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

The expected repair cost is $3.73.

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* 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* 0.09* 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* 0.4572)+0.36+2\\=3.7316\\\approx 3.73`

Thus, the expected repair cost is $3.73.