Question

The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000. What percentage of MBA's will have starting salaries of $34,000 to $46,000?

a. 38.49%
b. 38.59%
c. 50%
d. 76.98%

Answer 1

d. 76.98%

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 = (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 = 40000, \sigma = 5000`

What percentage of MBA's will have starting salaries of $34,000 to $46,000?

This is the pvalue of Z when X = 46000 subtracted by the pvalue of Z when X = 34000. So

X = 46000

`Z = (X - \mu)/(\sigma)`

`Z = (46000 - 40000)/(5000)`

`Z = 1.2`

`Z = 1.2` has a pvalue of 0.8849

X = 34000

`Z = (X - \mu)/(\sigma)`

`Z = (34000 - 40000)/(5000)`

`Z = -1.2`

`Z = -1.2` has a pvalue of 0.1151

0.8849 - 0.1151 = 0.7698

So the correct answer is:

d. 76.98%