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

Answer:

27,361.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.65 = \frac{X - 20307}{4275}$

$X - 20307 = 1.65*4275$

$X = 27,361$

So the answer is 27,361.