Question

A computer store has 75 printers of which 25 are laser printers and 50 are ink jet printers. if a group of 10 printers is chosen at random from the store, find the mean and variance of the number of ink jet printers.

Answer 1

Mean = 6.7

Variance = 2.211

Explanation:

For each printer, there are only two possible outcomes. Either it is an ink jet printer, or it is not. This means that we can solve this problem using the binomial probability distribution.

Binomial probability distribution.

This is the probability of exactly x sucesses on n repeated trials, with p probability.

Has a mean of:

`E(X) = np`

Has a variance of:

`Var(X) = np(1-p)`.

In this problem, we have that:

Of the 75 printers, 50 are ink jet ones. This means that
`p = (50)/(75) = 0.67`.

10 printers are going to be chosen, so
`n = 10`.

The mean number of ink jet printers is:

`E(X) = np = 10*(0.67) = 6.7`

The variance of the number of ink jet printers is:

`Var(X) = np(1-p) = 10*0.67*0.33 = 2.211`.

Answer 2

A computer store has 75 printers of which 25 are laser printers and 50 are ink jet printers. if a group of 10 printers is chosen at random from the store, find the mean and variance of the number of ink jet printers.
The mean of the number of ink jet printers will be:
mean=np
n=10
p=50/75
mean=10*50/75
mean=6 2/3

Variance:
variance=npq
=6 2/3*25/75
=2 2/9