Question

A​ bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50. Find Upper P 60​, the score which separates the lower​ 60% from the top​ 40%. Round to one decimal place.

Answer 1

Answer:

Upper P60 = 212.8

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:

$\mu = 200, \sigma = 50$

Find Upper P 60​, the score which separates the lower​ 60% from the top​ 40%.

This is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.255.

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

$0.255 = \frac{X - 200}{50}$

$X - 200 = 0.255*50$

$X = 212.8$

Upper P60 = 212.8