4.56% of salaries are less than $68,000 or more than $92,000. A percentage lower than 5% is unusual, so it is unusual to find a professor earning less than $68,000 or more than $92,000

Step-by-step explanation:

Problems of normally distributed(bell-shaped) 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:

$\mu = 80000, \sigma = 6000$

What can be said about the percentage of salaries that are less than $68,000 or more than $92,000?

Less than 68,000

pvalue of Z when X = 68000

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

$Z = \frac{68000 - 80000}{6000}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

2.28% of salaries are less than 68,000.

More than 92,000

1 subtracted by the pvalue of Z when X = 92000

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

$Z = \frac{68000 - 80000}{6000}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% of salaries are more than 92,000.

Less than 68,000 or more than 92,000:

2*2.28 = 4.56% of salaries are less than $68,000 or more than $92,000. A percentage lower than 5% is unusual, so it is unusual to find a professor earning less than $68,000 or more than $92,000

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3 years ago