Question

A report published by the Federal Reserve Bank of New York in 2012 reported the results of a nationwide study of college student debt. Researchers found that the average student loan balance per borrower is $23,300. They also reported that about one-quarter of borrowers owe more than $28,000. Assuming that the distribution of student loan balances is approximately normal.
Requird:
Estimate the proportion of borrowers who owe more than 54,000.

Answer 1

Answer:

The proportion of borrowers who owe more than 54,000 is 0%.

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.

Researchers found that the average student loan balance per borrower is $23,300.

This means that $\mu = 23300$

They also reported that about one-quarter of borrowers owe more than $28,000.

This means that the 1 subtracted by the p-value of Z is 0.25 when X = 28000, that is, when X = 28000, Z has a p-value of 0.75, so when X = 28000, Z = 0.675. We use this to find $\sigma$

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

$0.675 = \frac{28000 - 23300}{\sigma}$

$0.675\sigma = 4700$

$\sigma = \frac{4700}{0.675}$

$\sigma = 6963$

Estimate the proportion of borrowers who owe more than 54,000.

This is 1 subtracted by the p-value of Z when X = 54000. So

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

$Z = \frac{54000 - 23300}{6963}$

$Z = 4.41$

$Z = 4.41$ has a p-value of 1

1 - 1 = 0

The proportion of borrowers who owe more than 54,000 is 0%.