Question

A particular sale involves four items randomly selected from a large lot that is known to contain 9% defectives. Let X denote the number of defectives among the four sold. The purchaser of the items will return the defectives for repair, and the repair cost is given by 2 C X X = + + 3 2 1. Find the expected repair cost

Answer 1

Answer:

The expected repair cost is $3.73.

Step-by-step explanation:

The random variable X is defined as the number of defectives among the 4 items sold.

The probability of a large lot of items containing defectives is, p = 0.09.

An item is defective irrespective of the others.

The random variable X follows a Binomial distribution with parameters n = 9 and p = 0.09.

The repair cost of the item is given by:

$C=3X^{2}+X+2$

Compute the expected cost of repair as follows:

$E(C)=E(3X^{2}+X+2)$

        $=3E(X^{2})+E(X)+2$

Compute the expected value of X as follows:

$E(X)=np$

         $=4\times 0.09\\=0.36$

The expected value of X is 0.36.

Compute the variance of X as follows:

$V(X)=np(1-p)$

         $=4\times 0.09\times 0.91\\=0.3276$

The variance of X is 0.3276.

The variance can also be computed using the formula:

$V(X)=E(Y^{2})-(E(Y))^{2}$

Then the formula of $E(Y^{2})$ is:

$E(Y^{2})=V(X)+(E(Y))^{2}$

Compute the value of $E(Y^{2})$ as follows:

$E(Y^{2})=V(X)+(E(Y))^{2}$

          $=0.3276+(0.36)^{2}\\=0.4572$

The expected repair cost is:

$E(C)=3E(X^{2})+E(X)+2$

         $=(3\times 0.4572)+0.36+2\\=3.7316\\\approx 3.73$

Thus, the expected repair cost is $3.73.