Question

The average high-school teacher annual salary is $43,000. Let teacher salary be normally distributed with a standard deviation of $18,000.a. What percent of high school teachers make between $40,000 and $50,000? Round to 3 decimals and keep '0' before the decimal point.b. What percent of high school teachers make more than $80,000? Round to 3 decimals and keep '0' before the decimal point.

Answer 1

Answer:

a) 0.219

b) 0.02

Step-by-step explanation:

Problems of normally distributed samples can be 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:

$\mu = 43000, \sigma = 18000$

a. What percent of high school teachers make between $40,000 and $50,000?

This is the pvalue of Z when X = 50,000 subtracted by the pvalue of Z when X = 40,000.

So

X = 50,000

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

$Z = \frac{50000 - 43000}{18000}$

$Z = 0.39$

$Z = 0.39$ has a pvalue of 0.6517.

X = 40,000

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

$Z = \frac{40000 - 43000}{18000}$

$Z = -0.17$

$Z = -0.17$ has a pvalue of 0.4325

So 0.6517 - 0.4325 = 0.219 = 21.9% of high school teachers make between $40,000 and $50,000.

b. What percent of high school teachers make more than $80,000?

This is 1 subtracted by the pvalue of Z when X = 80000. SO

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

$Z = \frac{80000 - 43000}{18000}$

$Z = 2.06$

$Z = 2.06$ has a pvalue of 0.9803.

So 1-0.9803 = 0.0197 = 0.02 = 2% of high school teachers make more than $80,000.