Question

For the most recent year available, the mean annual cost to attend a private university in the United States was $50,900. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,500. Ninety-five percent of all students at private universities pay less than what amount

Answer 1

Answer:

95% of all students at private universities pay less than $58,302.5.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of $50,900, standard deviation of $4,500

This means that $\mu = 50900, \sigma = 4500$

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

Pay less than the 95th percentile, and the amount is given by X when Z has a pvalue of 0.95, so X when Z = 1.645.

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

$1.645 = \frac{X - 50900}{4500}$

$X - 50900 = 1.645*4500$

$X = 58302.5$

95% of all students at private universities pay less than $58,302.5.