Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 3 Exercise 3.3 Discrete Random Variables and Probability Distributions

Probability and Statistics for Engineering and the Sciences 8th Edition Chapter 3 Discrete Random Variables and Probability Distributions

Page 113 Problem 1 Answer

Given –

We will find E(Y).

We will use the formula E(Y)=∑yp(y).

Applying the formula of expectation for the given data, we get:

E(Y)=∑yp(y)

E(Y) =(0×0.6)+(1×0.25)+(2×0.1)+(3×0.05)

E(Y) =0+0.25+0.2+0.15

E(Y) =0.6

​We obtain: E(Y)=0.6

Page 113 Problem 2 Answer

Given –

​

We will calculate E(100Y²)

We note from previous part that E(Y²)=57.25.

We have:

E(Y²)=57.25

E(100Y²)=100E(Y²)

E(100Y²) =100⋅57.25

E(100Y²) =5725

We obtain : E(100Y²)=5725

Probability And Statistics For Engineering 8th Edition Solutions Exercise 3.3 Page 113 Problem 3 Answer

We are given that P(X=0)=1−p

P(X=1)=p

​We will find E(X²).We will useE(X²)=∑x²p(x).

Applying the formula, we get:

E(X²)=∑x²p(x)=[02×p(0)]+[12×p(1)]=[02×(1−p)]+[12×p]=p

​We obtain: E(X²)=p

Page 113 Problem 4 Answer

Given -P(X:0)=1−p

P(X:1)=p

​We will show that V(X)=p(1−p).

We know the formula V(X)=E(X²)−[E(x)]².

We know that V (X)

E(X)= E(X − [E(x)]2)²

= ∑xp(x)

Now, = [0 × p(0)] + [1 × p(1)]

= [0 × (1 − p)] + [1 × p]

= p

Putting these values in the above formula, we get:

V (X) = E(X − [E(x)]²)²

V (X)  = p − [p]²

V (X)  = p − p²

V (X)  = p(1 − p)

We showed that V(X)=p(1−p).

Page 113 Problem 5 Answer

Given: If X be the Bernoulli random variable with P(X:0)=1−p

P(X:1)=p

​We will find E(X79)

Using the definition, we obtain:

E(X79)=[079×p(0)]+[179×p(1)]=[079×(1−p)]+[179×p]=p

​We obtain the value:

E(X79)=p

Chapter 3 Exercise 3.3 Discrete Random Variables Solved Examples Page 113 Problem 6 Answer

Given – The data:

x           1          2            3            4           5           6

p(x)      1/15    2/15     3/15       4/15     5/15    6/15

We also know that owner bought copy for $2.00 and sells it for$4.00.

We will justify whether to buy 3 or 4 copies per week. We will find E(X),

If n=3, we note that owner bought 3copies.

E(x) is the expected value it is given by sum of the products ofx

value with p(x).

Hence, after x=3 multiply p(x) with 3.

Using E(x)=x=1

∑     x⋅p(x)    ∞, we get L E(x)=1(1/15)+2(1/15)+3(1/15)+3(4/15)+3(3/15)+3(2/15)

=41/15

=2.733

​The cost of 3 copies is given as 3(2)=6

We conclude that the owner sells those 3 copies at 4 each.

We suppose the net revenue, R=4(x)−6.

IT is in the form of W=a(x)+b.

Using the formula for expectation, we get:

​E(aX+b)=a(E(x))+b

Here E(x)=E[4(x)−6] here a=4 ,b=6

Hence E(W)=4

E(x)−6

​Using its value from the previous step , we get:

E(x)=41/15

E(x)= 2.733

Thus,E(W)=4(41/15)−6

≈4.933

​Now, we will suppose n=4.

let us assume that owner bought 4 copies .

We note the formula:

E(x)=n=0

∑ x⋅p(x)

∞ So , the expectation is:

E(x)=n=0

∑     x⋅p(x) ∞

E(x)=1(1/15)+2⋅(1/15)+3(1/15)+4(4/15)+4(3/15)+4(2/15)

E(x)= 50/15

≈3.33

The cost of 4 copies is given as 4(2.00)=8.00.

Hence, the owner sells those 4 copies at 8 each.

We know that the net revenue is R=4(x)−8.

It is in the form of W=a(x)+b.

Therefore, the expected value of this can be calculated as:

​E(x)=E[4(x)−8]

Here a=4 b=8

Hence E(W)=4

E(x)−8

​Using the previous step, we get:

​E(x)=10/3=3.733 approx

Hence E(W)=4(10/3)−8=5.33 3approx.

​If owner buys 3 copies he will get net revenue of 4.933.

If he if owner buys 4 copies he will get net revenue of 5.333

Clearly, it is better to buy 4 copies.

Chapter 3 Exercise 3.3 Study Guide Probability And Statistics Page 114 Problem 7 Answer

Given – A random variable x  and x is the random variable for both the batches. so we need to calculate both mean and variance of x .

After computing mean and variance we also need to compute  expected number of pounds after the next customer product is shipped and variance of pound left

We also have that a company has 100 lb of certain chemical in stock The customer orders in 5 lb batches do we are left with100−5X lb’s in total

We will find mean and variance for this.

Using the definition, we get:

E(x)=x=1

∑   xp(x) ∞

=1(0.2)+2(0.4)+3(0.3)+4(0.1)

=2.3

Also, the variance is:

V(x)E(x²)=[E(x²)−E(x)²]

=1(0.2)+2(0.4)+3(0.3)+4(0.1)

=6.1​

Putting the obtained values, we get:

V(x)=[E(x²)−[E(x)]²]

=6.1−(2.3)²

=0.81

​We observe that  100−5X  is in the form of a(x)+b

So, applying the rule:

E(aX+b)=a(E(x))+b=(−5)E(x)+100

Here a=−5

​​b=100

So −5(E(x))+100=−5(2.3)+100

=88.5.

​We also have the rule V(ax)=a²

v(x) for constant v(x)  is zero.

Hence,

V(100−5x)=(−5)²

V(x)=20.25

​We obtain:

E(x)=2.3

V(x)=0.81

E(100−5X)=88.5

V(100−5X)=20.25

​Page 113 Problem 8 Answer

We will plot a graph for pm f(x), We are given x and p(x) values.

For the data given , we obtain the graph as:

We note that for the given x values and p(x) values, in order to plot graph for negative values of x  we will multiply x with−1.

Clearly, the spread of both these graphs is same.

Therefore, we can say that V(X)=V(−X).

Using the spread in graphs, we showed that V (X) = V (−X).

Page 114 Problem 9 Answer

Using the formula we can prove that V(X)=V(−X)

We note that the variance is given as V(aX+b)=a{2}

(v(x)) for constant V(x) is zero

According to the formula V(aX+b)=a{2}

V(x) for constant V(x) is zero

Here, inV(−X)

we have: a=−1

b=0

Now, using the formula we can prove that V(x)=V(−x) , It is in the form of V(aX+b) here a=−1 and b=0

ThusV(−x)=(−1){2} V(x).

It is equal to V(x).

Hence,V(x)=V(−x)

With the help of proposition involving V(aX+b), we showed that V(x)=V(−x).

Discrete Random Variables Examples From Exercise 3.3 Engineering And Sciences Page 114 Problem 10 Answer

We will prove that V(aX+b)=a²⋅σX²

​We note that:

h(X)=aX+bV(aX+b)

=a²⋅σy²

​We also note that

h(X)=aX+bE[h(X)]

=aμ+b

Here μ=E(X).

The variance of a X+b is given by:V(aX+b)=[E(aX+b)²−(E(aX+b))²]

Using the given data, we obtain:
∞
V (aX + b) = ∑ (x − μ)2 p(x)V (aX + b)
∀

∞
=∑    [E ( (aX + b)2) − (E(aX + b))2]
∀

V (aX + b) = [aX − aμ] ⋅ p(x)

We see that a is constant so taking a out of the summation it becomes a²

Thus, V(aX+b)=a²

∀
∑          (x−(E(x))²)⋅p(x)
∞

V(aX+b)=a{2}

V(x) We showed that V(aX+b)=a² V(X).

Page 114 Problem 11 Answer

Given -a≤X≤b

We need to show that a≤E(X)≤b.

We will use: E(X)=μx

μx=∑x∈S x⋅p(x)

​Multiplying p(x) in the given inequality, we get:

a≤x≤b

So, a⋅p(x)≤x⋅p(x)≤b⋅p(x)

For all x∈S. This now implies that for the sum over all x∈S, the given inequality is satisfied.

x∈S

∑ a⋅p(x)≤x∈S

∑ x⋅p(x)≤x∈S

∑ b⋅p(x)

We obtain:

x∈S

∑ a⋅p(x) x∈S

∑ b⋅p(x) =a⋅x∈S

∑ p(x) =a⋅1=a;

=b⋅x∈S

∑ p(x) =b⋅1=b;

So, the sum over all x∈S of pmf p(x) is 1

We have that E(X)=x∈S

∑ x⋅p(x)

Hence, using the data obtained, we get: a≤E(X)≤b

Fora≤X≤b we showed that a≤E(X)≤b.