J.Susan Milton Introduction To Probability and Statistics Principles And Applications Chapter 6 Descriptive Distributions Exercises

J.Susan Milton Introduction To Probability And Statistics Principles And Applications Chapter 6 Descriptive Distributions

 

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.

 

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.

 

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 completed the job.

Thus, an architectural firm is to sublet a contract for a wiring project.

 

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.

 

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.

 

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).

 

Page 192  Exercise 6  Problem 7

Frist we need to find the random variable.

Then find the value of given.

Given:

The sample size be n = 5

The given sample are

x₁  =  45 , x₂  =  50 , x₃  =  62 , x₄  =  57 , x₅  =  70

Now calculate the statistic  Σ_iX_i

Simply sum the given values to get the values:

Σ_iX_i  = x₁ + x₂ + x₃ + x₄ + x₅

Now apply the value in the equation:

Σ_iX_i   = 45 + 50 + 62 + 57 + 70

Σ_iX_i  = 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 Σ_iX_i/n

Simply sum the given values to get the values:

 _ΣiX²i    = x²₁ + x²₂ + x²₃ + x²₄ + x²₅

Now apply the value in the equation

Σ_iX²_i  = 45² + 50² + 62² + 57² + 70²

Σ_iX²_i  = 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  Σ_iX²_i 

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 max_i {X_i},so that we to find the biggest value from the given sample.

By looking the values, we can easily identify them max_i {X_i}= x₅ = 70.

Now we going to calculate the statistic min_i {X_i}, so that we to find the biggest value from the given sample.

By looking the values, we can easily identify them min_i {X_i} = x₁ = 45.

Thus, the random variable value is Σ_iX_i  = 284

 Σ_iX_i =  16518

\(\Sigma_i \frac{X_i}{n}\) =   56.8

Max_i {X_i}= x₅ = 70.

Min_i {X_i} = x₁ = 45.

 

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

 

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.

 

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⟩

 

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.

 

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

 

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 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

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

 

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

 

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⟩

 

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:

The relative frequency histogram data

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:

 

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 ≤ p_0.25) ≥ 0.25

Then we can write as: P(X < p_0.25) ≤ 0.25 and the first quartile of a random variable is P(X ≤ p_0.25) ≥ 0.25

Thus, the first quartile of a random variable X is P(X ≤ p_0.25) ≥ 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 p_0.25 = 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 p_0.25 = 8.

 

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  ⇔  p_0.25 = −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 p_0.4

Such that P(X < p_0.4) ≤ 0.4

P (X ≤ p_0.4 ) ≥ 0.4

Then can write the fourth decile of the random variable as P(X ≤ p_0.4) ≥ 0.4

Thus, the 4th decile of random variable of a random variable X is P(X ≤ p_0.4) ≥ 0.4.

 

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 p_0.6 = 11.

 

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 :

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:

Thus, the approximate first quartile X value is p_0.25 = 0.0 415.

 

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 :

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:

Thus, the approximate fourth decile X value is p_0.4 = 7.7.