Question

An appliance dealer sells three different models of upright freezers having 13.5, 15.9, and 19.1 cubic feet of storage space, respectively. Let X = the amount of storage space purchased by the next customer to buy a freezer. Suppose that X has the following pmf.

x 13.5 15.9 19.1
p(x) 0.17 0.57 0.26

a. Compute E(X), E(X2), and V(X).
b. If the price of a freezer having capacity X cubic feet is 28X − 8.5, what is the expected price paid by the next customer to buy a freezer? (Round your answer to the nearest whole number.)
c. What is the variance of the price 28X − 8.5 paid by the next customer? (Round your answer to the nearest whole number.)
d. Suppose that although the rated capacity of a freezer is X, the actual capacity is h(X) = X − 0.02X2. What is the expected actual capacity of the freezer purchased by the next customer? (Round your answer to three decimal places.)

Answer 1

Answer:

a) $E(X) =13.5*0.17 + 15.9*0.57 + 19.1*0.26 = 16.324$

$E(X^2) =13.5^2*0.17 + 15.9^2*0.57 + 19.1^2*0.26 = 269.9348$

$Var(X) = E(X^2) -[E(X)]^2 = 269.9348-(16.324)^2 = 3.462$

b) $E(Y)= E(28X-8.5) = E(28X) - E(8.5) = 28 E(X) -8.5$

And replacing the result from part a we got:

$E(Y) = 28*16.324 -8.5= 448.572$

c) $Var(28X-8.5) = Var (28X)= 28^2 Var(X)= 784*3.462=2714.208$

d) $E(H) = E(X -0.02 X^2) = E(X) -0.02 E(X^2) = 16.324-0.02(269.9348)= 10.925$

Step-by-step explanation:

For this case we have the following probability function given:

x 13.5 15.9 19.1

p(x) 0.17 0.57 0.26

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

Part a

We can calculate the expected value with the following formula:

$E(X) = \sum_{i=1}^n X_i P(X_i)$

And if we replace we got:

$E(X) =13.5*0.17 + 15.9*0.57 + 19.1*0.26 = 16.324$

For the second moment we can use this definition:

$E(X^2) = \sum_{i=1}^n X^2_i P(X_i)$

And if we replace we got:

$E(X^2) =13.5^2*0.17 + 15.9^2*0.57 + 19.1^2*0.26 = 269.9348$

The variance is defined:

$Var(X) = E(X^2) -[E(X)]^2 = 269.9348-(16.324)^2 = 3.462$

Part b

For this case we define this new random variable Y = 28 X -8.5. And we want to find the expected value, so we have this:

$E(Y)= E(28X-8.5) = E(28X) - E(8.5) = 28 E(X) -8.5$

And replacing the result from part a we got:

$E(Y) = 28*16.324 -8.5= 448.572$

Part c

For the variance we can use the following property:

$Var(X+Y) = Var(X) + Var(Y) + 2 Cov(X,Y)$

And using this formula we have:

$Var(28X -8.5) = Var(28X)+ Var(8.5)+ 2 Cov(28X,-8.5)$

The variance for a constant is 0 so then Var(8.5)=0 and Cov(28X, -8.5) = 0 since by properties if X is a random variable and a represent a constant $Cov(X,a)=0$, so then we just have this:

$Var(28X-8.5) = Var (28X)$

Using the following property $Var(aX)= a^2 Var(X)$ we have:

$Var(28X-8.5) = Var (28X)= 28^2 Var(X)= 784*3.462=2714.208$

Part d

For this case we define $H = X -0.02 X^2$

And if we find the expected value we have this:

$E(H) = E(X -0.02 X^2) = E(X) -0.02 E(X^2) = 16.324-0.02(269.9348)= 10.925$