Question

For the most recent year available, the mean annual cost to attend a private university in the United States was $20,307. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,275.

Ninety-five percent of all students at private universities pay less than what amount? (Round z value to 2 decimal places and your final answer to the nearest whole number. Omit the "$" sign in your response.)

Answer 1

27,361.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 20307, \sigma = 4275`

Ninety-five percent of all students at private universities pay less than what amount?

This is the 95th percentile, that is, X when Z has a pvalue of 0.95. So X when Z = 1.65.

`Z = (X - \mu)/(\sigma)`

`1.65 = (X - 20307)/(4275)`

`X - 20307 = 1.65*4275`

`X = 27,361`

So the answer is 27,361.