem 17 Shaded plane

 

Page 16  Exercise 17  Problem 18

Given points U,V,W

To Shade the plane that contains the given points.

Using methods of geometry.

The possible plane is UVWX  which contains the given points.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 17 Problem 18 UVWX plane

The shaded plane contain points U,V,W .

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 17 Problem 18 Shaded plane

 

Page 16  Exercise 18  Problem 19

Given points U,X,X

To Shade the plane that contains the given points.

Using methods of geometry.

The possible plane is UXSR
Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 18 Problem 19 UXSR plane

In the figure the shaded plane contains points U,S,X.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 18 Problem 19 shaded plane

 

Page 16  Exercise 19  Problem 20

Given points T,U,V.

To Shade the plane that contains the given points.

Using methods of geometry.

The possible plane is STUV
Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 19 Problem 20 STUV plane

In the figure the Shaded plane contain points T,U,V

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.2 Points Lines and Planes Page 16 Exercise 19 Problem 20 Shaded plane

 

Page 17  Exercise 20  Problem 21

Given points S,U,V,Y

To determine whether the fourth point is in that plane.

Using methods of geometry.

The pointsS,U,V are in Blue plane.

The point Y is also in Blue plane.

Hence, point Y is co-planar to S,U,V.

The point Y is in Blue plane and co-planar to other points.

 

Page 17  Exercise 21  Problem 22

Given points X,Y,Z,U.

To determine whether the fourth point is in that plane.

Using methods of geometry.

Point Z lies in Yellow plane and other points are in Blue plane.

Hence non-coplanar.

The given points are non-coplanar.

 

Page 17  Exercise 22  Problem 23

Given points X,S,V,U

To determine whether the fourth point is in that plane.

Using methods of geometry.

All the 4 given points are in Blue plane.

Hence, the given points are co-planar.

The given points are co-planar.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry

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Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry

 

Savvas Learning Co Geometry Student Edition Chapter 1 Exercise 1.1 Nets And Drawings Solutions Page 7  Exercise 1  Problem  1

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 1 Problem 1 Isometric 1

To find an isometric drawing of the cube structure.

By using graph method.

The isometric drawing for the given figure should look like

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 1 Problem 1 Isometric 2

The isometric drawing for the given figure should look like

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 1 Problem 1 Isometric 3

 

Exercise 1.1 Nets And Drawings For Visualizing Geometry Savvas Geometry Answers Page 7  Exercise 2  Problem 2

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry

This shows a folded-out flat surface of a figure so it is a net drawing

 

Page 7  Exercise 2  Problem 3

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 2 Problem 3 Orthographic Method

To find the drawing is isometric, orthographic, a net, or none.

By using orthographic method.

This shows three separate views: a top view, a front view, and a right-side view so it is an orthographic drawing.

This shows three separate views: a top view, a front view, and a right-side view so it is an orthographic drawing.

 

Nets And Drawings For Visualizing Geometry Solutions Chapter 1 Exercise 1.1 Savvas Geometry Page 7  Exercise 2  Problem 4

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 2 Problem 4 Isomeric Method

To find the drawing is isometric, orthographic, a net, or none.

By using isomeric method.

This shows a three-dimensional figure using slanted lines to represent depth so it is an isometric drawing.

This shows a three-dimensional figure using slanted lines to represent depth so it is an isometric drawing.

 

Chapter 1 Exercise 1.1 Nets And Drawings Savvas Learning Co Geometry Explanation Page 7  Exercise 2  Problem 5

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 2 Problem 5 Graph Method

To find the drawing is isometric, orthographic, a net, or none.

By using graph method.

None of the three drawings’ description fit the figure.

None of the three drawings’ description fit the figure.

 

Solutions For Nets And Drawings Exercise 1.1 In Savvas Geometry Chapter 1 Student Edition Page 7  Exercise 3  Problem 6

Given: Isometric drawing and an orthographic drawing.

To find the differences and similarities between the given drawing.

By using isometric method.

In an isometric drawing, you see three sides of a figure from one corner view. In an orthographic drawing, you see three separate views of the figure.

In both drawings, you see the same three sides of the figure (top, front, and right).

Also both drawings represent a three-dimensional object in two-dimensions.

In an isometric drawing, you see three sides of a figure from one corner view. In an orthographic drawing, you see three separate views of the figure. In both drawings, you see the same three sides of the figure (top, front, and right). Also both drawings represent a three-dimensional object in two-dimensions.

 

Exercise 1.1 Nets And Drawings Savvas Learning Co Geometry Detailed Answers Page 7  Exercise 4  Problem 7

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 4 Problem 7 Three Dimensional 1(A)

To find the three-dimensional figure with its net.

By using hexagons method.

Looking at this shape, we see that it has bases of triangles and three sides that are rectangles. Therefore knowing this, the only one that fits is A.

Looking at this shape, we see that it has bases of triangles and three sides that are rectangles. Therefore knowing this, the only one that fits is A.

Similarly

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 4 Problem 7 Three Dimensional 1(B)

To find the three-dimensional figure with its net.

By using hexagons method.

Looking at the shape we see that the base and all of the sides are triangles and thus the only net that fits is B)

Looking at the shape we see that the base and all of the sides are triangles and thus the only net that fits is B).

 

Geometry Chapter 1 Nets And Drawings For Visualizing Geometry Savvas Learning Co Explanation Guide Page 7  Exercise 5  Problem 8

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 5 Problem 8 Dimensions

To find the net with its dimensions.

By using isometric method.

Start with the square that is shown on bottom.

First imagine the front side being lain down flat in front of the bottom.

Do the same with both sides and the back.

Then put the top square on any of the side that you have drawn. By looking at the book, label the sides appropriately.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 5 Problem 8 Square sides 1

Start with the square that is shown on bottom. first imagine the front side being lain down flat in front of the bottom. Do the same with both sides and the back. Then put the top square on any of the side that you have drawn. By looking at the book, label the sides appropriately.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 5 Problem 8 Square sides 2

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7  Exercise 6  Problem 9

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 6 Problem 9 Net Dimensions

To find the net with its dimensions.

By using isometric method.

A net is a pattern made when the surface of a three-dimensional figure is laid out flat showing each face of the figure.

The net for the figure should look like

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 6 Problem 9 Net Dimensional 1

The net for the figure should look like

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 6 Problem 9 Net Dimensional 2

 

Page 7  Exercise 7  Problem 10

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 7 Problem 10 Net Dimensions

To find the net with its dimensions.

By using isometric method.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 7 Problem 10 Net Dimensional 1

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 7 Exercise 7 Problem 10 Net Dimensional 2

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 8  Problem 11

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 8 Problem 11 Isometric 1

To find an isometric drawing of each cube structure on isometric dot paper.

By using isometric method.

The isometric drawing for the given figure should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 8 Problem 11 Isometric 2

The isometric drawing for the given figure should look like

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 8 Problem 11 Isometric 3

 

Page 8  Exercise 9  Problem 12

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 9 Problem 12 Isometric

To make an isometric drawing of each cube structure.

Using the method of isometric dot paper.

Let , An isometric drawing of each cube structure.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 9 Problem 12 Isometric of cube 1

Isomeric dot paper is used to draw cube structure.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 9 Problem 12 Isometric of cube 2

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 10  Problem 13

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 10 Problem 13 Isometric
To make an isometric drawing of each cube structure.

Using the method of isometric dot paper.

Let, An isometric drawing of each cube structure

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 10 Problem 13 Isometric of cube 1

Isomeric dot paper is used to draw cube structure.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 10 Problem 13 Isometric of cube 2

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 11  Problem 14

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 11 Problem 14 Isometric

To make an isometric drawing of each cube structure.

Using the method of isometric dot paper.

Let, An isometric drawing of each cube structure

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 11 Problem 14 Isometric of cube 1

Isomeric dot paper is used to draw cube structure.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 11 Problem 14 Isometric of cube 2

 

Page 8  Exercise 12  Problem 15

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 12 Problem 15 Orthographic

To make an orthographic drawing.

Using the method of isometric function.

Let, In an orthographic drawing, remember that:

An orthographic drawing shows the same three views of the isometric drawing (top, front, and right).Solid lines show visible edges.

Dashed lines show hidden edges.

The orthographic drawing for isometric drawing should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 12 Problem 15 Orthographic Top

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 12 Problem 15 Orthographic Front

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 12 Problem 15 Orthographic Right

An orthographic drawing for isometric drawing is drawned.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 13  Problem 16

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 13 Problem 16 Orthographic

To make an orthographic drawing.

Using the method of isometric function.

In an orthographic drawing, remember that

An orthographic drawing shows the same three views of the isometric drawing (top, front, and right).

Solid lines show visible edges.

Dashed lines show hidden edges.

The orthographic drawing for isometric drawing should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 13 Problem 16 Orthographic Top

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 13 Problem 16 Orthographic Front

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 13 Problem 16 Orthographic Right
An orthographic drawing for isometric drawing is drawned.

 

Page 8  Exercise 14  Problem 17

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 14 Problem 17 Orthographic

To make an orthographic drawing.

Using the method of isometric function.

Let , In an orthographic drawing, remember that:

An orthographic drawing shows the same three views of the isometric drawing (top, front, and right).

Solid lines show visible edges.Dashed lines show hidden edges.

The orthographic drawing for isometric drawing should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 14 Problem 17 Orthographic Top

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 14 Problem 17 Orthographic Front

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 14 Problem 17 Orthographic Right

An orthographic drawing for isometric drawing is drawned.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 15  Problem 18

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Solid

To draw eight different nets for the solid given.

Using the method of multiple representation.

Let

1.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 1

2.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 2

3.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 3

4.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 4

5.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 5

 6.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 6

7.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 7

8.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 15 Problem 18 Different nets of Solid 8
The eight different nets for the solid is represented.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 16  Problem 19

Given:  8 cubes

To make an isometric drawing of a structure.

Using the method of isometric function.

Let, A possible structure using 8 cubes is given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 16 Problem 19 cubes 8

The corresponding isometric drawing is given below

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 16 Problem 19 Isometric of cubes 1

The isometric drawing is drawn.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 16 Problem 19 Isometric of cubes 2

 

Page 8  Exercise 16  Problem 20

Given: 8 cubes

To make an orthographic drawing of the structure.

Using the method of orthographic function.

Let

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 16 Problem 20 Orthographic

In an orthographic drawing, remember that:

An orthographic drawing shows the same three views of the isometric drawing (top, front, and right).

Solid lines show visible edges.Dashed lines show hidden edges.

The orthographic drawing for isometric drawing should look like:


Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 16 Problem 20 Orthographic Top

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 16 Problem 20 Orthographic Front

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 16 Problem 20 Orthographic Right

The orthographic drawing is drawned.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 17  Problem 21

Given: Draw a net of the can at the right.

To explain what shape are the top and bottom of the can and uncurl the body.

Using the method of rectangular function.

Let

The shape of the top and bottom of the can are circles.

If we uncurl the body of the can, the shape is a rectangle

which has a length equal to the circumference of each circle and a width equal to the height of the can.

So, a possible net should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 17 Problem 21 Circle 1

The shape at top and bottom of can is a circle. A possible net is described.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 17 Problem 21 Circle 2

 

Page 8  Exercise 18  Problem 22

Given: German printmaker Albrecht Dürer first used the word net.

To find Why he choose the word net.

Using the method of three dimensional function.

Let, Maybe he used the word net because when you use a real net it can wrap around a solid object.

If it’s wrapped tightly the net can take on the overall shape of the wrapped object.

So similarly the printed pattern when you fold it up is like a net wrapping around an object.

The printed pattern when he folds it up is like a net wrapping, so he chooses the word net.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 19  Problem 23

Given:


Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 19 Problem 23 Package of Net

To match the package with its net.Using the method of cubic function.

Let

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 19 Problem 23 Cube

Visualize the right hand side and the left hand side each folding into a half of a cube.

These are the top and bottom.

The piece in the middle then holds them on top of each other.

The Above diagram matches correctly for the package.

 

Page 8  Exercise 20  Problem 24

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 20 Problem 24 Package of Net

To match the package with its net.

Using the method of cubic function.

Let

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 20 Problem 24 Triangular shape

In the above case,looks like opening the triangular shape.

The above diagram matches correctly for the package.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8  Exercise 21  Problem 25

Given:

To match the package with its net.

Using the method of cubic function.

Let

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 21 Problem 25 Triangular of squares
If you break down a Chinese takeout box it wouldn’t have mainly triangles but squares.

The Above diagram matches correctly for the package.

 

Page 9  Exercise 22  Problem 26

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 22 Problem 26 Orthographic

To make an orthographic drawing for the isometric drawing.

Using the method of isometric function.

Let

In an orthographic drawing, remember that:

An orthographic drawing shows the same three views of the isometric drawing (top, front, and right).

Solid lines show visible edges.

Dashed lines show hidden edges.

The orthographic drawing for isometric drawing should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 22 Problem 26 Orthographic Top

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 22 Problem 26 Orthographic Front

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 22 Problem 26 Orthographic Right

An orthographic drawing for isometric drawing is drawn.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9  Exercise 23  Problem 27

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 23 Problem 27 Orthographic

To make an orthographic drawing for the isometric drawing.

Using the method of isometric function.

Let

In an orthographic drawing, remember that

An orthographic drawing shows the same three views of the isometric drawing (top, front, and right).Solid lines show visible edges.

Dashed lines show hidden edges.

The orthographic drawing for isometric drawing should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 12 Problem 15 Orthographic Top

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 12 Problem 15 Orthographic Front

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 8 Exercise 12 Problem 15 Orthographic Right

An orthographic drawing for isometric drawing is drawn.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9  Exercise 24  Problem 28

Given:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 24 Problem 28 Orthographic

To make an orthographic drawing for the isometric drawing.

Using the method of isometric function.

In an orthographic drawing, remember that:

An orthographic drawing shows the same three views of the isometric drawing (top, front, and right).Solid lines show visible edges.

Dashed lines show hidden edges.

The orthographic drawing for isometric drawing should look like:

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 24 Problem 28 Orthographic Top

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 24 Problem 28 Orthographic Front

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise 1.1 Nets and Drawings for Visualizing Geometry Page 9 Exercise 24 Problem 28 Orthographic Right

An orthographic drawing for isometric drawing is drawn.

Savvas Learning Co Geometry Student Edition Chapter 1 Tools Of Geometry Exercise

by

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Exercise

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools Of Geometry Exercise Answers Page 1  Exercise 1  Problem 1

Given: 112

To simplify the above equation.

Using the method of the geometry.

The simplified equation is   112 = 121

 

Page 1  Exercise 2  Problem 2

Given: 2(7.5) + 2(11)

To simplify the above expression.

Using the method of geometry.

2(7.5) + 2(11) = 2(7.5) + 2(11)

2(7.5) + 2(11) = 15 + 22

2(7.5) + 2(11) = 37.

The simplified expression is 2.7.5 + 2.11 = 37.

 

Savvas Learning Co Geometry Student Edition Chapter 1 Tools of Geometry Page 1  Exercise 3  Problem 3

Given:  π(5)2

To simplify the above expression.

Using the method of geometry.

= π(5)2

π = 3.14

π(5)2 = 3.14(5) 2

π(5)2 = 3.14(25)

π(5)2 = 78.5

The simplified expression is π(5)2 = 78.5.

Read and Learn More Savvas Learning Co Geometry Student Edition Solutions

Page 1  Exercise 4  Problem 4

Savvas Learning Co Geometry Student Edition Chapter 1 Tools Of Geometry Exercise

Chapter 1 Tools Of Geometry Exercise Solutions Savvas Learning Co Geometry Page 1  Exercise 5  Problem 5

Given: a = 4, b = −2

To find \(\frac{a−7}{3−b}\)

Using the method of geometry.

⇒ \(\frac{a−7}{3−b}\)

a = 4, b = −2

\(=\frac{4-7}{3-(-2)}\)

 

= \(\frac{-3}{5}\)

The evaluated expression is  \(\frac{a−7}{3−b}\) = \(\frac{-3}{5}\)

 

Page 1  Exercise 6   Problem 6 

Given:  a =  4,b = −2

To find \(\sqrt{(7-a)^2+(2-b)^2}\)

Using the method of geometry

⇒ \(\sqrt{(7-a)^2+(2-b)^2}\)

= \(\sqrt{(7-4)^2+(2+2)^2}\)

= \(\sqrt{3^2+4^2}\)

= \(\sqrt{9+16}\)

= \(\sqrt{25}\)

= 5

The evaluated expression is \(\sqrt{(7-a)^2+(2-b)^2}\) = 5.

 

Savvas Geometry Student Edition Chapter 1 Solutions Tools of Geometry Page 1  Exercise 7  Problem 7

Given: ∣−8∣

To find the absolute value expression.

Using the method of geometry.

⇒ ∣−8∣

= 8

The absolute value is 8.

The absolute value of ∣−8∣ = 8.

 

Page 1  Exercise 8   Problem 8

Given:  ∣2−6∣

To find the absolute value expression.

Using the method of geometry.

⇒ |2-6|

|2-6| = |− 4|

|2-6| =  4

The absolute value is  4

The absolute value of ∣2−6∣ = 4.

 

Tools Of Geometry Savvas Learning Co Chapter 1 Explanation Page 1  Exercise 9  Problem 9

Given: 2x + 7 = 13

To solve the equation.

Using the method of algebra.

2x = 6

x = \(\frac{6}{2}\)

x = 3

The solution of the equation is x = 3.

 

Page 1  Exercise 10   Problem 10

Given: 5x − 12 = 2x + 6

To solve the equation.

Using the method of algebra.

5x − 12 = 2x + 6

5x − 2x = 6 + 12

3x = 18

x = \(\frac{18}{3}\)

x = 6

The solution of the equation is x = 6.

 

Savvas Learning Co Geometry Chapter 1 Tools Of Geometry Explanation Guide Page 1  Exercise 11  Problem 11

Given: 2(x + 3) − 1 = 7x

To solve the equation.

Using the method of algebra.

2(x + 3)−1 = 7x

2x + 6 − 1 = 7x

2x + 5 = 7x

5 = 7x − 2x

5 = 5x

x = 1

The solution of the equation is x = 1.

 

Page 1  Exercise 12  Problem 12

Given: A child can construct models of buildings by stacking and arranging colored blocks.

To find the term construction mean in geometry.

Using the method of geometry.

Construction in Geometry means to draw shapes, angles, or lines accurately.

Construction in Geometry means to draw shapes, angles, or lines accurately. 

 

Savvas Geometry Chapter 1 Answers For Tools Of Geometry Page 1  Exercise 13  Problem 13

Given:  Artists often use long streaks to show rays of light coming from the sun.

To find the properties of a ray .

Using the method of geometry.

A ray is a line with a single endpoint (or point of origin) that extends infinitely in one direction.

A ray is a line with a single endpoint (or point of origin) that extends infinitely in one direction. 

J Susan Milton Introduction To Probability and Statistics Chapter 8 Inferences On The Mean And Variance Of A Distribution Exercise

by

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution

 

Introduction To Probability And Statistics Chapter 8 Exercises Solutions Page 263  Exercise 1  Problem 1

Given problem statement, when programming from a terminal, one random variable response time was recorded in seconds.

These data are tabled also a table was given.

Next draw the stem and leaf diagram and assume the normality is reasonable or not

Stem and leaf plot response time N = 30

Leaf unit = 0.010

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 263 Exercise 1 Problem 1 Stem and leaf 1

Therefore, the step plot shows the data is equally distributed on both sides. So, the assumptions of normality appear reasonable.

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 263 Exercise 1 Problem 1 Stem and leaf 2

Read and Learn More J Susan Milton Introduction To Probability And Statistics Solutions

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 263  Exercise 1  Problem 2

J.Susan Milton Introduction To Probability and Statistics Chapter 8 Inferences On The Mean And Variance Of A Distribution Exercise

Given: When programming from a terminal, one random variable response time was recorded in seconds.

These scenarios can be represented in X.

n = 30

Determine the \(\bar{X}\) value

\(\overline{X_n}=\frac{\sum X_i}{n}\)

 

\(\bar{X}\)n = \(\left(\begin{array}{l}
1.48+1.26+1.52+1.56+1.48+1.46+1.30+1.28+ \\
1.43+1.43+1.55+1.57+1.51+1.53+1.68+1.37+ \\
1.47+1.61+1.49+1.43+1.64+1.51+1.60+1.65+ \\
1.60+1.64+1.51+1.51+1.53+1.74
\end{array} 30\right.\)

Therefore, an unbiased point estimate for σ2  is 0.0129

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 263  Exercise 1  Problem 3

From previous problem the point estimate of σ2 is s2 was obtained and using this value to find a 95 confidence interval for σ2

First determine the value of α

α = 1 − confidence level

Given :

From previous problem the point estimate of σ2 is s2 = 0.0129

Find the value of α is

​α = 1 − 95

α =  0.05
​
n =  30

Find a 95 confidence interval for σ2

Formula is , L1 ≤ σ2 L2

\(\frac{(n-1) S^2}{\chi_{\frac{\alpha}{2}}^2} \leq \sigma^2 \leq \frac{(n-1) S^2}{\chi_{1-\frac{\alpha}{2}}^2}\)

 

Using chi-square distribution table to find a probability value with corresponds to degrees of freedom.

Probability value0.025 that corresponds to 29 degrees of freedom is 45.7

Probability value 0.0975  that corresponds to 29 degrees of freedom is 16

Determine the confidence interval for σ2

\(\frac{(30-1)(0.0129)}{45.7} \leq \sigma^2 \leq \frac{(30-1)(0.0129)}{16.0}\) \(\frac{0.3741}{45.7} \leq \sigma^2 \leq \frac{0.3741}{16.0}\)

 

0.00082 ≤ σ2  0.0234

Hence, 95 confidence interval for σ2 is (0.0082,0.0234)

Therefore,95 confidence interval for σ2 is (0.0082,0.0234)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 263  Exercise 1  Problem 4

From previous problem the point estimate for σ2

Was obtained and using this value to find a 95 confidence interval for σ

First determine the value of α

α = 1 − confidence level

Given :

From previous problem the point estimate for σ2 is 0.0129

Find the value of α is  α = 1−95

α = 0.05

n = 30

Find a 95 confidence interval for σ formula is

\(\sqrt{L_1} \leq \sigma \leq \sqrt{L_2}\)

 

\(\sqrt{\frac{(n-1) S^2}{\chi_{\frac{\alpha}{2}}^2}} \leq \sigma \leq \sqrt{\frac{(n-1) S^2}{\chi_{1-\frac{\alpha}{2}}^2}}\)

 

Using chi-square distribution table to find a probability value with corresponds to degrees of freedom.

Probability value0.95 that corresponds to 29 ,degrees of freedom is 45.7

Probability value 0.025 that corresponds to 29 degrees of freedom is 16

Determine the confidence interval for σ

​\(\sqrt{\frac{(30-1)(0.0129)}{45.7}} \leq \sigma \leq \sqrt{\frac{(30-1)(0.0129)}{16.0}}\)

\(\sqrt{0.0082} \leq \sigma \leq \sqrt{0.0234}\)

 

0.091 0.091 ≤ σ ≤0.153

Hence, 95 confidence interval for σ is (0.091,0.153)


Therefore, 95 confidence interval for σ is (0.091,0.153)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 263  Exercise 2  Problem 5

Given problem statement, highway engineers have found a sign at night and it depends on its surround luminance.

These scenarios can be represented in X.

These surround luminance data are tabled also a table was given.

Estimate the value of X

For determine an unbiased estimate for σ2

Formula for find the point estimate of σ2 is s2

s2 \(=\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

Given : Highway engineers have found a sign at night and it depends on its surround luminance.

These scenario can be represented in X.

n = 30

Determine the \(\bar{X}\) value

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 263 Exercise 2 Problem 5 Solution 1

\(\bar{X}\) = \(\frac{258.6}{30}\)

\(\bar{X}\) = 8.62

The point estimate for σ2  is

\(s^2=\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

 

s2 = \(
\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 263 Exercise 2 Problem 5 Solution 2

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 263 Exercise 2 Problem 5 Solution 3

s2 = \(\frac{592.428}{29}\)

s2 = 20.428

Therefore, an unbiased point estimate for σ2  is, 20.428

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 263  Exercise 2  Problem 6

From the previous problem the point estimate for σ2

was obtained and using this value to find a 90 confidence interval for σ.

First determine the value of α, α = 1− confidence level

Given :

From previous probelm the point estimate for σ2 is

20.428

Find the value of α is

​α = 1 − 90

α = 0.10

n = 30

Find a 90 confidence interval for σ2

Formula is

L1 ≤ σ2 ≤ L2

\(\frac{(n-1) S^2}{\chi_{\frac{a}{2}}^2} \leq \sigma^2 \leq \frac{(n-1) S^2}{\chi_{1-\frac{a}{2}}^2}\)

 

Using chi-square distribution table to find a probability value with corresponds to degrees of freedom.

Probability value 0.05 that corresponds to 29 degrees of freedom is  42.557

Probability value 0.95 that corresponds to 29 degrees of freedom is  17.7084

Determine the confidence interval for σ2

​\(\frac{(30-1)(20.428)}{42.557} \leq \sigma^2 \leq \frac{(30-1)(20.428)}{17.7084}\)
​

\(\frac{5924.12}{42.557} \leq \sigma^2 \leq \frac{5924.12}{17.7084}\)

13.9204 ≤ σ2 ≤ 33.4537

Hence, 90 confidence interval for σ2 is (13.9204,33.4537)

Determine the confidence interval for σ

\(\sqrt{L_1} \leq \sigma \leq \sqrt{L_2}\) \(\sqrt{\frac{(n-1) S^2}{\chi_{\frac{a}{2}}^2}} \leq \sigma \leq \sqrt{\frac{(n-1) S^2}{\chi_{1-\frac{\alpha}{2}}^2}}\)

 

\(\sqrt{13.9204} \leq \sigma \leq \sqrt{33.4537}\)

 

3.731 ≤ σ2 ≤ 5.784

Hence, 90 confidence interval for σ is  (3.731,5.784)

Therefore, 90 confidence interval for σ is  (3.731,5.784)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 264   Exercise 3   Problem 7

Given problem statement, Two voltage technique is used to analyze the crystals. Using electron microprobe to measure both quantitative and qualitative measurements.

These data are tabled also a table was given.

Next draw the stem and leaf diagram and assume the normality is reasonable or not.

Given :

Stem and leaf plot measurement  N = 27

The values have been multiplied by 100

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 264 Exercise 3 Problem 7 Stem and leaf 1
Therefore, the step plot shows the data is equally distributed on both sides. So, the assumptions of normality appear reasonable.

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 264 Exercise 3 Problem 7 Stem and leaf 2

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 264  Exercise 3  Problem 8

Given problem statement, two voltage technique is used to analyze the crystals.

Using electron microprobe to measure both quantitative and qualitative measurements. These scenarios can be represented in X.

These data are tabled also a table was given.

Estimate the value of \(\bar{X}\) for determine an unbiased estimate for σ2

Formula for find the point estimate for σ2 is

\(s^2=\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

 

Given: Two voltage technique is used to analyze the crystals. Using electron microprobe to measure both quantiative and qualiitiative measurements.

These scenario can be represented in X.

n = 27

Determine the \(\bar{X}\) value

\(\bar{X}\) = \(\frac{\sum X_i}{n}\)

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 264 Exercise 3 Problem 8 Solution 1

\(\bar{X}\) = \(\frac{663.9}{27}\)

\(\bar{X}\) = 24.59

The point estimate for σ2 is

\(s^2=\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

 

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 264 Exercise 3 Problem 8 Solution 2

s2 = \(\frac{63.8267}{26}\)

s2 = 2.455

Therefore, an unbiased point estimate for σ2  is, 2.455

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 264  Exercise 4  Problem 9

Given problem statement, find the one-sided confidence interval for the upper bound.

Also, an interval in the form of [0,L].

Finally prove that the upper bound confidence interval

L = \(\frac{(n-1) s^2}{\chi_{1-\alpha}^2}\)

Given:

Find an interval in the form of  P[σ2 ≤ L] = 1 − α

That means Confidence level = 1 − α

Form the diagram, the evidence is

Determine the confidence interval for σ2

P \(\left(\chi_{1-\alpha}^2 \leq \frac{(n-1) s^2}{\sigma^2}\right)\) = 1 − α
​

P \(\left(\sigma^2 \leq \frac{(n-1) s^2}{\chi_{1-\alpha}^2}\right)\) = 1 − α

Hence \(=\frac{(n-1) s^2}{\chi_{1-\alpha}^2}\)

Therefore, the confidence interval for upper bound is \(=\frac{(n-1) s^2}{\chi_{1-\alpha}^2}\) and its proved.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 265  Exercise 5  Problem 10

Given problem statement, Robotic technology was explained.

The Robots are used to apply adhesive to a specified location.

These location data are tabled also a table was given.

Next draw the stem and leaf diagram and assume the normality is reasonable or not.

Given :

Stem and leaf plot measurement N = 25

The values have been multiplied by 1000

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 5 Problem 10 Stem and Leaf measurements 1

Therefore, the step plot shows the data is equally distributed on both sides. So, the assumptions of normality appear reasonable.
Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 5 Problem 10 Stem and Leaf measurements 2

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 265  Exercise 5  Problem 11

Given problem statement, Robotic technology was explained.

The Robots are used to apply adhesive to a specified location. These scenarios can be represented in X.

These location data are tabled also a table was given.

Estimate the value of \(\bar{X}\) For determine an unbiased estimate for σ2

Formula for find the point estimate for σ2 is

\(s^2=\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

 

Given : Robotic technology was explained.

The Robots are used to apply adhesive to a specified location.

These scenarios can be represented in X

n = 27

Determine the \(\bar{X}\) value

\(\bar{X}\) = \(\frac{\sum X_i}{n}\)

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 5 Problem 11 Solution 1

\(\bar{X}\) =\(\frac{0.09}{25}\)

\(\bar{X}\) = 0.0036

The point estimate for  σ2 is

\(\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 5 Problem 11 Solution 2

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 5 Problem 11 Solution 3

s2 = \(\frac{0.00009}{24}\)

s2 = 0.00000375

Therefore, an unbiased point estimate for σ2 is, 0.00000375

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 265  Exercise 6  Problem 12

Initially understand the theorems and using the theorem to prove that mean and variance values such as E[S2 ]= σ2 and Var S2

= \(\frac{2 \sigma^4}{(n-1)}\)

Show that X be a random variable. The mean and variance is
​​E[S2 ]= σ2

Var S2 = \(\frac{2 \sigma^4}{(n-1)}\)

From S2 is an unbiased estimator for σ2. Hence? E[S2] = σ2

From using of formula \(\frac{(n-1) S^2}{\sigma^2} \sim \chi_{(n-1)}\)

​The variance of the chi squared distribution is n 2(n−1)

Now

Var [ \(\frac{(n-1) S^2}{\sigma^2}\)] = 2(n – 1)

\(\frac{(n-1)^2}{\sigma^4}\)Var S2 = 2(n – 1)

Var S2 = 2(n – 1)\(\frac{\sigma^4}{(n-1)^2}\)

Var s2 =\(\frac{2 \sigma^4}{(n-1)}\)

Therefore, X be a random variable then the mean and variance E[S2 ]= σ2,Var S2 \(\frac{2 \sigma^4}{(n-1)}\) and its proved.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 265  Exercise 7  Problem 13

Given Samples: χ0.005 and χ0.95

Formula for find chi squared points: \(\chi_r{ }^2=1 / 2\left[z_r+\sqrt{2 \gamma-1}\right]^2\)

Using above formula to determine the approximate points of the given samples.

Given : χ0.005 and χ0.95

Formula for approximate the chi squared points \(\chi_r{ }^2=1 / 2\left[z_r+\sqrt{2 \gamma-1}\right]^2\)

r is significance level and γ is the degrees of freedom

Approximate the points for  χ0.005

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 7 Problem 13 Solution 1

Approximate the points for χ0.95

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 7 Problem 13 Solution 2

Therefore, approximated points of  χ0.005  and  χ0.95  is 124.061  and 77.652

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 265  Exercise 7  Problem 14

Given: Standard deviation value is s = 7.5 and sample size is n = 150

Using above values to find the confidence interval on deviation.

Formula for find the confidence interval for deviation

\(\sqrt{L_1}=\sqrt{\frac{(n-1) S^2}{\chi_{\alpha / 2}^2}}\sqrt{L_2}=\sqrt{\frac{(n-1) S^2}{\chi_{1-\alpha / 2}^2}}\)

Given:   Standard deviation and sample sizes are included below

s = 7.5

n = 150

Formula for find interval

\(\sqrt{L_1}=\sqrt{\frac{(n-1) S^2}{\chi_{\alpha / 2}^2}} \sqrt{L_2}=\sqrt{\frac{(n-1) S^2}{\chi_{1-\alpha / 2}^2}}\)

 

Determine the interval for χ0.025
Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 7 Problem 14 Solution 1

Determine the interval for χ0.025
Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 7 Problem 14 Solution 2

Confidence interval is

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 265 Exercise 7 Problem 14 Solution 3

Hence, 95 % confidence interval on the standard deviation is (6.725,8.444)

Therefore,95 % the confidence interval on the standard deviation is(6.725,8.444)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 266  Exercise 8  Problem 15

Given: Standard deviation value is s = 0.01 and sample size is n = 100

Using above values to find the confidence interval for deviation One sided confidence interval

\(L=\frac{(n-1) s^2}{\chi_{1-\alpha}^2}\)

 

Given : Standard deviation and sample size is

​s = 0.01

n = 100

Formula for find interval

\(L=\frac{(n-1) s^2}{\chi_{1-\alpha}^2}\)

 

Chi squared points can be approximated by the formula

\(\chi_{1-r}^2=1 / 2\left[z_{1-r}+\sqrt{2 \gamma-1}\right]^2\)

 

Approximate the points for χ0.95

Introduction to Probability and Statistics Principles and Applications Chapter 8 Inferences On The Mean And Variance Of A Distribution Page 266 Exercise 8 Problem 15 Solution

Determine the confidence interval

\(L=\frac{(n-1) s^2}{\chi_{1-\alpha}^2}\)

 

\(\sqrt{L}=\sqrt{\frac{99 \times(0.01)^2}{77.652}}\)

 

= 0.0113

Hence, 95 % confidence interval on the standard deviation is (0,0.0113)

Therefore, 95 % confidence interval on the standard deviation is (0,0.0113)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 266  Exercise 9  Problem 16

Given: t.05 (γ = 8)

In T  Distribution table, the cumulative probability values are given.

If find a critical value, look up the confidence interval in the bottom row of the table.

From T distribution table the row locates 8 and the column of P[Tr ≤ t]  locate 0.95 which corresponds to the critical value is 1.8595

Hence, the value of t.05 (γ = 8) is 1.8595

Therefore, the value of  t.05(γ = 8) is 1.8595

 

J. Susan Milton Chapter 8 Inferences On Mean And Variance Answers Page 266  Exercise 9  Problem 17

Given: t.95 (γ = 8)

In T Distribution table, the cumulative probability values are given.

If find a critical value, look up the confidence interval in the bottom row of the table.

From T distribution table the row locates 8 and the column of P[Tr ≤ t] locate 0.95 which corresponds to the critical value is −1.8595

Therefore, the value of t.95 (γ = 8)  is−1.8595

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 266  Exercise 9  Problem 18

Given:  t 0.975 (γ = 12)

In T Distribution table, the cumulative probability values are given.

If find a critical value, look up the confidence interval in the bottom row of the table.

From T distribution table the row locates 12 and the column of P[Tr  ≤ t] locate 0.975 which corresponds to the critical value is −2.1788

Hence, the value of  t.975 (γ = 12) is −2.1788

Therefore, the value of t 975 (γ = 12) is −2.1788

 

Solutions To Inferences On Mean And Variance Exercises Chapter 8 Susan Milton Page 265  Exercise 10  Problem 19

In this given question, t value is .05.

In this given question, γ value is 50

Have to find a probability for given t value with given γ value.

Point degree of freedom and search for a given t value.

Given t value = .05.

γ = 50.

By using the t table, t⋅05 (γ = 50) = 1.6759

Hence, the probability of given t value ⋅05 with degree of freedom γ = 50 is 1.6759.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 265  Exercise 10  Problem 20

In this given question, t value is .025.

In this given question, y value is 75.

Have to find a probability for given t value with given γ value.

Point degree of freedom and search for a given t value.

Given t

value  t  = .025

γ = 75

By using the t table, t  = .025

(γ = 75) = 1.9921

Hence, the probability of given t value .025 with degree of freedom γ = 75 is 1.9921

 

Chapter 8 Inferences on Mean and Variance examples and answers Susan Milton Page 265  Exercise 10  Problem 21

In this given question, t value is 0.1.

In this given question, γ value is 200.

Have to find a probability for given t value with given γ value.

Point degree of freedom and search for a given t value.

Given

t value = 0.1

γ = 200

By using the t table

t 0⋅1 (γ = 200) = 1.2858

Hence, the probability of given t value 0.1 with degree of freedom γ = 200 is 1.2858

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 266 Exercise 11 Problem 22

In this question, the given data is \(\sum_{i=1}^{20} x_i\) = 25.792 and

\(\sum_{i=1}^{20} x_i^2\) = 33. 261596

Have to find \(\bar{X}\) , s2 , s

\(\bar{X}\) = \(\frac{\sum x_i}{n}\)

s2 = \(\frac{1}{n-1}\left(\sum x_i^2-n(\bar{X})^2\right)\)

s = \(\sqrt{s^2}\)

Given

\(\sum_1^{15} x_i\)  =  0.07

\(\sum_1^{15} x_i^2\) = 0.0489

n = 15

\(\bar{X}\) = \(\frac{\sum x_i}{n}\)

\(\bar{X}\)  = \(\frac{0.07}{15}\)

\(\bar{X}\)  = 0.00467

\(\frac{1}{n-1}\left(\sum x_i^2-n(\bar{X})^2\right)\)

 

= \(\frac{1}{15-1}\left(0 \cdot 0489-15(0 \cdot 00467)^2\right)\)

=  \(\frac{1}{14}(0 \cdot 0489-0 \cdot 000327)\)

= \(\frac{0.048573}{14}\)

=   0.0034695

s = \(\sqrt{s^2}\) = \(\sqrt{0.0034695}\)

s  =  0.0589

Hence, the Value of \(\bar{X}\),s 2, s are 0⋅00467, 0⋅0034695,0⋅0589 respectively.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 266  Exercise 11  Problem 23

To find 95 % confidence interval 100(1 − α)

\(\bar{X} \pm t_{\frac{\alpha}{2}, \frac{n-1 s}{\sqrt{n}}}\) = 0.00467 \(\pm t_{0.05, \frac{15-10.0589}{\sqrt{5}}}\)

​=  0⋅00467 ± 1.7693 × 0.0152

=  0.00467 ± 0.0269

=  (0⋅00467 − 0.0269, 0.00467 + 0.0269)

=  (−0.02223, 0.03157)
​
Hence,95 % confidence interval on the mean outside diameter of the pipes is(−0.02223,0.03157)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 8 Page 266  Exercise  11  Problem 24

The makers of this pipe claim that the mean outside diameter is 1.29, so an average overestimate is 0.05.

The average overestimates the distance by 0.05 which is not reasonable.

Because the overestimates do not lie within 90 confidence interval.

Hence, the confident interval does not lead to suspect this responded.

J Susan Milton Introduction To Probability and Statistics Chapter 5 Joint Distributions Exercises

by

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Joint Distributions 

 

Introduction To Probability And Statistics Chapter 5 Exercises Solutions Page 169  Exercise 1  Problem 1

In Given problem, is a hypergeometric distribution.

Hypergeometric distribution: A random variable X has a hypergeometric distribution with parameters N,n and r if its density is given by

f(x) = \(\frac{\left(\begin{array}{l}
r \\
x
\end{array}\right)\left(\begin{array}{l}
N-r \\
n-x
\end{array}\right)}{\left(\begin{array}{l}
N \\
n
\end{array}\right)}\)

Max [0,n−(N−r)] ≤ x ≤ min(n,r)

Given table
Introduction to Probability and Statistics Principles and Applications Chapter 5 Joint Distributions Page 169 Exercise 1 Problem 1 Hypergeometric function

Find the probability for hypergeometric function

Using given statement to get a required probability such as, N = 7,r = 3

P(X = x) = \(\frac{\left(\begin{array}{l}
r \\
x
\end{array}\right)\left(\begin{array}{l}
N-r \\
n-x
\end{array}\right)}{\left(\begin{array}{l}
N \\
n
\end{array}\right)}\)

Putx = 0 in above equation

P(X = x) = \(\frac{\left(\begin{array}{l}
3 \\
0
\end{array}\right)\left(\begin{array}{l}
7-3 \\
4-0
\end{array}\right)}{\left(\begin{array}{l}
7 \\
4
\end{array}\right)}\)

P(X = x) = \(\frac{1}{35}\)

Therefore, the probability for a hypergeometric function value is \(\frac{1}{35}\) and the table values are verified.

Read and Learn More J Susan Milton Introduction To Probability And Statistics Solutions

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 169  Exercise 1  Problem 2

J.Susan Milton Introduction To Probability and Statistics Chapter 5 Joint Distributions Exercises

Introduction to Probability and Statistics Principles and Applications Chapter 5 Joint Distributions Page 169 Exercise 1 Problem 2 marginal density

Hence, the marginal density was obtained and the variable Y is the Continuous Random Variable.

Therefore, the marginal density was obtained and the variable Y is the Continuous Random Variable.

 


J. Susan Milton Joint Distributions Chapter 5 Answers Page 169  Exercise 1 Problem 3

If two random variables are independent then it satisfies the following conditions,

​1. P(x/y) = P(x)

2. P(x ∩ y) = P(x) ∗ P(y)
​
Also, the joint distribution of a function is fxy (x,y) = fx (x) fy(y)

Two random variables are independent, if the value of one variable does not change the probability value of another variable.

Therefore, If two random variables for independent then satisfies a condition 1. P(x∣y) = P(x) , 2. P(x∩y) = P(x)∗ P(y)

​

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 169   Exercise 2   Problem 4

Given problem, fxy(x,y) = 1/n2

If determine a function has to be joint density function satisfies the below condition,\(\sum_x \sum_y f_{X Y}(x, y)\) = 1

The values of X and Y between

​x = 1,2,3 ,…., n

y = 1,2,3, …., n

​Given: fxy(x,y) = 1/n2

Find joint density function

\(\sum_x \sum_y f_{X Y}(x, y)=\sum_{x=1}^n \sum_{y=1}^n \frac{1}{n^2}\)

 

\(\sum_x \sum_y f_{X Y}(x, y)=\frac{1}{n^2} \sum_{x=1}^n \sum_{y=1}^n 1\)

 

\(\sum_x \sum_y f_{X Y}(x, y)=\frac{n}{n^2} \sum_{y=1}^n 1\)

 

\(\sum_x \sum_y f_{X Y}(x, y)=\frac{n}{n^2}(n)\)

 

\(\sum_x \sum_y f_{X Y}(x, y)\) = 1

Hence, a given function is discrete joint density function and the condition is satisfied.

Therefore, the function is a discrete joint density function and the condition x \(\sum_x \sum_y f_{X Y}(x, y)\) = 1 is satisfied.

 

Solutions To Joint Distributions Exercises Chapter 5 Susan Milton Page 169  Exercise 2  Problem 5

Given problem, fxy(x,y) = 1/n2

If determine a function has to be joint density function satisfies the below condition,\(\sum_x \sum_y f_{X Y}(x, y)\)= 1

The values of X and Y between

​x = 1,2,3, …., n

y = 1,2,3, …., n

Given: fxy(x,y)=1/n2

Find joint density function

Using given function to get a marginal density

Find the marginal density of X

\(f_X(x)=\sum_y f_{X Y}(x, y)\)

 

\(f_X(x)=\sum_1^n \frac{1}{n^2}\)

 

\(f_X(x)=\frac{1}{n^2} \sum^n 1\)

 

\(f_X(x)=\sum_1^n \frac{1}{n^2}\)

 

\(f_X(x)=\frac{1}{n^2} \sum_1^n 1\)

 

Determine the marginal density of Y

\(f_Y(y)=\sum_1^n \frac{1}{n^2}\)

 

\(f_Y(y)=\frac{1}{n^2} \sum_1^n 1\)

 

 \(f_Y(y)=\frac{1}{n^2}(n)\)

 

 \(f_Y(y)=\frac{1}{n}\)

Therefore, the marginal densities of a given function both X and Y is  \(\frac{1}{n}\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 169  Exercise 2  Problem 6

If two random variables are independent then it satisfies the following conditions,

​1. P(x∣y) = P(x)

2. P(x ∩ y) = P(x) ∗ P(y)
​
Also, the joint distribution of a function is

fxy(x,y) = fx(x) fy(y)

Two random variables are independent, if the value of one variable does not change the probability value of another variable.

Given problem,{{f}{XY}}(x,y) = 1/n2

The marginal densities of a given function both X
and Y is \(f_Y(y)=\frac{1}{n}\).

Hence, the values are independent.

Therefore, the given function marginal densities of both X and Y is \(f_Y(y)=\frac{1}{n}\) and independent.

 

Chapter 5 Joint Distributions Examples And Answers Susan Milton Page 169  Exercise 3  Problem 7

Given problem ,fxy(x,y) = 2/n(n+1)

If determine a function has to be joint density function satisfies the below condition \(f_Y(y)=\frac{1}{n}\).

The values of X and Y between

​x = 1,2,3,….,n

y = 1,2,3,….,n
​
Given: fxy (x,y) = 2/n(n + 1)

Find joint density function

\(\sum_{y=1}^n \sum_{x=1}^n f_{X Y}(x, y)=\sum_{y=1}^n \sum_{x=1}^n \frac{2}{n(n+1)}\)

 

\(\sum_{y=1}^n \sum_{x=1}^n f_{X Y}(x, y)=\frac{2}{n(n+1)} \sum_{y=1}^n \sum_{x=1}^n \)

 

Sum of first n integers is given by \(\frac{n(n+1)}{2}\)

\(\sum_{y=1}^n \sum_{x=1}^n f_{X Y}(x, y)=\frac{2}{n(n+1)} \times \frac{n(n+1)}{2}\)

 

\(\sum_{y=1}^n \sum_{x=1}^n f_{X Y}(x, y)\) =  1

 

Hence, a given function is discrete joint density function and the condition is satisfied.

Therefore, the function is a discrete joint density function and the condition \(\sum_{y=1}^n \sum_{x=1}^n f_{X Y}(x, y)\)= 1 is satisfied.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 169  Exercise 3  Problem 8

Given problem, fxy (x,y) = 2/n(n + 1)

If determine a function has to be joint density function satisfies the below condition \(f_Y(y)=\frac{1}{n}\).

The values of X and Y between

​x = 1,2,3,….,n

y = 1,2,3,….,n

Using given function to get a marginal density

Find the marginal density of X,

\(f_X(x)=\sum_y f_{X Y}(x, y)\)

 

\(f_X(x)=\sum_{y=1}^n \frac{2}{n(n+1)}\)

 

\(f_X(x)=\frac{2}{n(n+1)} \sum_{y=1}^n 1\)

 

\(f_X(x)=\frac{2}{n(n+1)}(n)\)

 

\(f_X(x)=\frac{2}{(n+1)}\)

 

Determine the marginal density of Y

\(f_X(x)=\sum_y f_{X Y}(x, y)\)

 

\(f_X(x)=\sum_{y=1}^n \frac{2}{n(n+1)}\)

 

\(f_X(x)=\frac{2}{n(n+1)} \sum_{y=1}^n 1\)

 

\(f_X(x)=\frac{2}{n(n+1)}(n)\)

 

\(f_X(x)=\frac{2}{(n+1)}\)

 

Therefore, the marginal densities of a given function both X and Y \(f_X(x)=\frac{2}{(n+1)}\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 169  Exercise 3  Problem 9

If two random variables are independent then it satisfies the following conditions,

​1. P(x∣y) = P(x)

2. P(x∩y) = P(x) ∗ P(y)
​
Also, the joint distribution of a function is

fxy(x,y) = fx(x) fy(y)

Two random variables are independent, if the value of one variable does not change the probability value of another variable.

Given problem, fxy (x,y) = 2/n(n + 1)

The marginal densities of a given function both X and Y is \(f_X(x)=\frac{2}{(n+1)}\)

Hence, the values are independent.

Therefore, the given function marginal densities of both X and Y is \(f_X(x)=\frac{2}{(n+1)}\) and its independent.

 

Probability And Statistics J. Susan Milton Chapter 5 Solved Step-By-Step Page 170  Exercise 4  Problem 10

In Given table,X represents the number of syntax errors and Y represents the number of errors in logic.

Problem statement: Determine the probability for selected program have neither of these errors.

Given table
Introduction to Probability and Statistics Principles and Applications Chapter 5 Joint Distributions Page 170 Exercise 4 Problem 10 Logic syntax errors

In above table,X represents the number of syntax errors and Y represents the number of errors in logic.

Find the probability

Using given statement to get a required probability such as, p(x = 0,y = 0)

Hence, the value of p(x = 0,y = 0) is 0.4

Hence, the probability that selected program have neither these types of errors as 0.4

Therefore, the probability that selected program have neither these types of errors as 0.4

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 170  Exercise 4  Problem 11

In Given table,X represents the number of syntax errors and Y represents the number of errors in logic.

Problem statement: Determine the probability for selected program at least one syntax error and at most one error in logic.

Given table,In above table,X represents the number of syntax errors and Y
Introduction to Probability and Statistics Principles and Applications Chapter 5 Joint Distributions Page 170 Exercise 4 Problem 11 Logic Syntax errors

Find the probability

Using given statement to get a required probability such as,

P[X ≥ 1 and Y ≤ 1]

P[X ≥ 1and Y ≤ 1]

[P(X = 1,Y = 0) +  P(X = 2,Y = 0) + P(X = 3,Y = 0)

+ P(X = 4,Y = 0) + P(X = 5,Y = 0) + P(X = 1,Y = 1)

+ P(X = 1,Y = 2) +  P(X = 1,Y = 3) + P(X = 1,Y = 4)

+ P(X = 1,Y = 5)]

P[X ≥ 1and Y ≤ 1]​ = 0.300 + 0.040 + 0.009 + 0.008 + 0.005 + 0.040 + 0.010+ 0.008 + 0.007 + 0.002

P[X ≥ 1and Y ≤ 1] = 0.429

Hence, the probability for selected at least one syntax error and at most one error in logic is 0.429

Therefore, the probability for selected at least one syntax error and at most one error in logic is 0.429

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 170  Exercise 4  Problem 12

Using given table values for determine the marginal density of the function.

If two random variables with joint density fXY then the marginal density for X denoted as fx given by

\(f_X(x)=\sum_y f_{X Y}(x, y)\)

 

The mariginal density for y denoted as

\(f_Y(y)=\sum_y f_{X Y}(x, y)\)

 

Given table
Introduction to Probability and Statistics Principles and Applications Chapter 5 Joint Distributions Page 170 Exercise 4 Problem 12 Logic Syntax errors

In above table, X represents the number of syntax errors and y represents the number of errors in logic.

Using Given table, sum all values for determine a marginal density
Introduction to Probability and Statistics Principles and Applications Chapter 5 Joint Distributions Page 170 Exercise 4 Problem 12 marginal density

Hence, the marginal density was obtained

Therefore, the marginal density for both values are obtained in above table.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 170  Exercise 5  Problem 13

On previous example to get a function is, fxy(x,y)= \(\frac{1.72}{x}\)

If determine a function has to be joint density function satisfies the below condition

\(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X Y}(x, y) d x d y=1\)

 

The values of $X$ and $Y$ between 27 ≤ y ≤ x ≤ 33

Given: \(f_{X Y}(x, y)=\frac{1.72}{x}\)

Use continuous joint density function to find the value of P[X ≤ 30 and Y ≤ 28]

P[X ≤ 30 and Y≤ 28]= \(\int_{27}^{30} \int_{27}^{28} f_{X Y}(x, y) d x d y\)

\(=\int_{27}^{30} \int_{27}^{28} \frac{1.72}{x} d x d y\)

 

Integrate depends on y and apply the limit values in given function

\( = 1.72 \int_{27}^{30} \frac{1}{x}[y]_{27}^{28} d x\)

 

\( = 1.72 \int_{27}^{30} \frac{1}{x} d x\)

 

Integrate depends on x and apply the limit values in given function

P[X ≤ 30 and Y≤28] = 1.72× \([\ln x]_{27}^{30}\)

P[X ≤ 30 and Y≤28]  ​= 1.72(3.4012 − 3.2958)

P[X ≤ 30 and Y≤28]  =  0.1813
​
Therefore, using previous example to find a function as fxy (x,y)= \(\frac{1.72x}{x}\) and the value of P[X ≤ 30 and Y ≤ 28] is 0.1813.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 170  Exercise 6  Problem 14

On previous example to get a function is,f xy(x,y) = \(\frac{c}{x}\)

If determine a function has to be joint density function satisfies the below condition \(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X Y}(x, y) d x d y\) = 1

The values of X and Y between 27 ≤ y ≤ x ≤ 33

Given: fxy (x,y) = \(\frac{c}{x}\)

Use continuous joint density function to find the value of c

\(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X Y}(x, y) d y d x\) = 1

​\(\int_{27}^{33} \int_{27}^x \frac{c}{x} d y d x\)= 1

Integrate depends on y and apply the limit values in given function

\(\int_{27}^{33}\left(\frac{c}{x} y\right)_{27}^x d x\) = 1

\(\int_{27}^{33}\left(c-\frac{c}{x}(27)\right) d x\) = 1

Integrate depends on x and apply the limit values in given function

\(\int_{27}^{33} c d x-27 \int_{27}^{33} \frac{c}{x} d x\)= 1

6c − 27c (ln(33)−3ln(3)) = 1

6c − 5.4181c = 1

c = \(\frac{1}{0.5819}\)

c  = 1.72
​
Therefore, using previous example to find a function as f XY(x,y) = \(\frac{1.72}{x}\)with 27 ≤ y ≤ x ≤ 33 and the value of c is 1.72

 

Online Help For J. Susan Milton Joint Distributions Chapter 5 Exercises Page 170  Exercise 6  Problem 15

On previous example to get a function is \(f_{X Y}(x, y)=\frac{1.72}{x}\)

If determine a function has to be joint density function satisfies the below condition, \(f_X(x)=\int_{-\infty}^{\infty} f_{X Y}(x, y) d y\) 27 ≤ y ≤ x ≤ 33

Given: \(f_{X Y}(x, y)=\frac{1.72}{x}\)

Using given function to get a marginal density

Find the marginal density of X

\(f_X(x)=\int_{-\infty}^{\infty} f_{X Y}(x, y) d y\) \(f_X(x)=\int_{27}^{28} \frac{1.72}{x} d y\)

 

Integrate depends on Y and apply the limit values in given function

\(f_X(x)=1.72\left(\frac{1}{x}\right)[y]_{27}^{28}\)

 

\(f_X(x)=1.72\left(\frac{1}{x}\right)[28-27]\)

 

\(f_X(x)=1.72\left(\frac{1}{x}\right)\)

 

Therefore, Therefore, the marginal density for X and the value of P[X≤28] is \(f_{X Y}(x, y)=\frac{1.72}{x}\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 170  Exercise 7  Problem 16

Given problem, fxy (x,y) = c(4x + 2y + 1)

If determine a function has to be joint density function satisfies the below condition

\(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X Y}(x, y) d x d y\) = 1

The values of X and Y between

​0 ≤ x ≤ 40

0 ≤ y ≤ 2
​
Given: fxy (x,y) = c(4x + 2y + 1)

Use continuous joint density function to find the value of c

\(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f_{X Y}(x, y) d x d y\) = 1

 

\(\int_0^{40} \int_0^2 c(4 x+2 y+1) d x d y\) = 1

 

Integrate depends on y and apply the limit values in given function

\(c \int_0^{40}\left(4 x y+2 \frac{y^2}{2}+y\right)_0^2 d x\) = 1

 

\(c \int_0^{40}\left(4 x(2)+(2)^2+2\right) d x=\) 1

 

Integrate depends on x and apply the limit values in given function,

\(c \int_0^{40}(8 x+6) d x\)= 1

\(c\left(\frac{8 x^2}{2}+6 x\right)_0^{40}\)= 1

\(c\left(\frac{8(40)^2}{2}+6(40)\right)\)= 1

6640c  = 1

c = \(\frac{1}{6640}\)

Therefore, the given function fXY (x,y)=c(4x+2y+1) and the value of c is \(\frac{1}{6640}\)= 1

 

Step-By-Step Guide To Joint Distributions Exercises Chapter 5 Milton Page 170  Exercise 7  Problem 17

To solve this, we need to integrate the PDF where X and Y defined as follow.

​P(x > 20,y ≥ 1)

20 < x ≤ 0,1 ≤ y ≤ 2

P(x > 20,y ≥ 1) = 1 \(\int_1^2 \int_{20}^{40} \frac{1}{6640}(4 x+2 y+1) d x d y\)

P(x > 20,y ≥ 1)  =  \(\frac{1}{6640} \int_1^2\left(\frac{4 x^2}{2}+2 y x+x\right) \int_{20}^{40} d y\)

P(x > 20,y ≥ 1)  = \(\frac{1}{6640} \int_1^2(40 y+2420) d y\)

P(x > 20,y ≥ 1)  = \(\frac{1}{6640} \int_1^2(40 y+2420) d y\)

P(x > 20,y ≥ 1)  = \(=\frac{1}{6640}(2480)\)

P(x > 20,y ≥ 1)  = \(\frac{38}{83}\)

The probability of temperature is P(x > 20,y ≥ 1) =\(\frac{38}{83}\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 171  Exercise 8  Problem 18

Note that the integral of a valid joint density is equal to 1.

To verify joint density for a two-dimensional random variable.

f xy (x,y) =  \([\frac{1}{x}\),o < y < x < 1

That the integral of a valid joint density is equal to 1.

1= \(1\iint_R f_{X, Y}(x, y), o<y<x<1\)

= \(\int_0^1 \int_0^x \frac{1}{x} d y d x\)

= \(\left.\int_0^1\left(\frac{y}{x}\right)\right|_0 ^x d x\)

= \(\left.(x)\right|_0 ^1\)

1 = 1

It is a joint density variable.

 

Exercise Solutions For Chapter 5 Susan Milton Joint Distributions Page 171   Exercise 8  Problem 19

We can get the probability by integrating it into the following regions.

To find P(X ≤ 0.5 and Y ≤ 0.25)

We can get the probability by integrating it on the following regions

0 < y ≤ 0.25

0.25≤ x < 0.5

Thus, we get that

P(X ≤ 0.5 and Y ≤ 0.25) = \(\int_{0.25}^{0.5} \int_0^{0.25} \frac{1}{x} d y d x\)

P(X ≤ 0.5 and Y ≤ 0.25)  = \(\left.\int_{0.25}^{0.5}\left(\frac{y}{x}\right)\right|_0 ^{0.25} d x\)

P(X ≤ 0.5 and Y ≤ 0.25)  = \(\left.0.25[\ln (x)]\right|_0 ^{0.25}\)

P(X ≤ 0.5 and Y ≤ 0.25)  = 0.25 [(ln(0.5)] − [ln(0.25)]

P(X ≤ 0.5 and Y ≤ 0.25)  = 0.1733

Hence, we have found the answer for P(X ≤ 0.5and Y ≤ 0.25) = 0.1733

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 171  Exercise 9  Problem 20

The marginal density distribution of a subset of a collection of random variables is the probability distribution.

To find the marginal densities for X and Y.

\(f_{X, Y}(x, y) = \frac{x^2 y^2}{16}, 0 \leq x, y \leq 2\)

 

To calculate the marginal density for X, by definition, we have to calculate the integral

\(\left.f_X(x)=\int \mathbb{r} f_{(} X, Y\right)(x, y) d y\)

 

By plugging in the values where y is defined and the expression for joint density function, we obtain.

\(f_X(x)=\int_0^2 \frac{x^3 y^3}{16} d y=\frac{x^3}{64}(16-0)=\frac{x^3}{4}\), 0 ≤ x ≤ 2
​

To calculate the marginal density for Y, by definition, we have to calculate the integral

\(\left.f_Y(y)=\int \mathbb{r} f_{(} X, Y\right)(x, y) d x\)
​

By plugging in the values where y is defined and the expression for joint density function, we obtain.

\(f_X(x)=\int_0^2 \frac{x^3 y^3}{16} d y\)
​
\(=\frac{x^3}{64}(16-0)\)

\(=\frac{x^3}{4}\), 0 ≤ x ≤ 2

\(\frac{x^3}{4}\) 0 ≤ x ≤ 2

\(\frac{x^3}{4}\) 0 ≤ x ≤ 2

⇒ \(f_Y(y)=\frac{y^3}{4}, 0 \leq y \leq 2\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 171  Exercise 9  Problem 21

The marginal density distribution of a subset of a collection of a random variables is the probability distribution.

To explain if X and Y are independent?

To check if X and Y are independent, we can simply check whether or not the following equation is fulfilled.

fx,y(x,y) =  fx(x) fy(y)

By substituting the calculated expression , we obtain

\(f_{X, Y}(x, y)=\frac{x^3 y^3}{16}=\frac{x^3}{4} \frac{y^3}{4}=f_X(x) f_Y(y)\)

 

Hence X and Y are indeed independent.

Hence X and Y are indeed independent.

 

Page 171  Exercise 9  Problem 22

The marginal density distribution of a subset of a collection of random variables is the probability distribution.

To find P(X ≤ 1)

To find the probability P(X≤1), by definition, we have to calculate integral

P(X≤1) \(=\int_{-\infty}^1 f_X(x) d x\)

By substituting the given expression for the marginal density function and the values where x is defined, we proceed to calculate

P(X ≤ 1) \(\left.=\int_0^{\frac{x 3}{4}} d x=\frac{1}{16}-(1-0)=\frac{(}{1}\right)(16)\)

Therefore, the joint density for (X,Y) is given by \(\frac{1}{6}\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 171  Exercise 10  Problem 23

fx,y(x,y) = c,   20 < x < y < 40 The integral of the PDF should always be equal to l, where x and y is defined.

Thus we get

\(=\iint_R f_{X, Y}(x, y) d x d y\)

 

\(=\int_{20}^{40} \int_{20}^y c d x d y\)
​
\(=c \int_{20}^{40}(y-20) d y\)

\(=\left.c\left(\frac{y^2}{2}-20 y\right)\right|_{20} ^{40}\)

= c (200)

= c\(\frac{1}{200}\)

The value of c is c\(\frac{1}{200}\)

 

Page 171  Exercise 10  Problem 24

The integral of the PDF should always be equal to l, where x and y are defined, then integrating the PDF.

To find the probability that the carrier will pay at least $25 per barrel and the refinery will pay at most $30 per barrel for the oil.

We can express the problem

P(X ≥ 25 and Y ≥ 30)

By integrating the PDF, we get that
​
​P(X ≥ 25 and Y ≥ 30)

​​P(X ≥ 25 and Y ≥ 30\(=\int_{30}^{40} \int_{25}^y \frac{1}{200} d x d y\)

​P(X ≥ 25 and Y ≥ 30) = \(\frac{1}{200} \int_{30}^{40}(y-25) d y\)

​P(X ≥ 25 and Y ≥ 30)= \(\left.\frac{1}{200}\left(\frac{y^2}{2}-25 y\right)\right|_{30} ^{40}\)

​P(X ≥ 25 and Y ≥ 30)= \(\left.\frac{1}{200}\left(\frac{y^2}{2}-25 y\right)\right|_{30} ^{40}\)

​P(X ≥ 25 and Y ≥ 30)= \(\frac{1}{200}\)

​P(X ≥ 25 and Y ≥ 30)= \(\frac{1}{2}\)

P(X ≥ 25 and Y ≥ 30) = \(\frac{1}{2}\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 171  Exercise 10  Problem 25

The integral of the PDF should always be equal to l, where x and y are defined, then integrating the PDF.

To find the probability that the price paid by the refinery exceeds that of the carrier by at least $10 per barrel.

We can express the problem

P(Y > X + 10)

Now if we graph the domain, we get

Introduction to Probability and Statistics Principles and Applications Chapter 5 Joint Distributions Page 171 Exercise 10 Problem 25 Domain

 

Where

Point A = (20,40)

Point B = (30,40)

Point C = (20,30)

CB  = y − 10

By integrating the PDF
​
​P(Y>X+10)= \(\int_{30}^{40} \int_{25}^y \frac{1}{200} d x d y\)

\(=\left.\frac{1}{200} \int_{30}^{40}(x)\right|_{20} ^{y-10} d y\)

 

\(=\frac{1}{200} \int_{30}^{40}(y-10-20) d y\)

 

\(=\frac{1}{200} \int_{30}^{40}(y-30) d y\)

 

\(=\left.\frac{1}{200}\left(\frac{y^2}{2}-30 y\right)\right|_{30} ^{40}\)

 

= \(\frac{50}{200}\)

 

= 0.25

The probablity is P(Y > X + 10) = 0.25

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 5 Page 172  Exercise 11 Problem 26

From Continuous Joint density we get that the three properties and identify a function as a density (X1,X2,X3,….. .Xn) f X1, X2, X3 ,. ….. Xn(x1,x2,x3, ….. .x1 )≥0]−∞<Xi< ∞

For all Xi where i is from 1 to n

\(\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \cdots \cdots \int_{-\infty}^{\infty}\) f X1, X2, X3 ,…… X1 (x1 ,x2,x3,……x1 ) dx1 dx2 dx3……..

P[a ≤ X1 ≤ b,c ≤ X ≤ d,e ≤ X3 ≤ f,…..,g ≤ X1 ≤ h

\(\int_a^b \int_c^d \int_e^f \cdots \cdots \int_g^h\)  f X1, X2, X3 ,…… X(x1,x2,x3,……xn) dx1dx2 dx3 ………..;. Where a, b, c, …., h are real

Hence, a, b, c, …., h is called the joint density for (X1 ,X2,X3,…..,Xn)

J Susan Milton Introduction To Probability and Statistics Chapter 7 Estimation Descriptive Distributions Exercises

by

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Estimation Descriptive Distributions

 

Introduction To Probability And Statistics Chapter 7 Exercises Solutions Page 221  Exercise 1  Problem 1

Given problem, X1,X2,X3 ,…..,X20

In the given information, the population mean μ and variance σ were given.

Using the given values to find the sample mean and variance value.

Given: The population mean(μ)is 8 and the variance (σ)is 5.

Determine the mean of \((\bar{X})\)

The sample mean \((\bar{X})\) is an unbiased estimator of the population mean(μ)

The mean \((\bar{X})\) of the sample m X1,X2,X3 ,…..,X20   is the population mean.

Hence, the sample mean value is \((\bar{X})\) = 8

Determine the variance of \((\bar{X})\)

var \((\bar{X})\) = \(\frac{\sigma^2}{n}\)

Substitute n = 20 and σ2 = 5 in previous term

Var \((\bar{X})\) = \(\frac{5}{20}\)= 0.25

Therefore, the mean and variance of \((\bar{X})\) is Mean: 8 Variance: 0.25

Read and Learn More J Susan Milton Introduction To Probability And Statistics Solutions

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 221  Exercise 2  Problem 2

J.Susan Milton Introduction To Probability and Statistics Chapter 7 Estimation Descriptive Distributions Exercises
​
Therefore, an unbiased estimator for the given parameter λs is X. X is the unbiased estimator of λs

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 221  Exercise 3  Problem 3

Given problem statement, X is the number of paint defects in a square yard section of car body painted by robot.

A random sample from Poisson distribution with parameter λs.

Poisson distribution: P(X = x) = \(\frac{e^{-\lambda}(\lambda)^x}{x !}\)

Given : The random variable X is the number of paint defects in a square yard section of car body painted by robot.

Also X is the Poisson distribution with parameter λs.

Find the unbiased estimate for λs

Hence, the sample mean is the unbiased estimator of the population mean.

Then the estimator is

\(\hat{\mu}=\bar{X}\)  ,\(\widehat{\lambda s}=\bar{X}\)

Therefore, an unbiased estimator for λs \(\hat{\mu}=\bar{X}\) ,\(\widehat{\lambda s}=\bar{X}\)

 

J. Susan Milton Estimation Chapter 7 Descriptive Distributions Answers Page 221 Exercise 3  Problem 4

Given problem statement,X is the number of paint defects in a square yard section of car body painted by robot.

A random sample from Poisson distribution with parameter λs.

Determine the unbiased estimate for the average number of flaws per square yard.

Given: The random variable X is the number of paint defects in a square yard section of car body painted by robot.

Also X is the Poisson distribution with parameter λs.

Find the unbiased estimate for λs

Hence, the sample mean is the unbiased estimator of the population mean. Then the estimator is

\(\hat{\mu}=\bar{X}\), \(\widehat{\lambda s}=\bar{X}\)

Determine the unbiased estimate for the average number of flaws per square yard is

ΣiXi = 8 + 0 + 2 + 5 + 3 + 7 + 0 + 1 + 9 + 10 + 12 + 6 = 63

λ8 = \(\frac{63}{12}\)

λ8 = 5.25
​
Therefore, the unbiased estimate for the average number of flaws per square yard is, 5.25

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 221  Exercise 3  Problem  5

Given problem statement,X is the number of paint defects in a square yard section of car body painted by robot.

A random sample from Poisson distribution with parameter λs.

Determine the unbiased estimate for the average number of flaws per square foot.

​Initally,Determine the unbiased estimate for the average number of flaws per square yard is
​

ΣiXi = 8 + 0 + 2 + 5 + 3 + 7 + 0 + 1 + 9 + 10 + 12 + 6 = 63

​λ8 = \(\frac{63}{12}\)

​λ8 = 5.25

Determine the unbiased estimate for the average number of flaws per square foot is

One yard = Three feet

Hence, unbiased estimate

5.25 ×  3 = 15.75

Therefore, the unbiased estimate for the average number of flaws per square foot is 15.75

 

Solutions To Estimation And Descriptive Distributions Exercises Chapter 7 Milton Page 221  Exercise 4  Problem  6

Given problem statement,X is the number of requests for the system received per hour.

A random sample from Poisson distribution with parameter λs.

Poisson distribution: P(X = x) = \(\frac{e^{-\lambda}(\lambda)^x}{x !}\)

Given : The random variable X is the number of requests for the system received per hour.

Also X is the Poisson distribution with parameter λs.

Find the unbiased estimate for λs

Hence, the sample mean is the unbiased estimator of the population mean.

Then the estimator is  \(\hat{\mu}=\bar{X}\), \(\widehat{\lambda s}=\bar{X}\)

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 ​Page 221  Exercise 4  Problem  7

Given problem statement,X is the number of requests for the system received per hour.

A random sample from Poisson distribution with parameter λs.

Determine the unbiased estimate for the average number of request received per hour.

Given : The random variable X is the number of requests for the system received per hour.

Also X is the Poisson distribution with parameter λs.

Find the unbiased estimate for λs

Hence, the sample mean is the unbiased estimator of the population mean. Then the estimator is

\(\hat{\mu}=\bar{X}\) , \(\widehat{\lambda s}=\bar{X}\)

Determine the unbiased estimate for the average number of requests received per hour is

 ΣiXi = 25 + 30 + 10 + 20 + 24 + 23 + 20 + 15 + 4 = 171

λ8 = \(\frac{171}{9}\)

λ8 = 19

​Therefore, the unbiased estimate for the average number of flaws per square yard is 19.

 

Chapter 7 Estimation And Distributions Examples And Answers Susan Milton ​Page 221  Exercise 4  Problem  8

Given problem statement,X is the number of requests for the system received per hour.

A random sample from Poisson distribution with parameter λs.

Determine the unbiased estimate for the average number of request received per quarter hour.

Determine the unbiased estimate for the average number of requests received per hour is
​
 ΣiXi  = 25 + 30 + 10 + 20 + 24 + 23 + 20 + 15 + 4 = 171

λ8 = \(\frac{171}{9}\)

λ8 = 19

Determine the unbiased estimate for the average number of requests received per quarter hour is

1hour = 60minutes

45 Minutes = \(\frac{45}{60}\) hour

Hence, unbiased estimate is

19 × \(\frac{45}{60}\)

= 14.25

Therefore, the unbiased estimate for the average number of requests received per quarter hour is,14.25

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 222  Exercise 5  Problem  9

Given problem statement,X is the distance in inches from the anchored end of rod to the crack location.

A random sample from binomial distribution with interval (0,b).

Determine the unbiased estimator for the average distance.

Given: The random samples X defines the distance in inches from the anchored end of rod to the crack location.

Also X follows the binomial distribution with interval(0,b).

Hence, the sample mean value is the unbiased estimator for the population mean.

Find the unbiased estimator for the average distance \(\hat{\mu}=\bar{X}\)

Therefore, the unbiased estimator for the average distance in inches from the anchored end of rod to the crack location is \(\hat{\mu}=\bar{X}\)

 

Probability And Statistics J. Susan Milton Chapter 7 Solved Step-By-Step Page 222  Exercise 5  Problem 10

Given problem statement is the distance in inches from the anchored end of rod to the crack location.

A random sample from binomial distribution with interval (0,b).

Next determine the estimate for p at approximately.

Given: The random samples X defines the distance in inches from the anchored end of rod to the crack location..

Also X follows the binomial distribution with interval (0,b).

Hence, the sample mean value is the unbiased estimator for the population mean.

Find the unbiased estimator for the average distance \(\hat{\mu}=\bar{X}\)

Determine the variance for X is s2

s2 \(=\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

s2 =  \(\frac{1}{10 – 1}\)  (​(10−9.7)2+(8−9.7)2+(7−9.7)2 + (9−9.7)2 + (11−9.7)2 + (10−9.7)2 + (12−9.7)2 +(9−9.7)2 + (8−9.7)2 +(13−9.7)2)

s2  =\(\frac{0.09+2.89+7.29+0.49+1.69+0.09+5.29+0.49+2.89+10.89}{9}\)

​s2 = \(\frac{32.1}{9}\)

s2 = 3.567
​

Therefore, an estimate for p based on the given data it is approximately 3.567.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 222  Exercise 5  Problem 11

Given problem statement,X is the distance in inches from the anchored end of rod to the crack location.

A random sample from binomial distribution with interval (0,b).

Next determine the estimate for b.

Given: The random samples X defines the distance in inches from the anchored end of rod to the crack location..

Also X follows the binomial distribution with interval(0,b).

Hence, the sample mean value is the unbiased estimator for the population mean.

Find the unbiased estimator for the average distance \(\hat{\mu}=\bar{X}\)

Equating \(\bar{X}\) and E(x)

\(\frac{b}{2}\) = 9.7

b = 9.7 × 2

b = 19.4

Therefore, an estimate for b based on the given data is 19.4.

 

Online Help For J. Susan Milton Estimation Chapter 7 Exercises Page 222  Exercise 6  Problem 12

Given problem statement, showS is not unbiased estimator for σ.

Assume that E(S)=σ and using the contradiction method to show that S is not unbiased estimator for σ.

Given: Hence, the sample variance value is the unbiased estimator for the population variance.

Show that S is not unbiased estimator

E(S)= σ2

Assume,E(S) = σ

Based on theorem 3.3.2

V(X) = E(X2) −(E(X))2

Hence

​V(S)=E(S2) − (E(S))2

V(S) = σ2 − σ2

V(S) = 0
​
Therefore,S is not unbiased estimator for σ because the variance value is zero.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 222  Exercise 7  Problem 13

Given problem statement, k independent random samples and it generates unbiased estimators for the mean value.

Just show that the arithmetic average is an unbiased for μ.

So proof \(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\mu\)

Given: show that arithmetic average of the estimator is unbiased for μ.

Prove that  \(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\mu\)

Mean value of the estimator is, E[Xi] =  μ

\(E\left(\frac{X_1+X_2+\ldots+X_k}{k}\right)=E\left[\frac{1}{k}\left(\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k\right)\right]\)

 

\(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\frac{1}{k} E\left[\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k\right]\)

 

\(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\frac{E\left[\bar{X}_1\right]+E\left[\bar{X}_2\right]+\ldots+E\left[\bar{X}_k\right]}{k}\)

 

\(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\frac{\mu+\mu+\ldots \mu}{k}\)

 

\(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\frac{k \mu}{k}\)

 

\(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\mu\)

 

Therefore, the arithmetic average of the estimator \(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\mu\) is an unbiased forμ and its proved.

 

Step-By-Step Guide To Estimation Exercises Chapter 7 Milton Page 222  Exercise 7  Problem 14

iGiven problem:

\(\bar{X}\) = 0.8, n1 = 9

\(\bar{X}\) = 0.95, n2 = 3

\(\bar{X}\) = 0.7, n3 = 200

Using previous value, for determine the averaging of three values to get the unbiased estimator for μ.

On previous lesson, the arithmetic average of the estimator is unbiased for μ.

\(E\left(\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{k}\right)=\mu\)

 

Given:

\(\bar{X}\) = 0.8, n1 = 9

\(\bar{X}\) = 0.95, n2 = 3

\(\bar{X}\) = 0.7, n3 = 200

Determine average of three values

\(\widehat{\mu}=\frac{\bar{X}_1+\bar{X}_2+\ldots+\bar{X}_k}{-k}\)

 

⇒ \(\widehat{\mu}=\frac{\bar{X}_1+\bar{X}_2+\bar{X}_3}{3}\)

⇒  \(\widehat{\mu}=\frac{0.8+0.95+0.7}{3}\)

⇒  \(\widehat{\mu}=\frac{2.45}{3}\)

⇒  \(\widehat{\mu}\) = 0.8167

Therefore, Averaging the three values to get the estimate for μ is 0.8167.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 222  Exercise 7  Problem 15

Given problem: \(=\frac{n_1 X_1+n_2 X_2+\ldots+n_k X_k}{n_1+n_2+\ldots+n_k}\)

Just show that given is an un biased for μ .so proof, E(\(\widehat{\mu_w}\)) = μ

Given: \(=\frac{n_1 X_1+n_2 X_2+\ldots+n_k X_k}{n_1+n_2+\ldots+n_k}\)

Then prove , E(\(\widehat{\mu_w}\)) = μ

Introduction to Probability and Statistics Principles and Applications Chapter 7 Estimation Page 222 Exercise 7 Problem 15 Solution

Therefore, given \(\widehat{\mu_w}\) Is ann unbiased estimator for μ and its proved

 

Exercise Solutions For Chapter 7 Susan Milton Estimation And Distributions Page 222  Exercise 7  Problem 16

Previous problem: \(\widehat{\mu_w}\) = \(=\frac{n_1 X_1+n_2 X_2+\ldots+n_k X_k}{n_1+n_2+\ldots+n_k}\)

Using previous problem data and next deterine the estimate for based on the previous problem.

Given: \(=\frac{n_1 X_1+n_2 X_2+\ldots+n_k X_k}{n_1+n_2+\ldots+n_k}\)

Introduction to Probability and Statistics Principles and Applications Chapter 7 Estimation Page 222 Exercise 7 Problem 16 Solution

Therefore compared to the previous problem then the weighted average mean is less than mean.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 223  Exercise 8  Problem 17

Given problem statement,X defines the number of heads obtained when a coin is tossed four times.

Determine the expected value E[X] and variance Var[X] for the variable X.

Given: The random variable X defines the number of heads obtained when a coin is tossed four times.

Also X follows the binomial distribution with parameters

n = 4

p = \(\frac{1}{2}\)

Determine the expected value

​​E[X] = np

E[X]= 4 × \(\frac{1}{2}\)

E[X] = 2

​Find the variance
​
​Var[X] = np(1 − p)

Var[X] = 4 × \(\frac{1}{2}\) (1−\(\frac{1}{2}\))

​Var[X]= 4 ×\(\frac{1}{2}\) × \(\frac{1}{2}\)

Var[X] = 1

​Therefore, the expected value and variance for the variable X is Expected Value: E[X] = 2,  Variance: Var[X] = 1

 

Page 223  Exercise 8  Problem 18

Given problem statement,X defines the number of heads obtained when a coin is tossed.

The number of heads obtained at 10 times. That means X = 10 and binomial distribution with parameters

n = 4

p = \(\frac{1}{2}\)

Given: The random variableX defines the number of heads obtained when a coin is tossed four times.

Also X follows the binomial distribution with parameters,

n = 4

p = \(\frac{1}{2}\)

When heads obtained at ten times.

That means X = 10 and apply all values in the binomial distribution formula

Introduction to Probability and Statistics Principles and Applications Chapter 7 Estimation Page 223 Exercise 8 Problem 18 Solution

Therefore, the probability for obtain the number of heads at 10  times, 0.0379.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 223  Exercise 8  Problem 19

Given problem statement,X defines the number of heads obtained when a coin is tossed.

Determine the expected value E[X] and variance Var[X]
for based on 10 observations of the variable X.

Given: The random variable defines the number of heads obtained when a coin tossed.

That means [X = 10] and binomial distribution with parameters

n = 4

p = \(\frac{1}{2}\)

The mean and variance was given
​
​E[X] = 2

Var[X] = 1

​Then determine the mean and variance for ten observations.

Mean is summation of all observations divided by number of observations.

The means value is 2.

Variance is sum of squares of each observation subtracted by the mean.

Hence, the variance value is 1

Therefore, based on 10 observations the expected value and variance for the variable X is  Expected Value: E[X] = 2 Variance: Var[X] = 1

 

Page 223  Exercise 8  Problem 20

Given problem statement, X defines the number of heads obtained when a coin is tossed.

The variance Var[X] of the variable X is an unbiased estimate for s2.

Next determine the variance of the value \(\bar{X}\)

\(\bar{X}=\frac{\sum\left(X_i-\bar{X}\right)^2}{n-1}\)

 

Given: The random variable defines the number of heads obtained when a coin tossed.

Apply n = 10 for the observations

The unbiased estimate for the variance of X is s2.

Then the \(\bar{X}\)

Introduction to Probability and Statistics Principles and Applications Chapter 7 Estimation Page 223 Exercise 8 Problem 20 Solution

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 7 Page 223  Exercise 9  Problem 21

Given: The Random Samples are X1,X2,X3 ,…..,Xn with mean and variance.

Using mean μ and the variance σ2

Value for prove that the below function

\(\frac{\sum\left(X_i-\bar{X}\right)^2}{n}=\frac{(n-1) s^2}{n}\)

 

Given: The Random Samples are X1,X2,X3 ,…..,Xn with mean μ and variance σ2.

Prove that \(\frac{\sum\left(X_i-\bar{X}\right)^2}{n}=\frac{(n-1) s^2}{n}\) and  E(s2 ) = σ2

Consider LHS

Introduction to Probability and Statistics Principles and Applications Chapter 7 Estimation Page 223 Exercise 9 Problem 21 Solution

Therefore, the Random Samples X1,X2,X3 ,…..,Xn with mean μ Value and variance σ2  Value for \(\frac{\sum\left(X_i-\bar{X}\right)^2}{n}=\frac{(n-1) s^2}{n}\) and its proved

 

Page 223  Exercise 10  Problem 22

Given: The Random Sample  X1,X2,X3 ,…..,Xm  with size m and parameter n.

Using method of moments technique to Determine the first moment of the given sample.

Given: The Random Sample X1,X2,X3 ,…..,Xm with size m and parameter n.

Also, it follows the Binomial Distribution X1,X2,X3 ,…..,Xm

Use method of moments to find the estimator for p,

M1 = \(\frac{\sum_i X_i}{m}=\bar{X}\)

Find the first moment of the sample, E(X) = np

\(n \hat{p}=\bar{X}\) \(\hat{p}=\frac{\bar{X}}{n}\)

 

Therefore, using Method of Moments technique for find the estimator of p is \(\hat{p}=\frac{\bar{X}}{n}\) and its proved.

J Susan Milton Introduction To Probability and Statistics Chapter 6 Descriptive Distributions Exercises

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J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Descriptive Distributions

 

Introduction To Probability And Statistics Chapter 6 Exercises Solutions  Page 192  Exercise 1  Problem 1

In this case a statistical study is appropriate.

The population of interest is consisted of the wind speed per day.

An engineer can measure the speed over some period of time and then determine various statistics of that sample including

1. Mean

2. Minimal

3. Maximal value

4. Sample deviance etc..

The draw necessary conclusions about the population based on observing the given sample.

Thus, In this case, a statistical study is appropriate because of various statistics of that sample including Mean, Minimal, maximal value, sample deviance, etc. Hence, the maximum wind speed per day at all sites can be designed.

Read and Learn More J Susan Milton Introduction To Probability And Statistics Solutions

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 192  Exercise 2  Problem 2

In this case study:

A statistical study is appropriate. Because the population of interest is consisted of two groups of cuttings:

1. A control group

2. A test group

Various static test can be used for this kind of problem to determine whether or not there is a statistical significant between group or in this case whether or not indoleacetic acid really is effective.

Thus, the botanist thinks that indoleacetic acid is effective in stimulating the formation of roots in cutting from the lemon tree.

 


Solutions To Descriptive Distributions Exercises Chapter 6 Susan Milton Page 192  Exercise 3  Problem 3

In this case study:

A statistical study is not appropriate.

Because the sample might be too small to correctly approximate the average time and cost required to complete the job.

Thus, an architectural firm is to sublet a contract for a wiring project.

 

Chapter 6 Descriptive Distributions Examples And Answers Susan Milton Page 192  Exercise 4  Problem 4

In this case study:

A statistical study is not appropriate. Because the length of the sessions does not tell us anything about the occupancy of the terminal.

Thus, A statistical study is not appropriate in a computer system that has a number of remote terminals attached to it. Because the length of the sessions does not tell us anything about the occupancy of the terminal.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 192  Exercise 5  Problem 5

In this case study:

A statistical study is appropriate.

Because the population of interest is consisted of the affected workers. We can also draw a random sample out of those 50000 workers, because the original population might be too large to study in its entirety.

Sampling the people from population would help us draw necessary conclusion about the population.

Thus, the statistical study is appropriate. Because the population of interest consists of the affected workers. prior to changing from the traditional is appropriate and also drew a random sample out of those 50,000 workers, because the original population might be too large to study entirety.

 

Probability And Statistics J. Susan Milton Chapter 6 solved Step-By-Step Page 192  Exercise 6  Problem 6

Frist we need to find the random variable.

Then identify with know or unknown mean.

Given:

Random variable  = X1

Particular level =  24 hours period.

Now , Let X1  be the random variable for the particular level for the first 24-hour period.

The random variable is normally distributed with unknown mean μ and also unknown variance is σ, so that we get  X1 ≈ N(μ,σ2)

Thus, the distribution of this random variable is  X1 ≈ N(μ,σ2).

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 192  Exercise 6  Problem 7

J.Susan Milton Introduction To Probability and Statistics Chapter 6 Descriptive Distributions Exercises

Now apply the value in the equation:

ΣiXi   = 45 + 50 + 62 + 57 + 70

ΣiXi  = 284

The sample size be n = 5

The given sample are

x1  =  45 , x2  =  50 , x3  =  62 , x4  =  57 , x5  =  70

Now calculate the statistic ΣiXi/n

Simply sum the given values to get the values:

 ΣiX2i    = x21 + x22 + x23 + x24 + x25

Now apply the value in the equation

ΣiX2i  = 452 + 502 + 622 + 572 + 702

ΣiX2i  = 16518

The sample size be n = 5

The given sample are

x1  =  45 , x2  =  50 , x3  =  62 , x4  =  57 , x5  =  70

Now calculate the statistic  ΣiX2i 

Simply sum the given values to get the values:

\(\Sigma_i \frac{X_i}{n}=\frac{x_1+x_2+x_3+x_4+x_5}{n}\)

Now apply the value in the equation

\(\Sigma_i \frac{X_i}{n}=\frac{45+50+62+57+70}{5}\)

 

⇒  \(\Sigma_i \frac{X_i}{n}\) = 56. 8

Now we going to calculate the statistic maxi {Xi},so that we to find the biggest value from the given sample.

By looking the values, we can easily identify them maxi {Xi}= x5 = 70.

Now we going to calculate the statistic mini {Xi}, so that we to find the biggest value from the given sample.

By looking the values, we can easily identify them mini {Xi} = x1 = 45.

Thus, the random variable value is ΣiXi  = 284

 ΣiXi =  16518

\(\Sigma_i \frac{X_i}{n}\) =   56.8

Maxi {Xi}= x5 = 70.

Mini {Xi} = x1 = 45.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 192  Exercise 6  Problem 8

Frist we need to find the random variable.

Then find the value of given.

The sample size be n = 5

The given sample are

x1  =  45 , x2  =  50 , x3  =  62 , x4  =  57 , x5  =  70

The random variables X5and \(\frac{X_5-\mu}{\sigma}\) are not a statistic.

Since this random variable we can’t determine its numerical value from a random sample.

Thus, this random variable we can’t determine its numerical value from a random sample.

 

Page 193  Exercise 7  Problem 9

Given:

The number is = 02,03,04,05,06,07

First, we need to find the length of the categories.

To construct a Stem and leaf diagram, we have to use the initial two digits as stems, in this case 02,03,04,05,06,07

The third digit will be representation a leaf.

By applying the logic to the given sample, so we get that easily construct the given stem and leaf diagram:

02 ∣ 0

03 ∣ 079909

04 ∣ 407549262

05 ∣ 75012268431

06 ∣ 1612120

07 ∣ 0

Thus, the construct Stem and leaf diagram. By applying the logic to the given sample, so we get that easily construct the given stem and leaf diagram:

02 ∣ 0

03 ∣ 079909

04 ∣ 407549262

05 ∣ 75012268431

06 ∣ 1612120

07 ∣ 0

 

Online Help For J. Susan Milton Descriptive Distributions Chapter 6 Exercises Page 193  Exercise 7  Problem 10

By turning the stem and leaf diagram to the side , we can very clearly see that the diagram has a notorious bell shape, thus confirming the persons suspicious of it being normally distributed.

Hence we should not be surprised if we hear someone claim such thing.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 193  Exercise 7  Problem 11

First we need to find the biggest value.

Then we are going to find the smallest value.

Given:

First half unit = \(\frac{1}{1000}\)

Other Half unit= .0005

To break the given data into six category , first we have to find the length of the interval convering the data.

The biggest value from the sample which is 0.070

The smallest value from the sample which is 0.020

0.070 − 0.050  =  0.020

To find the length of the categories for that we going to divide the length of the whole interval by the number of categories.

Now, Split data into 6 categories and we calculate the length of 0.02 units.

To divide those number and round it up to the nearest number that has the same number of decimals as the original data.

\(\frac{0.02}{6}\) = 0.00333333

The data has three decimals, so we going to round the calculated number to 0.010.

Hence the length of each category.

The lower boundary for the first category is obtained by subtracting 0.0005 from the lowest value of the sample which is 0.02.

0.02 − 0.0005 = 0.0195

Hence the lower boundary for the first category is 0.0195.

To find the remaining categories, we have to successively add the length of each category starting from the lowest boundary, which is 16.25.

0.0195 + 0.010 = 0.0295 ⇒ [0.0195,0.0295⟩

0.0295 + 0.010 = 0.0395 ⇒ [0.0295,0.0395⟩

0.0395 + 0.010 = 0.0495 ⇒ [0.0395,0.0495⟩

0.0495 + 0.010 = 0.0595 ⇒ [0.0495,0.0595⟩

0.0595 + 0.010 = 0.0695 ⇒ [0.0595,0.0695⟩

0.0695 + 0.010 = 0.0795​ ⇒ [0.0695,0.0795⟩

Thus, the method outlined in this section breaks these data into six categories and also add the length of each category starting from the lowest boundary, which is 16.25

0.0195 + 0.010 = 0.0295 ⇒ [0.0195,0.0295⟩

0.0295 + 0.010 = 0.0395 ⇒ [0.0295,0.0395⟩

0.0395 + 0.010 = 0.0495 ⇒ [0.0395,0.0495⟩

0.0495 + 0.010 = 0.0595 ⇒ [0.0495,0.0595⟩

0.0595 + 0.010 = 0.0695 ⇒ [0.0595,0.0695⟩

0.0695 + 0.010 = 0.0795​ ⇒ [0.0695,0.0795⟩

 

Step-By-Step Guide To Descriptive Distributions Exercises Chapter 6 Milton Page 193  Exercise 8  Problem 12

Since the stem and leaf diagram is slightly skewed to the left instead of having a normal bell shape it can be suggested that the data comes from a family of X2 distribution.

Thus, the stem and leaf diagram is slightly skewed to the left instead of having a normal bell shape.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 193  Exercise 9   Problem 13

Given:

The number is = 5,6,7,8,9,10

First, we need to find the length of the categories and then find the value.

Now

To construct Stem and leaf diagram, we have to use initial digits as “stems”, in this case 5,6,7,8,9,10

The second digits will be representation a leaf.

By applying the logic to the given sample, so we get that easily construct the given stem and leaf diagram:

5 ∣ 3

6 ∣ 12728257

7 ∣ 6467816399117494

8 ∣ 8728710275241

9 ∣ 056127563

10 ∣ 0

Thus, the construct for the Stem and leaf diagram is given below:

5 ∣ 3

6 ∣ 12728257

7 ∣ 6467816399117494

8 ∣ 8728710275241

9 ∣ 056127563

10 ∣ 0

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 193  Exercise 9  Problem 14

To construct Stem and leaf diagram, we have to use initial digits as “stems”, in this case, 5,6,7,8,9,10

The second digit will be a representation a leaf.

By applying the logic to the given sample, so we get easily construct the given stem and leaf diagram:

​5 ∣ 3

​6 ∣ 12728257

7 ∣ 6467816399117494

8 ∣ 8728710275241

9 ∣ 056127563

10 ∣ 0

By turning this stem and leaf diagram to the side, we can very clearly see that the diagram has a notorious bell shape, thus confirming the assumption of it being normally distributed.

Thus, the assumption X is normally distributed. The given stem and leaf diagram is

5 ∣ 3

6 ∣ 12728257

7 ∣ 6467816399117494

8 ∣ 8728710275241

9 ∣ 056127563

10 ∣ 0

 

Exercise Solutions For Chapter 6 Susan Milton Descriptive Distributions Page 194  Exercise  10  Problem 15

Given:

The number is = 0,1,2,3,4,5

First, we need to find the length of the categories and then find the value.

Now

To construct Stem and leaf diagram, we have to use initial digits as “stems”, in this case 0,1,2,3,4,5

The second digits will be representation a leaf.

By applying the logic to the given sample, so we get that easily construct the given stem and leaf diagram:

0 ∣ 578

1 ∣ 19358620972988457

2 ∣ 06443483575730231

3 ∣ 6281007541

4 ∣ 05

5 ∣ 0

Thus, the construct for Stem and leaf diagram is given below:

0 ∣ 578

1 ∣ 19358620972988457

2 ∣ 06443483575730231

3 ∣ 6281007541

4 ∣ 05

5 ∣ 0

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 194  Exercise 10  Problem 16

Given:

The assumption that X is not normally distributed.

There might be a slight reason that X is not normally distributed as by turning the stem and leaf diagram to the sight.

We can see that it does not exactly resemble perfect symmetrical bell shape like a normal distribution, but rather slightly skewed to the left.

Thus, we can be suspicious about the data not being normally distributed.

Thus, the assumption that X mis not normally distributed.

0 ∣ 578

1 ∣ 19358620972988457

2 ∣ 06443483575730231

3 ∣ 6281007541

4 ∣ 05

5 ∣ 0

 

Page 194   Exercise 10  Problem 17

First we need to find the biggest value.

Then we going to find the smallest value.

To break the given data into seven category , first we have to find the length of the interval covering the data.

The biggest value from the sample which is 5.0

The smallest value from the sample which is0.5

5.0 − 0.5 = 4.5

Whole interval by the number of categories.

Now

Split data into 7 categories and we calculate the length of 4.5 units.

To divide those number and round it up to the nearest number that has the same number of decimals as the original data.

\(\frac{4.5}{7}\) = 0.642857

The data has one decimals, so we going to round the calculated number to0.6.

Hence the length of each category.

The lower boundary for the first category is obtained by subtracting0.7

0.5−0.05 = 0.45

Hence the lower boundary for the first category is0.45.

To finding the remaining categories, we have to successively add the length of each category starting from the lowest boundary, which is n 0.45.

0.45 + 0.7 = 1.15 ⇒ [0.45,1.15⟩

1.15 + 0.7 = 1.85 ⇒ [1.15,1.85⟩

1.85+0.7 = 2.55 ⇒ [1.85,2.55⟩

2.55 + 0.7 = 3.25 ⇒ [2.55,3.25⟩

3.25 + 0.7 = 3.95 ⇒ [3.25,3.95⟩

3.95 + 0.7 = 4.65 ⇒ [3.95,4.65⟩

4.65 + 0.7 = 5.35 ⇒ [4.65,5.35⟩

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 194  Exercise 10  Problem 18

Given:

Let one variable be X

A random sample of 50 mosses yields under observation

First, we need to find the frequency table.

Then we going to find the histogram of data.

The frequency table constructed by observing and counting how many of the values from the sample are located inside of each of those six categories.


The table is:

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 194 Exercise 10 Problem 18 Frequency values 1

The relative frequency histogram data

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 194 Exercise 10 Problem 18 Histogram Data 1

Thus, the looking the shape of the histogram, we can clearly see that it resembles the bell shapes, but is slightly skewed to the left, it is just like the stem and leaf.

Diagram we calculated easier. Hence having characteristics is not normal density.

The table is given below:

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 194 Exercise 10 Problem 18 Frequency values 2

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 194 Exercise 10 Problem 18 Histogram Data 2

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 195  Exercise 11  Problem 19

Given:

The 25th, 50th,75th, and 100th percentiles for X

First, we need to find the point X value.

Then we going to find the binomial value.

The first quartile of a random variable X is a point p 0.25

Such that P(X < p 0.25) ≤ 0.25
​
P(X ≤ p0.25) ≥ 0.25

Then we can write as: P(X < p0.25) ≤ 0.25 and the first quartile of a random variable is P(X ≤ p0.25) ≥ 0.25

Thus, the first quartile of a random variable X is P(X ≤ p0.25) ≥ 0.25.

 

Page 195  Exercise 11  Problem 20

Given:

n = 20

p = 0.5

First we need to find the point X value .

Then we going to find the binomial value.

If X is binomial variable with n = 20 and p = 0.5, then its first quartile is obtained by using the definition, and table from appendix or a mathematical software such as R.

The first percentile of this distribution is p0.25 = 8

P(X < 8) = P(X ≤ 7) = 0.132 ≤ 0.25

P(X ≤ 8) = 0.2517 ≥ 0.25

Thus, the first quartile of a random variable p0.25 = 8.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 195  Exercise 11  Problem 21

Given: \(\int_0^p e^{-x} d x\) = 0.25

First we need to find the point X value.

Then we going to find the binomial value

To calculate exponential random variables with β = 1,

We have to solve the given equation:

\(\int_0^p e^{-x} d x\) = 0.25

Let integrate the given equation:

\(\int_0^p e^{-x} d x\)= \(\left.\left(-e^{-x}\right)\right|_0 ^p\)

\(\int_0^p e^{-x} d x\)= \(-\left(e^{-p}-e^0\right)\)

Where e0= 1

\(\int_0^p e^{-x} d x\) = \(-\left(e^{-p}-1\right)\)

\(\int_0^p e^{-x} d x\)= \(1-e^{-p}\)  ……………………….. (1)

To solve equation(1) so that we get

1 − e − p = 0.25

⇔  0.75  =  e − p

⇔ ln 0.75 =  −p

⇔p  = −ln 0.75 ≈ 0.2877

Thus, the pointp value is  ⇔  p0.25 = −ln0.75.

 

Page 195  Exercise 12  Problem 22

Given:

(Deciles.) The 10th,20th,30th,40th,50th,60th,70th,80th,90th,and100th

First, we need to find the point 40th deciles X value.

Then we going to find the binomial value.

The 4th decile of a random variable X is a point p0.4

Such that P(X < p0.4) ≤ 0.4

P (X ≤ p0.4 ) ≥ 0.4

Then can write the fourth decile of the random variable as P(X ≤ p0.4) ≥ 0.4

Thus, the 4th decile of random variable of a random variable X is P(X ≤ p0.4) ≥ 0.4.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 195  Exercise 12  Problem 23

Given: λ = 20

First we need to find the point X value

Then we going to find the Poisson value.

If X is Poisson random variable with λ = 20, then its first 6th decile is obtained by using the definition, and table from appendix or a mathematical software such as R.

The first percentile of this distribution is p 0.6= 11

P(X<11) = P(X ≤ 10) = 0.5830 ≤ 0.6

P(X ≤ 11) = 0.6968 ≥ 0.6

Thus, the 6th decile X value is p0.6 = 11.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 195  Exercise 13  Problem 24

First we need to find the point X value .

Then diagram the relative cumulative frequency.

The relative cumulative frequency ogive from the data :

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 195 Exercise 13 Problem 24 Relative cumulative frequency ogive 1

By using projection method , approximate the first quartile for X by drawing the horizontal line at he height of 25 then add a perpendicular line.

That passes through the previously obtained point on the relative cumulative frequency ogive to read the approximate value for it.

The related graph is:

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 195 Exercise 13 Problem 24 Relative cumulative frequency ogive 2

Thus, the approximate first quartile X value is p0.25 = 0.0 415.

 

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Page 195  Exercise 13  Problem 25

First we need to find the point X value .

Then diagram the relative cumulative frequency.

The relative cumulative frequency ogive from the data :

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 195 Exercise 13 Problem 25 Relative cumulative frequency ogive 1

By using projection method , approximate the first quartile for X by drawing the horizontal line at he height of 40 (it means 40%)then add a perpendicular line.

That passes through the previously obtained point on the relative cumulative frequency ogive to read the approximate value for it.

The related graph is:

Introduction to Probability and Statistics Principles and Applications Chapter 6 Descriptive Distributions Page 195 Exercise 13 Problem 25 Relative cumulative frequency ogive 2

Thus, the approximate fourth decile X value is p0.4 = 7.7.

J.Susan Milton Introduction To Probability and Statistics Chapter 4 Continuous Distributions Exercises

by

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Continuous Distributions

 

Introduction To Probability And Statistics Chapter 4 Exercises Solutions Page 127   Exercise 1  Problem 1

Given:

The function  f(x) = kx where 2 ≤ x ≤ 4

To find – The value of k

Method: The method here used is probability and continuous random variable.

The function f(x) = kx.

For, 2 ≤ x ≤ 4 variable.

Assume x = 3.

f(3) = 3k

Consider f(3) = 1 for the function f(x).

​1 = 3k

k = \(\frac{1}{3}\)

Hence, it is verified that the value of k is k = \(\frac{1}{3}\)

Read and Learn More J Susan Milton Introduction To Probability And Statistics Solutions

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 127  Exercise 1  Problem 2

​Given:

The function f(x) = kx

Where  2 ≤ x ≤ 4.

To find:

Find the value of probabilities of P(2.5 ≤ X ≤ 3).

Method: The method here used is probability and continuous random variable.

The function f(x) = kx where 2 ≤ x ≤ 4

Substitute x = 3

f(3) = 3k

The function of P[2.5≤X≤3]

Here, there is no function that occurs between the X = 2.5 and X = 3 X.

Hence, it is verified that the function f(x) is not possible for the values P(2.5 ≤ X ≤ 3).

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4  Page 127  Exercise 1  Problem 3

J.Susan Milton Introduction To Probability and Statistics Chapter 4 Continuous Distributions Exercises

J. Susan Milton Chapter 4 Continuous Distributions Answers Page 127  Exercise 1  Problem 4

Given: 

The function f(x) = kx

where 2 ≤ x ≤ 4.

To find – Find the probability of P(2.5 < X ≤ 3)

Method: The method here used is probability and continuous random variable.

The function f(x) = kx

For P (2.5 < X ≤ 3)

The equation is, f(x) = 3k

Assume k = 2

The function

​f(x) = 3 × 2

f(x) = 6
​
Hence, it is verified that the function f(x) at P(2.5 < X ≤ 3) is f(x) = 6

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 128  Exercise 2  Problem 5

Given:

The function f(x) = \(\left(\frac{1}{10}\right) e^{\frac{-x}{10}}\)

To find – The function is a continuous random variable

Method : The method used is a probalility and continuous random variable

The given function f(x) =\(\left(\frac{1}{10}\right) e^{\frac{-x}{q p}}\)

Substitute 1 or x

f(1) \( = \frac{1}{10} e^{\frac{-x}{10}}\)

f(1)  =  0.0904

Hence, it is verified that the density of the function is f(1) = 0.0904

 

Solutions To Continuous Distributions Exercises Chapter 4 Susan Milton Page 128  Exercise 2  Problem 6

Given:

The function f(x)= \(\frac{1}{10} e^{\frac{-x}{10}}\)

To find –  Find the density at 7 minutes

Method: The method used here are probability and continuous random variable.

The given function

f(x) = \(\frac{1}{10} e^{\frac{-x}{10}}\)

For the density of 7 minutes

​f(7)  = \(\frac{1}{10} e^{\frac{-7}{10}}\)

f(7)  =  0.0496

Hence, it is verified that the density of the function at 7 minutes is f(7) = 0.0496

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 128  Exercise 2  Problem 7

Given:

The function f(x) = \(\frac{1}{10} e^{\frac{-x}{10}}\)

To find –  The probability of density of call last between 1 to 2 minutes.

Method: The method used here is probability and continuous random variable.

The given function is.

f(x) = \(\)

For the call last one minute

​f(1) = \(e^{\frac{-1}{10}}\)

​f(1) = 0.0904

For the call lasts two minutes

f(2) = 0.1 × \(e^{\frac{-2}{10}}\)

f(2) = 0.0607

The probaability of the call lasts one minuteto two minutes, the density of the function gradually increases to show the calls gain more and more density by increasing the call time.

Hence, it is verified that the call lasts from one to two minutes then the density of the call is also increasing.

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 128  Exercise 3  Problem 8

Given:

The graph of bird moving in θ.

To find –  The angle of the bird moving

Method: The method used in this problem is probability, continuous random variable, and graphical method.

Let the function f for the interval [0,2Π].

f(θ)\(=\int_0^{2 \Pi} \theta\)

Reduce the equation.

​f(θ) =  2Π − 0

f(θ) = 2Π
​
Hence, it is verified that the density of the function f with the interval [0,2Π] is f(θ) = 2Π

 

Chapter 4 Continuous Distributions Examples And Answers Susan Milton Page 128  Exercise 3  Problem 9

Given:

The graph of moving bird denoted in θ.

To find – Sketch the graph of the moving bird in uniform motion with interval [0,2Π].

Method: The method used in this problem is probability, continuous random variable, and graphical method.

The function of the moving bird in uniform distribution interval [0,2Π].

Let \(=\int_0^{2 \Pi} \theta\)

Reduce the equation.

​f(θ) = Π + Π

f(θ) = 2Π
​
The graph of the moving bird.

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 128 Exercise 3 Problem 9 Moving bird 1

Hence, it is verified that the graph of a uniform distribution over the interval.

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 128 Exercise 3 Problem 9 Moving bird 2

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 128  Exercise 3  Problem 10

Given:

The function f of moving bird in the angle θ.

To find – Sketch the graph of the function fin the interval [0,2Π].

Method: The method used in this problem is a probability, continuous random variable, and graphical method

Graph the function f and shade the orient within \(\frac{\Pi}{4}\)radians of home.

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 128 Exercise 3 Problem 10 Radians 1

The graphs shows, the bird flying in the direction of \(\frac{\Pi}{4}\) radians from home in the straight direction from home.

Hence, it is verified that the 

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 128 Exercise 3 Problem 10 Radians 2

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 128  Exercise 3  Problem 11

Given:

The function f mof the interval [0,2Π].

To find –  Find the probability of the function f with the orient within the
\(\frac{\Pi}{4}\) radians of home.

Method: The method used in this problem is a probability, continuous random variable, and graphical method.

The function for orient within the \(\frac{\Pi}{4}\) radians of the home .

​f(θ)\(  =\int_0^{2 \Pi} \theta+\frac{\Pi}{4}\)

Reduce the equation.

​f(θ) = \(2 \Pi+\frac{\Pi}{4}\)

​f(θ) = \(\frac{9 \Pi}{4}\)

Hence, it is verified that the possibilities of the moving bird within the
\(\frac{\Pi}{4}\) radians of home is \(=\frac{9 \Pi}{4}\)

 

Probability And Statistics J. Susan Milton Chapter 4 Solved Step-By-Step Page 129  Exercise 4  Problem 12

Given:

The graph of the probabilities.

To find: Show probabilities in terms of the cumulative distribution function F.

Method: The method used in this problem is a probability, continuous random variable, and cumulative distribution function.

The graph f(x) of the cumulative distribution function F(x) has the possible intervals of 0 ≤ x ≤ 20.

For graph A

F(x) = \(\int_0^{20} x+1 d x\)

Reduce the equation.

​F(x) = 20 + 1−(0 + 1)

F(x) = 20
​
The probability at graph A X≤ 5

Similarly, for the graph B, C, D, and E

​F(x)= \(\int_0^{20} x+1 d x\)

Reduce the equation

​F(x) = 20 + 1−(0 + 1)

F(x) = 20
​
The probability of graph B, X > 5.

The probability of graph C, X = 10.

The probability of graph D, 5 ≤ X < 10.

The probability of graph E, 5 < X < 10

Hence, it is verified.

The probability of the graph A, X ≤ 5.

The probability of the graph B, X > 5.

The probability of the graph C, X = 0.

The probability of the graph D, 5 ≤ X < 10.

The probability of the graph E, 5 < X < 10.

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129  Exercise 5  Problem 13

Given:

The function f(x) = kx where 2 ≤ x ≤ 4.

To find –  Find the cumulative function of F(x).

Method:

The method used in this problem is a probability, continuous random variable, and cumulative distribution function.

The given function is f(x) = kx where 2 ≤ x ≤ 4.

The cumulative distribution function is

F(X) = \(\int_2^4(k x) d x\)

Reduce the equation.

​F(X) = 4k − 2k

F(X) = 0 2k
​
Hence, it is verified that the cumulative distribution function of F(x) = 2k.

 

Step-By-Step Guide To Continuous Distributions Exercises Chapter 4 Milton Page 129  Exercise 5  Problem 14

Given:

The cumulative distribution function , F(X) = \(=\int_2^4 k X d X\)

To find – Find the cumulative distribution function, P[2.5 ≤ X ≤ 3].

Method: The method used in this problem is a probability, continuous random variable, and cumulative distribution function.

The given function.

​F(X) = \(\int_2^4 k x d x\)

The cumulative distribution function for the interval P[2.5 ≤ X ≤ 3].

​F(X) = \(\int_{2.5}^3 k x \mathrm{dx}\)

Reduce the equation.

​F(X) = 3k − 2.5k

F(X) = 0.5k

Hence, it is verified that the cumulative distribution function of the interval P[2.5 ≤ X ≤ 3] is F(X) = 0.5k

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129  Exercise 5  Problem 15

Given:

The function F with limits \(\lim _{n \rightarrow \infty} F(x)\) and \(\lim _{n \rightarrow \infty} F(x)\).

To find – Draw the graph of a function with limits \lim _{n \rightarrow \infty} F(x)

Method: The method used in this problem is a probability, continuous random variable, and cumulative distribution function.

The graph of the cumulative distribution function

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 129 Exercise 5 Problem 15 Commutative distribution 1

Here, the graph of the function shows the function F is the increasing function.

This function is a non-decreasing function up to the interval \(\lim _{n \rightarrow \infty} F(x)\) and \(\lim _{n \rightarrow \infty} F(x)\) is possible in the graph of the function.

Hence, it is verified that graph of the cumulative distributive function Fand the function is non-decreasing.

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 129 Exercise 5 Problem 15 Commutative distribution 2

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129   Exercise 6  Problem 16

Given:

The function f(x) = \(\frac{1}{b-a}\)

To find – Find the uniform distribution over the interval (a,b).

Method: The method used in this problem is a probability, continuous random variable, and cumulative distribution function.

The function f(x) = \(\frac{1}{b-a}\)

The uniform distribution over the interval (a,b).

The cumulative distribution function

f(x) = \(\int_a^b \frac{1}{b-a} d x\)

f(x) = \(\left[\frac{x}{b-a}\right]_a^b\)

f(x) = \(\frac{b}{b-a}-\frac{a}{b-a}\)

f(x) = \(\frac{b-a}{b-a}\)

f(x) = 1

Hence, it is verified that the uniform distribution over the interval (a,b) is f(x)=1

 

Exercise Solutions For Chapter 4 Susan Milton Continuous Distributions Page 129  Exercise 7  Problem 17

Given: The function f(θ) \(=\int_0^{2 \Pi} \theta d \)

To find – 

Find the uniform distribution of f.

Method – The methods used here are a probability, cumulative random variable, and uniform distribution.

The function f(θ) \(=\int_0^{2 \Pi} \theta d \theta\)

For the cumulative distribution function, θ = Π.

f(θ) \(=\int_0^{2 \Pi} \theta d \theta\)

f(θ) = 2 Π

Hence, it is verified that the uniform distribution of cumulative function is f(θ) = 2Π.

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129  Exercise 7  Problem 18

Given:

The function f(θ) = \(\int_0^{2 \Pi} \theta d \)

To find – Graph the function F and F is non-decreasing or not.

Method: The method used here is a probability, cumulative distributive function and uniform distribution.

The function.

f(θ) =  \(\int_0^{2 \Pi} \theta d \)

Reduce by uniform Distribution.

​F(θ) = [θ2]02π

F(θ) = 4Π2

The graph of the function

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 129 Exercise 7 Problem 18 Non decreasing 1

Hence, it is verified that the uniform function is F(θ)=4Π2 and the function is non-decreasing. The graph of the function F

Introduction to Probability and Statistics Principles and Applications Chapter 4 Continuous Distributions Page 129 Exercise 7 Problem 18 Non decreasing 2

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129   Exercise 8  Problem 19

Given:

The function f(x)=\(\frac{1}{10} e^{\frac{-x}{10}}\).

To find  – Find the Cumulative distribution f.

Method: The methods used here are probability, cumulative distribution function.

The function

f(x) = \(\frac{1}{10} e^{\frac{-x}{10}}\).

For the interval P[1 ≤ X ≤ 2] .

The cumulative distribution function X = 1

f(1) = \(\frac{1}{10} e^{\frac{-1}{10}}\)

f(1) = 0.906 × 0.1

f(1) = 0.906

The cumulative distribution function X = 2

​f(1) = \(\frac{1}{10} e^{\frac{-2}{10}}\)

f(2) = 0.1 × 0.818

f(2) = 0.0818
​
Hence, it is verified that the continuous random variable has only one possibilities of probability, but the cumulative distribution function has two probability values.

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129  Exercise 9  Problem 20

Given:

The function f(x) = \(\frac{1}{\ln 2} \frac{1}{x}\).

To find – Find the cumulative distribution of function f.

Methods: The methods used here is the probability, cumulative distribution function

The given function f(x)= \(\frac{1}{\ln 2} \frac{1}{x}\)

The cumulative distribution of interval P[30 ≤ X ≤ 40].

For X = 30

f(x) = \(\frac{1}{\ln 2} \frac{1}{30}\)

f(x) = 1.44 × 0.33

f(x) = 0.475

Hence, it is verified that the cumulative distributive function of function f is f(x) = 0.475
​

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129  Exercise 10  Problem 21

Given: The function

F(x) = \(\left\{\begin{array}{c}0, \mathrm{X}<-1 \\X+1,-1 \leq x \leq 0 \\1, x>0\end{array}\right.\)

To find – Find the cumulative distribution function.

Method: The methods used here are probability and cumulative distributive function.

The given function.

F(x) = \(\left\{\begin{array}{c}
0, \mathrm{X}<-1 \\
X+1,-1 \leq x \leq 0 \\
1, x>0
\end{array}\right.\)

The cumulative distribution function.

For, X < − 1.

F(x) = 0

For, −1 ≤ x ≤ 0.

F(x) = 1

For, x > 0.

F(x) = 1

Hence, it is verified that the cumulative distributive function is F(x)=1 and the function is non-decreasing for the limit \(\lim _{x \rightarrow-\infty} F(x)\) = 0 and \(\lim _{x \rightarrow-\infty} F(x)\) = 1.

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 129  Exercise 10  Problem 22

Given:

The function F(x)= \(\left\{\begin{aligned}
0, \mathrm{X} & \leq 0 \\
x^2, 0<x & \leq \frac{1}{2} \\
\frac{1}{2} x, \frac{1}{2}<x & \leq 0 \\
1, \mathrm{x} & >1
\end{aligned}\right.\)

​To find –  Find the cumulative distribution function.

Method: The methods used here are probability, cumulative distributive function.

The given function.

F(x) = \(\left\{\begin{array}{r}
0, \mathrm{X} \leq 0 \\
x^2, 0<x \leq \frac{1}{2} \\
\frac{1}{2} x, \frac{1}{2}<x \leq 0 \\
1, \mathrm{x}>1
\end{array}\right.\)

For, x ≤ 0.

F(x) =  0

For, 0 < x ≤ \(\frac{1}{2}\)

F(x) = \(\frac{1}{4}\)

For, \(\frac{1}{2}\) <x≤0.

F(x)= \(\frac{1}{4}\)

For, x > 1.

F(x) = 1

Hence, it is verified that the cumulative distributive function is F(x)= \(\frac{1}{4}\) and function is ono-decreasing with the limit F(x) = 0.

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 130  Exercise 11  Problem 23

Given:

The function f(x) = \(\frac{1}{6}\) (x) where 2 ≤ x ≤ 4.

To find – Find the function E(X) f(x) =\(\frac{1}{6}\)(x).

Method:

The method used here is probability, cumulative distributive function.

The function f(x)= \(\frac{1}{6}\)

The uniform distribution.

For, 2 ≤ x ≤ 4.

E(X) = \(\int_2^4 f(x) d x\)

Reduce the equation.

E(X) = \(\int_2^4 \frac{1}{6}(x) d x\)

E(X) = \(\frac{1}{6}\left[x^2\right]_2^4\)

E(X) =\(\frac{1}{6}\)(16 – 4)

E(X) = \(\frac{1}{6}\)(12)

E(X) = 2

Hence, it is verified that the uniform distribution of the function f(x) = \(\frac{1}{6}\) (x) E(x) = 2

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 130  Exercise 11  Problem 24

Given:

The function f(x) = \(\frac{1}{6}\) (x) where 2 ≤ x ≤ 4.

To find – 

Find E(X2).

Method: The methods used here are probability, Cumulative distributive function, and Uniform distribution.

The function f(x) = \(\frac{1}{6}\) (x).

The cumulative distribution of function.

E(X2) = \(\int_2^4(f(x))^2 d x\)

Reduce the equation.

​E(X2) = \(\int_2^4 \frac{1}{36}\left(x^2\right) d x\)

​E(X2) = ​\(\left.\frac{1}{36} \frac{x^3}{3}\right]_2^4\)

​E(X2) = \(\left(\frac{16}{27}\right)-\left(\frac{2}{27}\right)\)

Hence, it is verified the uniform distribution of the function f(x) = \(\frac{1}{6}x\) is ​E(X2) =\(\frac{14}{7}\).

 

Page 130  Exercise 12  Problem 25

Given:

The function f(x) = \(\frac{1}{10} e^{\frac{-x}{10}}\) where x>0.

To find – Find the moment generating function M​​X(t).

Method: The method used here is the cumulative distribution function and the uniform distribution.

The function f(x) = \(\frac{1}{10} e^{\frac{-x}{10}}\) where x>0

The expression for moment generating function M​X(t)

M​X(t)= E [ext]

Reduce the equation.

M​x(t) = \(\int_1^{\infty} e^{t x} \frac{1}{10} e^{\frac{-x}{10}} d x\)

M​x(t) = 0.1 [\(\left[e^{t x} \frac{1}{10} e^{\frac{-x}{10}}\right]_1^{\infty}\)

M​X(t) = 0.1 \(\left[e^{t x-\frac{x}{10}}\right]_1^{\infty}\)

M​x(t) = 0.1 \(\left[e^{t-\frac{1}{10}}-e^{\infty}\right]\)

M​x(t) = 0.1 \(0.1\left[e^{t-0.1}-\infty\right]\)

M​x(t) = ∞

Hence, it is verified that the moment generating function is M​x(t) = ∞.

 

J. Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 4 Page 130  Exercise 12  Problem 26

Given:

The function f(x)= \(\int_1^{\infty} e^{t x} \frac{1}{10} e^{\frac{-x}{10}} d x\) where x>0.

To find – Find the average length of such a call with the moment generating function.

Method: The method used here is the cumulative distribution function and the uniform distribution.

The function f(x)= \(\int_1^{\infty} e^{t x} \frac{1}{10} e^{\frac{-x}{10}} d x\).

The Expression for the moment generating function.

M​x(t) = E[etX]

M​x(t) = \(0.1\left[e^{t-\frac{1}{10}}-\infty\right]\)

For the average length of a call, assume t = 1.

M​x(1) = \(0.1\left[e^{t-\frac{1}{10}}-\infty\right]\)

M​x(t) = \(0.1\left[e^{t-\frac{1}{10}}-\infty\right]\)

M​x(t) = ∞

Hence, it is verified that the average length of such a call in moment generating function is M​x(t) = ∞.

J Susan Milton Introduction To Probability and Statistics Chapter 3 Discrete Distributions Exercises

by

Introduction To Probability And Statistics Principles And Applications Chapter 3 Discrete Distributions Exercises

 

Introduction To Probability And Statistics Chapter 3 Exercises Solutions Page 73  Exercise 1  Problem 1

We are asked to identify whether the given variable is discrete or not discrete.

Given that M as the number of meteorites hitting a satellite per day. It seems to be a count variable because it will take the value as 0,1,2…

Hence, the number of times a satellite gets hit by the meteorites is random and countable.

So M is a discrete random variable.

Therefore, we conclude that M is a discrete random variable as it holds the countable many values.

 

J. Susan Milton Discrete Distributions Chapter 3 Answers Page 73  Exercise 2  Problem 2

We are asked to identify whether the given variable is discrete or not discrete.

Given that N as the number of neutrons expelled per thermal neutron that is absorbed in the uranium fission−235.

This seems to be a count variable because it takes the value as 0,1,2…

Hence, the number of neutrons gets expelled is random and so N is a discrete random variable.

Therefore, we conclude that N is a discrete random variable as it holds the countable many values.

Read and Learn More J Susan Milton Introduction To Probability And Statistics Solutions

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 73  Exercise 3  Problem 3

We are asked to identify whether the given variable is discrete or not discrete.

Given that prompt neutrons holds for 99% of all emitted neutrons and are released within 10−4 of this fission.

Delayed neutrons are emitted for several hours.

Let us take D as the random variable which represents the time at which the emission of delayed neutron is continuous.

The feasible values of the random variable D will be the set of some intervals or continuous of real numbers.

Therefore, we conclude that the variableD is not discrete as the set of real numbers is neither finite nor countably infinite.

 

Solutions To Discrete Distributions Exercises Chapter 3 Susan Milton Page 74  Exercise 4  Problem 4

We are asked to identify whether the given variable is discrete or not discrete.

Given that the variable O is the random variable which represents the actual resistance of a bell selected at random.

From the given question we are able to understand that the value of O will be between 1.5 and 1.5.

Therefore, we conclude that the variable O is a continuous random variable as the value will be between 1.5 and 3.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 74  Exercise 5  Problem 5

We are asked to identify whether the given variable is discrete or not discrete.

Given that the variable X denotes the number of power failures per month in the Tennessee Valley power network.

This seems to be a count variable since it holds the value 0,1,2,…

So it consists of a countable many values.

Therefore, we conclude that the variable X is a discrete random variable as the power failure will be happening at a random value.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 74   Exercise 6  Problem6

J.Susan Milton Introduction To Probability and Statistics Chapter 3 Discrete Distributions Exercises

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 74  Exercise 6  Problem 7

In a blasting soft rock such as limestone, the holes bored to hold the explosives are drilled with a Kelly bar.

Given that the variable X will be the number of holes which can be drilled per bit and we are asked to find the table for F.

For the values x = 1,2,3,4,5,6,7,8, the value of F(x) will be
​
​F(1) = P[X ≤ 1]

​F(1)  = f(1)

​F(1)  =0.02
​

​F(2) = P[ X≤2 ]

​F(2)  = f(1) + f(2)

​F(2) = 0.02+0.03

​F(2) = 0.05
​

​F(3) = P[X ≤ 3]

​F(3) = f(1)+ f(2) + f(3)

​F(3) = 0.02 + 0.03 + 0.05

​F(3) = 0.1
​

​F(4) = P[X ≤ 4]

​F(4) = f(1) + f(2) + f(3) + f(4)

​F(4) = 0.02 + 0.03 + 0.05 + 0.2

​F(4)  = 0.3
​
​F(5) = P[X ≤ 5]

​F(5) =f(1) + f(2) + f(3) + f(4) + f(5)

​F(5)= 0.02 + 0.03 + 0.05 + 0.2 + 0.4

​F(5) = 0.7
​
​F(6) = P[X ≤ 6]

​F(6) = f(1) + f(2) + f(3) + f(4) + f(5) + f(6)

​F(6) = 0.02 + 0.03 + 0.05 + 0.2 + 0.4 + 0.2

​F(6) = 0.9
​

​F(7) = P[X ≤ 7]

​F(7)  = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7)

​F(7)  = 0.02 + 0.03 + 0.05 + 0.2 + 0.4 + 0.2 + 0.07

​F(7)  = 0.97
​

​F(8) = P[X ≤ 8]

​F(8)  = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) + f(7) + f(8)

​F(8) = 0.02 + 0.03 + 0.05 + 0.2 + 0.4 + 0.2 + 0.07 + 0.03

​F(8)  = 1
​
Therefore, the table for the value F will be

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 74 Exercise 6 Problem 7 values

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 74  Exercise 6  Problem 8

In a blasting soft rock such as limestone, the holes bored to hold the explosives are drilled with a Kelly bar.

Given that the variable X will be the number of holes which can be drilled per bit and we are asked to find out the probability that a bit can be used to drill between three and five holes inclusive.

Using the F table, we get the probability of drilling between three and five holes inclusive

​P [3 ≤ X ≤ 5] = P[X ≤ 5] − P[X > 3]

​P [3 ≤ X ≤ 5] = P[X ≤ 5]−P[X ≤ 2]

​P [3 ≤ X ≤ 5] = 0.7 − 0.05

​P [3 ≤ X ≤ 5]= 0.65
​
Therefore, by using the table F, we get the probability of drilling between three and five holes inclusive is 0.65.

 

Chapter 3 Discrete Distributions Examples And Answers Susan Milton Page 75  Exercise 7   Problem 9

Let X denote the number of computer systems operable at the time of the launch.

Assume that each system is operable is 0.9.

We have to use the tree of table to find the density table.

Obtain the density table by

From the table  sample space (S) is given below:

S={yyy,yyn,yny,ynn,nyy,nyn,nny,nnn}

Here, the probability of system is operable(y) is 0.9 and probability of system is not operable(n) is 0.1 =( 1−0.9).

Therefore, the probabilities are given below:

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 7 Problem 9 Probabilities 1

The density for X is given below

At x = 0, f(0) = (0.1)3

At x = 1

f(1) = (0.9)(0.1)2 + (0.9)(0.1)2 + (0.9)(0.1)2

f(1) =  (0.9)(0.1)2 (1 + 1 + 1)

f(1) = 3(0.9)(0.1)2

At x = 2

f(2) = (0.1)(0.9)2 + (0.1)(0.9)2 + (0.1)(0.9)2

f(2) = (0.1)(0.9)2 (1 + 1 + 1)

f(2) = 3(0.1)(0.9)2

At x = 3

f(3) = (0.9)3

The density table for X is given below

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 7 Problem 9 Probabilities 2

Hence the equation x 2+ 6x in the form of (x + k)2 + his (x + 3)2 − 9.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 75  Exercise 7  Problem 10

Let X denote the number of computer systems operable at the time of the launch.

Assume that each system is operable is 0.9.

There is a pattern to the probabilities in the density table.

In particular f(x) = k(x)(0,9)x (0.1)3 − x

Where k(x) gives the number of paths through the tree yielding a particular value for X.

We have to use F to find the probability that at least one system is operable at launch time.

We have to find the value of P(x ≥ 1).

Consider

⇒  P(x ≥ 1)

= 1−P(x < 1)

=  1−P(x ≤ 0)

=  1 − F(0)

From F table, the value of F(0) is 0.001.

Therefore

⇒ P(x ≥ 1) 

= 1 − F(0)

= 1 − 0.001

= 0.999

Thus the probability that at least one system is operable at launch time is 0.999.

Hence using F table, the probability that at least one system is operable at launch time is 0.999.

 

Probability And Statistics J. Susan Milton Chapter 3 Solved Step-By-Step Page 75  Exercise 8  Problem 11

Given the function is

F(x) = \( \begin{cases}0 & x<0 \\ .70 & 0 \leq x<1 \\ .90 & 1 \leq x<2 \\ .95 & 2 \leq x<3 \\ .98 & 3 \leq x<4 \\ .99 & 4 \leq x<5 \\ 1.00 & x \geq 5\end{cases}\)

To draw the graph of this function.

Cumulative distribution

Let X be a discrete random variable with density f.

The cumulative distribution function for X, denoted by F is defined by F(x) = P[X≤x]for x real.

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 8 Problem 11 density values 1

Yes. This function is called the step function.

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 8 Problem 11 density values 2

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 8 Problem 11 density values 3

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 8 Problem 11 density values 4

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 75   Exercise 8  Problem 12

Given that

F(x) = \(\begin{cases}0 & x<0 \\ .70 & 0 \leq x<1 \\ .90 & 1 \leq x<2 \\ .95 & 2 \leq x<3 \\ .98 & 3 \leq x<4 \\ .99 & 4 \leq x<5 \\ 1.00 & x \geq 5\end{cases}\)

Need to determine that it is a continuous function

A function is said to be continuous as the values of x increases the function also increases.

Here I have attached an example graph which clearly explain that the function is continuous function.

Example of graph:

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 8 Problem 12 Increase values

Similarly in our case also the graph increases as the value of x increases the function value also increases .

For example ,in our case consider any value to check that above said condition are satisfied.

1. f(a) Exists for any value of x

2. \(\lim _{x \rightarrow a} f(x)\) for each and every value of x from 0 to 5,the function exists.

3. \(\lim _{x \rightarrow a} f(x)\) = f(a) at last the function for every values

Therefore the given function is a continuous function.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 75  Exercise 8  Problem 13

We need to find the \(\lim _{x \rightarrow \infty} F(x)\) and \(\lim _{x \rightarrow \infty}\) F(x)

Cumulative distribution

Let X be a discrete random variable with density f.

The cumulative distribution function for X, denoted by F, is defined by F(x) = P[X≤x] for x real.

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 8 Problem 13 density values 1

The value of \(\lim _{x \rightarrow \infty} \) F(x) = 1 and the value of \(\lim _{x \rightarrow \infty}\) F(x) = 0

Introduction to Probability and Statistics Principles and Applications Chapter 3 Discrete Distributions Page 75 Exercise 8 Problem 13 density values 2
The value of \(\lim _{x \rightarrow \infty} \)F(x) = 1 and the value of \(\lim _{x \rightarrow \infty}\) F(x) = 0

 

Step-By-Step Guide To Discrete Distributions Exercises Chapter 3 Milton Page 76  Exercise  9  Problem 14

Given: In an experiment to graft Florida sweet orange trees to the root of a sour orange variety, a series of five trials is conducted.

Let X denote the number of grafts that fail.

The density for X is given in

To find –  Find E[X]

We can find the E[X] by the following formula

E[X] = \(\sum_z x f(x)\)

So the expectation of the random variable X wil be as foliows:
​
​E[X] = \(\sum_z x f(x)\)

​E[X]  = 0 × f(0) + 1 × f(1) + 2 × f(2) + 3 × f(3) + 4 × f(4) + 5 × f(5)

​E[X] = 0 × 0.7 + 1 × 0.2 + 2 × 0.05 + 3 × 0.03 + 4 × 0.01 + 5 × 0.01

​E[X] = 0 + 0.2 + 0.1 + 0.09 + 0.04 + 0.05

​E[X] = 0.48

​For the above two calculations we have used the R software.

Hence, from the above explanation value of E[X] is 0.48

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76   Exercise 9  Problem 15

Given: In an experiment to graft Florida sweet orange trees to the root of a sour orange variety, a series of five trials is conducted.

Let X denote the number of grafts that fail.

The density for X is given in

TO find – Find μx

Since, we know that μx =  E[X]

Therefore;So the expectation of the random variable X wil be as foliows:

E[X]  = \(\sum_z x f(x)\)

E[X]  = 0 × f(0) + 1 × f(1) + 2 × f(2) + 3 × f(3) + 4 × f(4) + 5 × f(5)

E[X]  = 0 × 0.7 + 1 × 0.2 + 2 × 0.05 + 3 × 0.03 + 4 × 0.01 + 5 × 0.01

E[X]  = 0 + 0.2 + 0.1 + 0.09 + 0.04 + 0.05

E[X]  = 0.48

Hence, μx E[X] = 0.48

Hence, from the above explanation the value of μX is equal to 0.48.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76  Exercise 9   Problem 16

In an experiment to graft Florida sweet orange trees to the root of a sour orange variety, a series of five trials is conducted.

Let X denote the number of grafts that fail.

The density for X is  

To Find μx

We can find the E[X2] by the following formula

E[X2] = \(\sum_z x f(x)\)

So the expectation of the random variable x will be as follows:
​
​E[X2] =  \(\sum_z x f(x)\)

f(x) =0 × f(0)+ 1 × f(1) + 22 × f(2) + 32 × f(3) + 42× f(4) + 52 × f(5)

f(x) = 0 × 0.7 + 1 × 0.2 + 4 × 0.05 + 9 × 0.03 + 16 × 0.01 + 25 × 0.01

f(x) = 0.210.210.27 + 0.1610.25

f(x) = 1.08
​
For the above two calculations we have used the R software.

Hence, from the above explanation the value of E[X2] is equal to 1.08.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76   Exercise 9  Exercise 17

Given: In an experiment to graft Florida sweet orange trees to the root of a sour orange variety, a series of five trials is conducted.

Let X denote the number of grafts that fail.

The density for X is

To find –  Find σ X2.

We can find the σX = Var X by the following formula

Var X = E[X2] − (E[X])(E[X])2……………. (1)

Using the above & from Equation 1

VarX = E[X2]−(E[X])2

VarX = 1.08 − 0.482

VarX = 1.077

Hence, from the above explanation the value of the σ​X2 is equal to 1.077.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76  Exercise 10  Exercise 18

Given: The density for X,the number of holes that can be drilled per bit while drilling into limestone is given in

To find –  Find E[X] and (E[X])2

In blasting soft rock such as limestone, the holes bored to hold the explosives are drilied with a Kelly bar. 

This drill is designed so that the explosives can be packed into the hole before the drill is removed. 

This is necessary since in soft rock the hole often collapses as the drill is removed. The bits for these drills must be changed fairly often. 

Let X denote the number of holes that can be drilled per bit (a) We can find the E[X] by the following formula−

So the expectation of the random variable will X be as foliows:
 
​​E[X] =  \(\sum_s x f(x)\)
 
​​E[X] = 1 × f(1) + 2 × f(2) + 3 × f(3) + 4 × f(4) + 5 × f(5) + 6 × f(6) + 7 × f(7) + 8 × f(8)
 
​​E[X] =1 × 0.02 + 2 × 0.03 + 3 × 0.05 + 4 × 0.2 + 5 × 0.4 + 6 × 0.2
 
​​E[X] = 7 × 0.07 + 8 × 0.03 = 0.02 + 0.06 + 0.15 − 0.8 + 2 + 1.2 + .49 + 0.24 = 4.96
​

For the above two calculations we have used the

We can find the E[X2] by the following formula− 

E[X2] \(-\sum_z x^2 f(x)\)

So the expectation of the random variable X wil be as follows:

​​E[X2] =  \(-\sum_z x^2 f(x)\)

​​E[X2] = 1 × f(1) + 2 × f(2) + 32 ×f (3) + 42 × f(4)+ 52 × f(5) +62 × f(6)× 72 × f(7)+ 82 × f(8)

​​E[X2] = 1 × 0.02 + 4 × 0.03 + 9 × 0.05 + 16× 0.2+ 25 × 0.4+ 36 × 0.2+ 19 × 0.07 + 61 × 0.03

​​E[X2] = 0.02 + 0.12 + 0.45 + 3.2 + 10 + 7.2 + 3.43 + 1.92

​​E[X2] = 26.34

For the above two calculations we have used the R software.

​Hence, from the above explanation the value of E[X] & E[X2] is equal to 4.96 & 26.34

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76   Exercise 10  Exercise 19

Given: The density for X, the number of holes that can be drilled per bit while drilling into limestone is given in

To find –  Find Var X and σx .

We can find the Var X by the following formula

Var X = E[X ]−  (E[X])2

Using the above we get

​Var X = E[X2] − (E[X])2

​Var X =  26.34 − 4.962

​Var X =  1.7384

Star dard deviation of X is giver by σx

= \(\sqrt{VarX}\)

Using the above we get

σx = \(\sqrt{VarX}\)

σx=  \(\sqrt{1.7384}\)

σx= 1.318
​
Hence, from the above explanation the value of Var X and σx is 1.7384 & 1.318.

​

Exercise Solutions For Chapter 3 Susan Milton Discrete Distributions Page 76  Exercise 10  Exercise 20

Given : The density for X ,the number of holes that can be drilled per bit while drilling into limestone is g

To find – What physical unit is associated with σX?

The physical unit associated with σx is the number of holes that can be drilled per bit.

Hence, from the above explanation the physical unit associated with σx is the number of holes that can be drilled per bit.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76  Exercise 11  Problem 21

Given: Let X be a discrete random variable with density f.

Let c be any real number.

To find – Show that E[c] = c

As f(x) is the density of X given in the table, it should satisty \(\sum_{a l k} f(x)\) = 1

We can find the E[X] by the following formula-

E[X] = \(\sum_z x f(x)\)

So the expectation of c will be as follows:

E ∣c∣ = \(\sum_z c f(x)-c \sum_x f(x)\)

Hence, from the above explanation showed that E[c]= c

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76  Exercise 11  Problem 22

Given: Let X be a discrete random variable with density f .

Let c be any real number.

To find – Show that E[cX] = cE[X]

As f(x)is the density of X given in the table, it should satisty

\(\sum_{a l k} f(x)\) = 1

We can find the E[X] by the following formula-

E[X] = \(\sum_z x f(x)\)

So the expectation of cX will te as follows:
​
​E[cX] = \(\sum_z\)cxf(x)

​E[cX] =  c \(\sum_z x f(x)\)
​
​E[cX] = cE[X]

Hence the above explanation we have showed that E[cX] = cE[X].

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76  Exercise 12  Problem 23

Given: Use the rules for expectation

To find – To verify that Varc = 0, and Varc X = c{2} Var X for any real number c

We can find the Var X by the following formula

Var X = E[X]2 − (E[X])2

In order to get Var c we need to caculate E[c2] and  E[c].

We can find the E[X] by the following formula

E[X] = \(\sum_z x f(x)\)

So the expectation of C will be as follows

E[c] = \(\sum_z x cf(x)\)

E[c] = c\(\sum_x f(x)\)

E[c] = c

We can find the E[X2] by the following formula

E[X2] = \(\sum_z c^2 f(x)\)

So the expectation of C will be as follows

E[c2] =  \(\sum c^2 f(x)\)

E[c2] = c2\(\sum_z c x f(x)\)

E[c2] = c2

Using the above we get

Varc = E[c2 ]−(E[c2 ])

Varc= c2− c2 = 0

In orderto get Var(cX) , we need to calculate E[(cX)2 ]and F[cX].

We can find the E[cX] by the following formula

E[cX]= \(\sum_z c x f(x)\)

So the expectation of C will be as follows
​
​E[c] =\(\sum_x c x f(x)\)

​E[c] = c\(\sum_x f(x)\)

​E[c] = cE[X]

We can find the F[X2]by the following formula

E[X2] = \(\sum_z x^2 f(x)\)

So the expectation of (cX)2 will be as follows:

​E[c2X2] = \(\sum_z c^2 x^2 f(x)\)

​E[c2X2] = c2 \(\sum_z x^2 f(x)\)

​E[c2X2] = c2E[X2]

Using the above we get, Var cX = E[c2X2]− (E[cX])2 = e2 E[X2] − c2(E[X])2

= c2(E[X2] − (E[X])2)

= c2 Var X

Hence, from the above explanation by using the rules for expectation we verify that Varc = 0 and Varc X = c2 Var X for any real number c

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76  Exercise 13  Problem 24

Given: Let X and Y be independent random variables with

E[X] = 3, E[X2] = 25, E[Y] = 10 and E[Y2] = 164

To find – Find E[3X + Y − 8]

Let X and Y be two independent random variables with

E[X] = 3, E[X2] = 25, E[Y] = 10 E[Y2] = 164

Using the above properties given in tip of expectation we can say:

​E[3X + Y − 8] ⇒ E[3X] + E[Y] + E[−8] (Using rule 3)

E [3X + Y − 8]  = 3E[X] + E[Y] + E[−8]​ (Using rule 2)

E [3X + Y − 8] = 3E[X] + E[Y] − 8​

E [3X + Y − 8] = 3 × 3 + 10 − 8

​E [3X + Y − 8] = 9 + 10 − 8

E [3X + Y − 8] = 11
​
So we find E [3X + Y − 8] = 11

Hence, from the above explnation the value of E[3X+Y−8]=11.

 

Page 76  Exercise 13  Problem 25

Given: Let X and Y be independent random variables with E[X]= 3, E[X2 ] = 25, E[Y] = 10 and E[Y2] = 164

To find – Find E [2X − 3Y + 7].

We will use some properties of expectation.

LetX and Y be random variables and C be any real number.

1. E[c] = c   (The expected value of any constant is that constant)

2. E[cX] = cE ∣X∣    (Constants can be fectored from expectat ons.)

3. E[X + Y] = E[X] − E[Y]   (The expected value of a sum is equal to the sum of the expected values.)

Using the above properties of expectation we can say-

​E[2X−3Y+7]= E[2X] − E[−3Y] + E[7]​    (Using rule 3)

E[2X − 3Y + 7]= 2E[X]−(−3)E[Y]+E[7]

E[2X − 3Y + 7]= 2E[X]−(−3)E[Y] + 7​   (Using rule 1)

E[2X − 3Y + 7]= 2 × 3 − 3 × 10 + 7

E[2X − 3Y + 7]= 6 − 30 + 7

E[2X − 3Y + 7]= −17
​
So we find E[2X − 3Y + 7] = −17

Hence, from the above explanation the value of E[2X − 3Y + 7] = −17

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 76  Exercise 15  Problem 26

Let X and Y be independent random variables with E[X] = 3

E[X2] = 25, E[Y] = 10 and  E[Y2] = 164.

We have to find VarX.

From the formulae of variance we know that

Var X =  E[X2] − (E[X])2

Using the above we get

Var X = E[X2] − (E[X])2

Var X= 25−32

Var X= 25−9

Var X= 16

Hence the value of V ar X is 16.

 

Page 76  Exercise 15  Problem 27

Let X and Y be independent random variables with E[X] = 3, E[X2 ] = 25
,E[Y] = 10 and E[Y2 ] =164.

We have to find σx.

Standard deviation of X is given by σx

σx=  \(\sqrt{Var X}\)

σx=  \(\sqrt{16}\)

σx= 4

Hence the value of σx .is 4..

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 77  Exercise 16  Problem 28

Given: The function f is defined by f(x) = (1/2)−∣x∣

Where x = ±1,± 2,± 3,± 4,….

To be found: Verify that the given function is the density for a discrete random variable X

We have, the function f is defined by f(x) = (1/2)2−∣x∣

where x = ± 1,± 2, ± 3,± 4,….

Using the first condition for all, we get

We know, x is a real number

​⇒ 2 − ∣x∣ ≥ 0

⇒ f(x) ≥ 0

​Using the second condition to verify, we get

⇒ \(\sum_x f(x)\) =\(\sum_x \frac{1}{2} 2^{-|x|}\)

⇒ \(\sum_x f(x)\) = \(\sum_x \frac{1}{2} 2^{-|x|}\) = \(\ldots+\sum_x \frac{1}{2} 2^{-|-2|}+\sum_x \frac{1}{2} 2^{-|-1|}+\sum_x \frac{1}{2} 2^{-|1|}+\sum_x \frac{1}{2} 2^{-|2|}+\)………

​⇒  \(\sum_x f(x)\) =\(\sum_x \frac{1}{2} 2^{-|x|}\)

=  2 − 1 + 2  − 2 + 2 − 3 +…………..

​⇒ \(\sum_x f(x)\) = \( \frac{2^{-1}}{1-2^{-1}}\)
​

​⇒ \(\sum_x f(x)\) = \(\sum_x \frac{1}{2} 2^{-|x|}\)

Finally, we get the required condition,\(\sum_x f(x)\) = 1

Hence, verified that the given function

f(x) = (1/2)2−∣x∣, where x = ± 1, ± 2, ± 3, ± 4,…. is the density for a discrete random variable X.

Hence, verified that the given function f(x)=(1/2)2−∣x∣ , where x = ±1, ± 2, ± 3, ± 4,…. is the density for a discrete random variable X.

 

Introduction To Probability And Statistics Principles And Applications Chapter 3 Page 77  Exercise 16  Problem 29

Given: The function f is defined by f(x) = (1/2)2−∣x∣

where  x = ±1,± 2, ± 3,± 4,….

Let g(X) = (−1)∣X∣−1 [2∣x∣/ 2 ∣X∣ − 1)]

To be found: Show that \(\sum_{\text {all }} x(x) f(x)<\infty\)

We have, g(X) =[(2 ∣X∣ /2 ∣X∣ − 1)] and f(x) = (1/2)2−∣x∣

Now, substituting the above values and expanding the series, we get \(\sum_{\text {all }} x g(x) f(x)\)

​⇒ \(\sum_{a l l} x(x) f(x)=\sum_x(-1)^{|x|-1}\left[\frac{2^{|x|}}{(2|x|-1)}\right]\)\(\frac{1}{2} 2^{-|x|}\)

​⇒ \(\sum_{a l l}{ }_x g(x) f(x)=\sum_x(-1)^{|x|-1}\left[\frac{2^{|x|}}{(2|x|-1)}\right] \frac{1}{2}\)

​\(\Rightarrow \sum_{\text {all }} x g(x) f(x)=\sum_x^{\infty}(-1)^{x-1}\left[\frac{2^x}{(2 x-1)}\right] \frac{1}{2}\)

Expanding the series, we get a final alternating series which converges

\(\Rightarrow \sum_{\text {all }}{ }_x g(x) f(x)=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots \ldots+\infty\)

 

Hence , proved that \(\sum_{a l l} x g(x) f(x)\)

By the method of expansion, it is shown that \(\sum_{a l l} x g(x) f(x) \) <∞

J Susan Milton Introduction To Probability And Statistics Chapter 1 Introduction To Probability And Counting Exercises

by

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Exercises

 

Introduction to Probability and Statistics Chapter 1 exercises solutions Page 14  Exercise 1 Problem 1

According to the question, A government study defines a “group 1” nuclear accident to be one involving severe core damage, melting of uranium fuel, essential failure of all safety systems, and a major breach of the reactor’s containment resulting in a large release of radioactivity into the atmosphere.

In1982, officials at Nuclear Regulatory commission estimated the probability of such an accident occurring in the United States before the year 2000 to be.02.

We need to tell which approach to probability is used to determine the value.

According to data given in the question it has dangerous consequences, this experiment definitely isn’t repeatable.

So, officials at the Nuclear Regulatory Commission estimated the given probability by using Classical Method.

Officials at Nuclear Regulatory commission estimated the probability of such an accident occurring in the United States before the year 2000 to be.02 by using Classical Method.

J. Susan Milton Probability And Counting Chapter 1 Answers Page 14  Exercise 2  Problem 2

According to the question, Hemophilia is a sex-linked hereditary blood defect of the males characterized by delayed clotting of the blood which makes it difficult to control bleeding even in the case of a minor injury.

When a woman is carrier of classical hemophilia there is 50%chance that a male child will inherit the disease.

We need to answer what will be the probability that the carrier gives birth to two sons and the approach we used to find the probability.

Read and Learn More J Susan Milton Introduction To Probability And Statistics Solutions

According to data given in the question, we have four possible outcomes for sons to have or not to have disease

J.Susan Milton Introduction To Probability and Statistics Chapter 1 Introduction To Probability And Counting Exercises

​The probability that the carrier gives birth to two sons is 0.25 and the approach we used to find the probability is Classical Method.

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 14  Exercise 3  Problem 3

According to the question, The probability of having a fatal accident in the work place is assessed using the fatal accident frequency rate (FAFR).

This rate is defined by

FAFR = Number of fatalities per1000workers during a working lifetime.

We need to find the approach to probability of an individual having fatal accident while at work.

Also, Find the approximate probability that a coal miner will suffer a fatal injury. Coal mining industry and is taken into account when computing the industry-wide FAFR of 4.

We need to explain how the rate could be so low while at least some of the components used in its computation are high.

Given that  FAFR = Number of fatalities per1000workers during a working lifetime.

Hence we use the relative frequency approach to approximate the given probability

Given that FAFR for coal mining occupation is 12

We use the relative frequency approach, hence we conclude that the probability that a coal miner will suffer a fatal injury is.

⇒ \(\frac{12}{1000}\)

=  0.012

The overall industry FAFR is equal to 4 because for a large number of occupations, FAFR is very low because those occupations are not dangerous and risky for human life.

The probability approach we used is relative frequency approach. The probability that a coal miner will suffer a fatal injury is 0.012. Coal mining industry and is taken into account when computing the industry-wide FAFR of 4, that is very low because those occupations are not dangerous and risky for human life.

Solutions To Probability And Counting Exercises Chapter 1 Susan Milton Page 15  Exercise 4  Problem 4

Questions explains that in ballistics studies conducted during World War II, it was found that inground-to-ground firing, artillery shells tended to fall in an elliptical pattern such as given in the question.

The probability that a shell would fall in the inner ellipse is 0.50 ; the probability that it would fall in the outer ellipse is 0.95.

A firing is considered to be a success (s)if the shell falls within the inner ellipse; otherwise, it is failure (f).

We need to construct a tree to represent the firing of four shells in succession.

For each of the four shells, we have two possible outcomes  (given shell falls within inner ellipse) or f (given shell doesn’t fall within inner ellipse).

Hence the tree diagram that represent the firing of four shells in succession as follows

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 15 Exercise 4 Problem 4 Firing Shells 1

The tree diagram that represent the firing of four shells in succession as follows

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 15 Exercise 4 Problem 4 Firing Shells 2

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 15  Exercise 5 Problem 5

Questions explains that in ballistics studies conducted during World War II, it was found that inground-to ground firing, artillery shells tended to fall in an elliptical pattern such as given in the question.

The probability that a shell would fall in the inner ellipse is 0.50 ; the probability that it would fall in the outer ellipse is 0.95.

We need to list the sample points generated by the tree.

With the help of tree diagram from  Page 15  Exercise 4  Problem 4 , We conclude the sample space and sample points are:

S = {ssss,sssf,ssfs,ssff,sfss,sfsf,sfs,sfff,fsss,fssf,fsfs,fsff,ffss,ffsf,ffs,ffff}

The sample space and the sample points are: S={ssss,sssf,ssfs,ssff,sfss,sfsf,sff,sff,fsss,fssf,fsfs,fsff,ffss,ffsf,fff,ffff}

 

Probability and Statistics J. Susan Milton Chapter 1 solved step-by-step Page 15  Exercise 5  Problem 6

Questions explains that in ballistics studies conducted during World War II, it was found that inground-to ground firing, artillery shells tended to fall in an elliptical pattern such as given in the question.

The probability that a shell would fall in the inner ellipse is 0.50 ; the probability that it would fall in the outer ellipse is 0.95.

Let Ai,i = 1,2,34 denote the event that the i−thfiring is successful. We need to list the sample points that constitute each of the events A1 ,A2,A3,A4 and check are these events mutually exclusive.

The sample points that constitute each of the events A1,A2,A3,A4 are

A1= the first firing is successfulsssss,sssf, ssfs,ssff,sfss,sfsf,sffs,sfff

A2 = The second firing is successful{ssss,sssf,ssfs,ssff,fsss,fssf,fsfs,fsff}

A3 = The third firing is successful{ssss,sssf,sfss,sfsf,fSSS,fssf,ffss,ffSf}

A4 = The fourth firing is successful{sssss, ssfs, sfss, sffs, fsss, fsfs, ffss, fffs }

A,A2,A3,A4 are not mutually exclusive events because event{ssss} is in every four events.

The sample points that constitute each of the events A1,A2 ,A3,A4  are

A1 = {ssss,sssf,ssfs,ssff,sfss,sfsf,sffs,sfff}

A2 = {ssss,sssf,ssfs,ssff,fsss,fssf,fsf,fsff

A3 = {ssss,sssf,sfss,sfsf,fsss,fssf,ffss,ffsf}

A4 = {ssss,ssfs,sfss,sffs,fsss,fsfs,ffs,fffs}

A1, A2, A3, A4 are not mutually exclusive events because event {ssss}is in every four events.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 15  Exercise 5  Problem 7

Questions explains that in ballistics studies conducted during World War II, it was found that inground-to ground firing, artillery shells tended to fall in an elliptical pattern such as given in the question.

The probability that a shell would fall in the inner ellipse is 0.50; the probability that it would fall in the outer ellipse is 0.95.

We need to list the sample points that constitute each of these events and describe the events verbally:
​
​A1′

A1∪A2

A1∩A2

A1∩A2∩A3∩A4

​A1∩A2∩A3∩A′4

(A1∪A2∪A3∪A4)A1∩A′1

​The sample points from the definition of complement, union and intersection

A1′= The first firing is not successful{fSSS,fSSf,fsfs,fsff,ffS,ffsf,fffs,fff}

A1∪A2 = The first or second firing is successful{ssss,sssf,ssfs,ssff,sfss,sfsf,sffs,sfff,fsss,fssf,fsfs,fsff}

A1∩A2  =  The first and second firing is successfulsssss,sssf,ssfs,ssff}

A1∩A2∩A3∩A4 = All four firing are successful {ssss}

A1∩A2∩A3∩A′4= The first three firings are successful and the last one is not {sssf}

(A1∪A2∪A3∪A4)′ = All firings are unsuccessful {fff}

A1∩A1′= The first firing is successful and the first firing is not successful =Φ

The sample points from the definition of complement, union and intersection

A1′ = {fSSS,fssf,fsfS,fSff,fSS,ffsf,ffS,fff}

A1∪A2 = {ssss,sssf,ssfs,ssff,sfs,sfsf,sffs,sfff,fsss,fssf,fsfs,fsff}

A1∩A2 = {ssss,sssf,ssfs,ssff}

A1∩A2∩A3∩A4 = {sSSS}

A1∩A2∩A3∩A4 = {sssf}

(A1∪A2∪A3∪A4) = {ffff}

A1∩A1′ = Φ

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 15  Exercise 5  Problem 8

Questions explains that in ballistics studies conducted during World War II, it was found that inground-to ground firing, artillery shells tended to fall in an elliptical pattern such as given in the question.

The probability that a shell would fall in the inner ellipse is 0.50 ; the probability that it would fall in the outer ellipse is 0.95.

We need to find the probability of each of the events of part(d) by classical probability and why is it true.

The probability for a single shell to fall when the inner ellipse is 0.5, then he probability for a single shell to fall outside of the inner ellipse is also 0.5.From the classical method we have:

P(A) = \(\frac{\text { Number of ways } A \text { can occur }}{\text { Number of ways the experiment can proceed }}\)

Each sample point can occur in one and only one way. Also, there are 16 possible outcomes in total. So, the probability of each sample point is: 1

= \(\frac{1}{16}\)

= 0.0625

The probability of each sample point is: 0.0625

 

Online help for J. Susan Milton Probability Chapter 1 exercises Page 16  Exercise 6  Problem 9

We need to evaluate the expression  9!

We have expression as  9!

​9! = 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

​9!  = 362880
​
The value of the expression 9! is 362880.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 16  Exercise 6   Problem 10

We need to evaluate the expression 6!

We have expression as 6!

​6! = 6 × 5 × 4 × 3 × 2 × 1

​6!  = 720

​The value of the expression 6! is 720.

 

Step-by-step guide to Probability and Counting exercises Chapter 1 Milton Page 16  Exercise 6  Problem 11

We need to evaluate the expression  7P3

We have expression as 7P3

​ 7P3  = \(\frac{7 !}{(7-3) !}\)

​ 7P3  =  \(\frac{7 !}{4 !}\)

​ 7P3  = \(\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\)

Cancilation of (4,3,2,1)

= 7 × 6 × 5

= 210
​
The value of the expression  7P3  is 210.

 

Page 16  Exercise 6  Problem 12

We have expression as  6p2

6p2 = \(\frac{6 !}{(6-2) !}\)

6p2 = \(\frac{6 !}{4 !}\)

6p2 =  \(\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\)

Cancilation of (4,3,2,1)

= 6 × 5

= 30

The value of the expression 6p2  is 30. 

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 16  Exercise 6  Problem 13

According to the question, in investigating the ideal Gas Law, experiment are to be run at four different pressures and three different temperatures.

We need to find the number of experimental conditions are to be studied.

The multiplication principle: Consider an experiment taking place in k stages.

Let ni denote the number of ways in which stage i can occur for  i = 1,2,3,……k.

Altogether the experiment can occur in ways.

\(\prod_{i=1}^k\) ni = n1 × n2  nk

Since we have four different pressures and three different temperatures, according to the multiplication principle, we conclude that  4 × 3 = 12 experimental conditions will be studied.

12 experimental conditions are to be studied.

 

Exercise solutions for Chapter 1 Susan Milton Probability and Counting Page 16  Exercise 6  Problem 14

According to the question, in investigating the ideal Gas Law, experiment are to be run at four different pressures and three different temperatures.

We need to find number of experiments will be conducted on the given gas if each experiment condition is replicated five times.

The multiplication principle: Consider an experiment taking place in k stages.

Let  ni denote the number of ways in which stage  i can occur for i = 1,2,3,……k.

Altogether the experiment can occur in

\(\prod_{i=1}^k\) ni = n1 × n2  nk

Since, each experimental condition is repeated five times, from the multiplication principle, we conclude that 5 × 12 = 60 experimental conditions will be conducted on a given gas.

60 experimental conditions will be conducted on a given gas.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 16  Exercise 6  Problem 15

According to the question, in investigating the ideal Gas Law, experiment is to be run at four different pressures and three different temperatures.

We need to find number of experiments will be conducted to obtain five replications on each experimental condition for each of the six different gases.

The multiplication principle: Consider an experiment taking place in k
stages.

Let ni denote the number of ways in which stage i can occur for i = 1,2,3,……k.

Altogether the experiment can occur in

\(\prod_{i=1}^k\) ni = n1 × n2  nk

Since, each of six gases, there will be 60 conducted experiments from the multiplication principle, we conclude that 6 × 60 = 360

360 experiments will be conducted to obtain five replications on each experimental condition for each of the six different gases.

 

Page 17  Exercise 7  Problem 16

Four artillery shells have been fired in succession, each firing is either considered as a success or a failure.

We need to use the multiplication rule to prove that the number of paths through the tree for this experiment is 16.

There are four artillery shells, k = 4

The experiment is either a success or failure for i = 1,2,3,4

Therefore n1 ,n2,n3 and n4 are equal to 2.

From the multiplication principle

\(\prod_{i=1}^4 n_i\) = 2.2.2.2.

= 16

Number of paths through the tree representing this experiment is 16.

Using the multiplication principle, the number of paths through the tree to represent the experiment which involved firing of four artillery shells is proven to be 16.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 17  Exercise 8  Problem 17

The Apollo mission consists of five components and each component is marked as operable or inoperable but for the proper functioning of this mission all the states must be operable.

We need to indentify the number of states that are operable.

In this mission the number of components are five, k = 5.

The components are either operable or inoperable for i = 1,2,3,4,5

Hence n1 ,n2,n3,n4 and n5 is equal to 2.

Therefore, from the multiplication principle;

\(\prod_{i=1}^5 n_i\) = 2.2.2.2.2

= 32

The number of states that are operable is 32.

For the Apollo mission that consists of five components that is marked as either operable or inoperable, the number of states that are operable is found to be 32.

 

Page 17  Exercise 8  Problem 18

The Apollo mission consists of five components and each component is marked as operable or inoperable but for the proper functioning of this mission all the states must be operable.

We need to indentify the number of states in which LEM is inoperable.

The Apollo mission has five components in which if LEM is inoperable then the remaining states are four, k = 4.

The components are either operable or inoperable for i = 1,2,3,4

Therefore n1 ,n2 ,n3 and n4 are equal to 2.

From the multiplication principle

\(\prod_{i=1}^4 n_i\) = 2.2.2.2

= 16

The number of states in which the LEM is inoperable is 16.

For the Apollo mission that consists of five stages, the number of states when LEM Is inoperable is 16.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 17  Exercise 8   Problem 19

The Apollo mission consists of five components and each component is marked as operable or inoperable but for the mission to be partially successful the first three components must be operable.

Therefore we need to find the number of states which represents, partially successful mission.

The mission has five components in which if the first three components are operable then the mission is considered as partially successful.

So we keep the first three components fixed and the last two components is either operable or inoperable hence according to the multiplication principle;

n4  = n5 ⇒ 2

= 2⋅2

= 4

The number of states that represent a partially successful mission

For the Apollo mission to be partially successful the first three components must be operable hence the number of states that represents a partially successful mission is 4.

 

Page 17  Exercise 8  Problem 20

The Apollo mission consists of five components and each component is marked as operable (o) or inoperable (i) but for the success of this mission all the states must be operable.

We need to in dentify the number of states when the mission is fully successful.

The operable state is denoted as (o) and the inoperable state is denoted as (i) For a complete successful mission all the five components must be operable and therefore the numbers of states that represent fully successful mission is ooooo.

For the Apollo mission to be fully successful the all five components must be operable hence the number of states that represents a fully successful mission is ooooo.

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 17  Exercise 9   Problem 21

Binary code consists of on (1) and off (0) values.

When each pixel is quantized to gray level using a binary code, we need to find how many gray levels can be quantized using a four bit binary code.

Four level binary code is there in which each binary digit is considered as a stage hence κ = 4

In binary system we have (0)’s and (1)’s hence  n1,n2,n3 and n4 are equal to 2.

From the multiplication principle

\(\prod_{i=1}^4 n_i\) = 2.2.2.2

The number of gray levels that are quantized using four bit binary system is 16.

The number of gray levels that are quantized using four bit binary system is 16.

 

Page 17  Exercise 9  Problem 22

Binary code consists of on (1) and off (0) values.

We need to find the number of bits required to code a pixel that is quantized to 32  grey levels

The number of bits required is the number of stages which is equal to k.

Since there are two possibilities0and 1 we have, 2⋅2⋅2⋅…⋅nk = 32

2k =  32

k = 5 since (25 = 32)

Therefore the number of bits of binary code that is required to code a pixel that is quantized to 32 grey levels is 5.

The number of bits of binary code that is required to code a pixel that is quantized to 32 grey levels is 5.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 17  Exercise 10  Problem 23

We know that nCr= \(\left(\begin{array}{l}n \\r\end{array}\right)\) = \(\frac{n !}{r !(n-r) !}\) , hence we have to prove  nCr = nCn-r

We know that    nCr  = \(\left(\begin{array}{l}n \\r \end{array}\right) \Rightarrow \frac{n !}{r !(n-r) !}\)

So , Cn-r = \(=\left(\begin{array}{l}n \\n-r\end{array}\right) \Rightarrow \frac{n !}{(n-r) !(n-(n-r)) !}\)

Cn-r  = \(\frac{n !}{r !(n-r) !} \Rightarrow{ }_n C_r\)

Hence  nCr = nCn-r

Since nCr=\(\left(\begin{array}{l}n \\r\end{array}\right) \Rightarrow \frac{n !}{r !(n-r) !}\) and also  Cn-r =\(\left(\begin{array}{l}n \\n-r\end{array}\right)\)  \(\Rightarrow \frac{n !}{(n-r) !(n-(n-r)) !}\) , we prove that nCr =  nCn-r.

 

Page 17  Exercise 11  Problem 24

Given five compilers, pair wise comparisons are done hence we need to find the combinations of these five compilers selected two at a time.

The number of compilers is five, n = 5

Two are compared, r = 2

Substituting in the equation  nCr = \(\left(\begin{array}{l}n \\r\end{array}\right)\) \(\Rightarrow \frac{n !}{r !(n-r) !}\)

5C2 = \(\left(\begin{array}{l}5 \\2\end{array}\right) \Rightarrow \frac{5 !}{2 !(5-2) !}\)

5C2  = \(\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{2 \cdot 1(3 \cdot 2 \cdot 1)}\)

5C2  = \(\frac{20}{2}\)

5C2  = 10

The number of pair wise comparisons made are 10.

Given five compilers, the number of pair wise comparisons made are 10.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 17  Exercise 12  Problem 25

We need to use the formula nCr = \(\frac{n !}{r !(n-r) !}\)

Where

n = 103 Is the pool of qualified applicants

r = 22 Is number of applications to select.

We need to find nCr  = \(\frac{n !}{r !(n-r) !}\)

Substituting values

nCr = \(\frac{n !}{r !(n-r) !}\)

Substituting values

nCr  ​⇒  \(\frac{103 !}{22 !(81) !}\)

nCr  ⇒ 1.51978828 × 1022

The number of ways 22 applications can be selected from a pool of 103 applications is 1.51978828 × 1022.

 

Page 17  Exercise 12  Problem 26

We need to use the formula = \(\frac{n !}{r !(n-r) !}\)

Considering that you are one of the applicants, we need to find the number of pools you will be included it.

Thus, of the 22 people, you are one of them and the other 21 people will be chosen from a pool of 102

Where

n = 102 is the pool of qualified applicants

r = 21 Is number of applications to select.

Substituting values

nCr  = \(\frac{n !}{r !(n-r) !}\)

nCr   ⇒ \(\frac{102 !}{21 !(81) !}\)

nCr   ​⇒ 3.25 × 10 21

The number of sub-groups you will be included in is  3.25 × 10 21.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 17  Exercise 12  Problem 27

We need to find the probability of being selected considering all candidates are equal.

The number of times you will be in the pool of selected applicant is  3.25 × 10 22.

The total number of ways pools can be formed is 1.52 × 10 22

We need to find the probability of getting selected.

Find the probability of being selected

P(A) = \(\frac{A}{B}\)

​⇒  \(\frac{3.25 \times 10^{21}}{1.52 \times 10^{22}}\)

​⇒ 0.21

The probability of getting selected from a pool of 103 applicants is 0.21.

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 18  Exercise 13  Problem 28

There are 128 – bit messages.

Each bit can be either correct or incorrect.

Hence, the total number of possible messages are 2 218

We need to find the number of cases where only two of these bits are wrong and the rest are correct.

For this we look at number of ways two bits can be selected of theone-twenty-eight.

Using that, we shall find the probability.

Find the number of events satisfying our condition

nCr = \(\frac{n !}{r !(n-r) !}\)

128 C2 = \(\frac{128 !}{2 !(128-2) !}\)

Find the probability

The parobability of two of the bits being wrong

⇒ \(\frac{2-b i t s}{\text { Total }}\)

⇒ \(\frac{\frac{128 !}{2 !(128-2) !}}{2^{128}}\)

⇒ 2.39 10– 35

⇒ \(\frac{127 \times 64}{2^{128}}\)

The probability of only two of the bits being wrong is 2.39 × 10– 35

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 18  Exercise 13  Problem 29

There are \(\frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !}\) experiments with 4 different temperatures, each three times.

So, comparing with

n = n1 + n2 +…+ nk

12  = 3 + 3 + 3 + 3

We need to find the number of ways the experiment can be conducted.\(\frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !}\)

Substituting into  \(\frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !}\)

⇒ \(\frac{12 !}{3 ! \times 3 ! \times 3 ! \times 3 !}\)

⇒ 369600

The number of ways the experiment can be conducted is 369600.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 18  Exercise 13  Problem 30

We need to prove that \(\left(\begin{array}{c}
n \\
n_1
\end{array}\right)\left(\begin{array}{c}
n-n_1 \\
n_2
\end{array}\right) \ldots\left(\begin{array}{c}
n-n_1-n_2 \ldots n_{k-1} \\
n_k
\end{array}\right)\) = \(\frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !}\)

⇒ \(\frac{n !}{n_{1} !\left(n-n_1\right) !} \times \frac{\left(n-n_1\right) !}{n_{2} !\left(n-n_1-n_2\right) !} \cdots \cdot \frac{\left(n-n_1-n_2 \ldots n_{k-1}\right) !}{n_{k} !\left(n-n_1-n_2 \ldots n_k\right) !}\)

Every denominator of the numerator is cancelled out by the denominator of the previous term’s part leaving us with

\(\frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !\left(n-n_1-n_2 \ldots n_k\right) !}\)

 

⇒ \(\frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !(n-n) !}\)

\(\Rightarrow \frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !}\)

Which is the RHS

We thus proved that \(\left(\begin{array}{c}n \\n_1\end{array}\right)\left(\begin{array}{c}n-n_1 \\n_2\end{array}\right) \ldots\left(\begin{array}{c}
n-n_1-n_2 \ldots n_{k-1} \\n_k\end{array}\right)\) =\( \frac{n !}{n_{1} ! \times n_{2} ! \ldots n_{k} !\left(n-n_1-n_2 \ldots n_k\right) !}\)

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 18  Exercise 14  Problem 31

We need to find n when

\(\left(\begin{array}{l}n \\2\end{array}\right)\) = 21 , \(\left(\begin{array}{l}n \\2\end{array}\right)\) = 105

We shall use the formula  \(\left(\begin{array}{l}
n \\
r
\end{array}\right)\) = \(\frac{n !}{r !(n-r) !}\)

For \(\left(\begin{array}{l}
n \\
2
\end{array}\right)\) = 21

Substituting in 

\(\left(\begin{array}{l}
n \\
r
\end{array}\right)=\frac{n !}{r !(n-r) !}\) 

= \(\frac{n !}{r !(n-r) !}\)

⇒ \(\frac{n !}{2 !(n-2) !}\) = 21

⇒ n(n – 1) = 21 (2)

⇒ n(n – 1) = 42

Thus we Know n = 7

For \(\left(\begin{array}{l}n \\2\end{array}\right)\) = 105

Substituting in 

⇒  \(\frac{n !}{r !(n-r) !}\)

⇒ \(\frac{n !}{2 !(n-2) !}\) 105

⇒ n(n – 1) = 105(2)

⇒  n(n – 1) = (15)(14)

Thus n = 15

For \(\left(\begin{array}{l}n \\2\end{array}\right)\)  = 21 , For \(\left(\begin{array}{l}n \\2\end{array}\right)\) = 105 , n = 15.

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 19  Exercise 15  Problem 32

There are 25 packages.

10 packages need to be chosen.

The number of ways this can be done is

⇒  \(\left(\begin{array}{l}25 \\10\end{array}\right)\).

Then we need to find ways to select 3 games  if  5 games packages exist within the given packages.

The number of ways this can be done is

⇒ \(\left(\begin{array}{l}20 \\7\end{array}\right)\) \(\left(\begin{array}{l}5 \\3\end{array}\right)\)

Finding \(\left(\begin{array}{l}20 \\7\end{array}\right)\) , \(\left(\begin{array}{l}5 \\3\end{array}\right)\)

⇒  \(\frac{20 !}{7 !(13) !} \times \frac{5 !}{3 !(2) !}\)

 = 775200

The number of ways 10 packages can be chosen is 3.27 × 106. The number of ways three of these can be games is 775200.

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 19  Exercise 16  Problem 33

There are four women and six men.

We need to find the total number of ways three employees can be chosen at random.

Then we need to find the ways no women is chosen.

The ration of the two gives us the probability.

Total number of ways three employees can be chosen is

\(\left(\begin{array}{1}10\\3\end{array}\right)=\frac{10 !}{3 !(7) !}\)

 

⇒ 240

Number of ways no women is chosen is the number of ways three men are chosen which is

\(\left(\begin{array}{1}6 \\3\end{array}\right)=\frac{6 !}{3 !(3) !}\)

 

⇒ 20

Probability that no women is chosen is

⇒ \(\frac{20}{240}\)

Which is \(\frac{1}{12}\).

Probability that no women is chosen is \(\frac{1}{12}\) As the probability is very low, the occurrence of such an event is suspicious as there is a god chance that at least one woman would be chosen if randomly chosen.

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 19  Exercise 17  Problem 34

We need to use five alphabets and one digit to make the password.

Any of the twenty six alphabets and ten digits can be used.

We need to find the total number of all possible passwords.

Repeating the alphabets is allowed.

The total number of passwords that can exist are
​
​(26)(26)(26)(26)(26)(10)

=118813760

​The total number of possible passwords are 118813760.

 

Page 19  Exercise 17  Problem 35

We need to use five alphabets and one digit to make the password.

Any of the twenty six alphabets and ten digits can be used.

We need to find the ways three As and two Bs can be used with an even digit.

There are 5 even digits.

Repeating the alphabets is allowed.

The total number of ways three As and two Bs can be used with an even digit

⇒  \(\left(\begin{array}{1}5 \\3\end{array}\right)\) 5

⇒ \(\frac{5 !}{3 ! 2 !}\) × 5 = 50

The total number of ways three As and two Bs can be used with an even digit is 50.

 

Page 19  Exercise 17  Problem 36

We need to find the ways three As and two Bs can be used with an even digit.

There are 5 even digits.

The total number of ways three As and two Bs can be used with an even digit is 50.

The probability of guessing the correct password of the fifty possibilities is \(\frac{1}{50}\).

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 19  Exercise  18  Problem 37

Given: An electrical control panel has three toggle switcheslabeled I, II, and III each of which can be either on (O)or off (F).

To find – Construct a tree to represent the possible configurations for these three switches.

For each switch we have two options on (O)and off (F).

Thus the three diagram is given with

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 19 Exercise 18 Problem 37 Electrical control plane Switches 1

 

Hence, a tree to represent the possible configurations for these three switches is as follow:
Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 19 Exercise 18 Problem 37 Electrical control plane Switches 2

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 19  Exercise  18  Problem 38

Given –  An electrical control panel has three toggle switches labeled I, II, and III each of which can be either on (O) or off(F).

To find –  List the elements of the sample space generated by the tree.

From the tree diagram we see that the sample space and sample points are S  {OOO,OOF,OFO,OFF,FOO,FOF,FFO,FFF}

Hence, from the elements of the sample space generated by the tree are S = {OOO,OOF,OFO,OFF,FOO,FOF,FFO,FFF}

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 19  Exercise  19  Problem 39

Given : An electrical control panel has three toggle switches labeled I, II, and III each of which can be either on (O) or off (F).

To find – What is the name given to an event such as D?

Event such as event D = 0 is called aa impossible event.

Hence, from the above explanation Event such as event D = 0 is called a impossible event.

 

Page 19  Exercise  19  Problem 40

Given:  Two items are randomly selected one at a time from an assembly line and classed as to whether they are of superior quality(+), average quality (0), or inferior quality (−)

To find – Construct a tree for this two-stage experiment.

For each of the two items we have tree oscillates. superior quality (+) average quality (0) or inferior quality (-) .

Thus the tree diagram a given with:

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 19 Exercise 19 Problem 40 Oscillates tree 1

Hence, tree for this two-stage experiment is as follow

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 19 Exercise 19 Problem 40 Oscillates tree 2

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 19  Exercise  19  Problem 41

Given: Two items are randomly selected one at a time from an assembly line and classed as to whether they are of superior quality (+), average quality (0), or inferior quality (-)

To find – List the elements of the sample space generated by the tree.

From the tree diagram, we concluded that to sample space and sample points are S = {++,+0,+−,0+,00,0−,−+,−0,−−}

Hence, the elements of the sample space generated by the tree is as follow S = {++,+0,+−,0+,00,0−,−+,−0,−−}

 

Page 19  Exercise 19  Problem 42

Given: Two items are randomly selected one at a time from an assembly line and classed as to whether they are of superior quality (+), average quality (0), or inferior quality(−)

To find – List the sample points that constitute the events

A: The first item selected is of inferior quality

B: The quality of each of the items is the same

C: The quality of the first item exceeds that of the second

Using Page 19  Exercise 19   Problem 40   and Page 19  Exercise 19   Problem 41  we have

A = The first item selected is of inferior equal = {−+,−0,−−}

B = The quality of each of the items is the same = {++,00,−}

C = The quality of the First term exceeds that of the second = {+0,+−,0−}

Hence, the sample points that constitute the event are as follow:

A: The first item selected is of inferior quality ={−+,−0,−}

B: The quality of each of the items is the same={++,00,−−}

C: The quality of the first item exceeds that of the second ={+0,+−,0−}

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 20  Exercise  20   Problem 43

Given: An experiment consists of selecting a digit from among the digits 0 to 9 in such a way that each digit has the same chance of being selected as any other.

We name the digit selected A.

These lines of code are then executed.

IF A < 2 THEN B = 12;  ELSE B = 17

IFB = 12 THEN C = A − 1;  ELSE C = 0

To find – Construct a tree to illustrate the ways in which values can be assigned to the variables A, B, and C

From above given coordinate we have drawn a diagram :

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 20 Exercise 20 Problem 43 Coordinate 1

Hence:  A tree to illustrate the ways in which values can be assigned to the variables A, B, and C

Introduction to Probability and Statistics Principles and Applications Chapter 1 Introduction to Probability and Counting Page 20 Exercise 20 Problem 43 Coordinate 2

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 20  Exercise 20  Problem 44

Given: An experiment consists of selecting a digit from among the digits 0 to 9 in such a way that each digit has the same chance of being selected as any other.

We name the digit selected A.

These lines of code are then executed.

IF A <2 THEN B = 12;  ELSE B = 17

IF B = 12 THEN C = A−1​; ELSEC = 0

To find –  Find the sample space generated by the tree.

From the tree diagraram, we see that the sample space and sample points are

S−{(0,12,−1),(1,12,0),(2,17,0),(3,17,0),(4,17,0),(5,17,0),(6,17,0),(7,17,0),(3,17,0),(9,17,0)\}

Hence, from the above explanation the sample space generated by the tree is as follow S−{(0,12,−1),(1,12,0),(2,17,0),(3,17,0),(4,17,0),(5,17,0),(6,17,0),(7,17,0),(3,17,0),(Officials at Nuclear Regulatory commission estimated the probability of such an accident occurring in the United States before the year to be by using Classical Method.9,17,0)\}

 

J Susan Milton Introduction To Probability And Statistics Chapter 1 Page 20  Exercise  20  Problem 45

Given: An experiment consists of selecting a digit from among the digits 0 to 9 in such a way that each digit has the same chance of being selected as any other. We name the digit selected A.

These lines of code are then executed.

IF A < 2 THEN B = 12;  ELSE B =  17

IF B = 12 THEN C = A-1; ELSE C = 0

To find –Find the probability that A is an even number.

The probability that A is an even number is the probability that A will be 0,2,4,6 or 8. so, there are five possibilities for A to be an even number and five to be an odd number.

Thus the probability that A is are even number is

5.\(\frac{1}{10}\) = \(\frac{1}{2}\)

Hence, from the above explanation the probability that A is an even number is 5.\(\frac{1}{10}\) = \(\frac{1}{2}\)