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
and standard deviation
, the z-score of a measure X is given by:

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 
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 





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



has a p-value of 1
1 - 1 = 0
The proportion of borrowers who owe more than 54,000 is 0%.