Answer to Question #343289 in Statistics and Probability for heylol

Question

For the most recent year available, the mean annual cost to attend a private university in the United States was $26,889. Assume the distribution of annual cost follows the normal distribution and the standard deviation $4,500. 95% of all students at private university pay less than what amount?

Answer 1

Problems of normally distributed samples can be 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:

$\mu=26889, \sigma=4500$

The probability of Ninety-five percent of all students at private universities is 0.95. Looking at the normal distribution table, the z is score corresponding to 0.95 is 1.64.

So, $1.64=\frac {X-26889}{4500}$

$X-26889=1.64*4500$

$X=34269$

Ninety-five percent of all students at private universities pay less than $34269.