Question

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was $20.50, with a standard deviation of $3.50. Assume the distribution of hourly wages follows the normal probability dis- tribution. If we select a crew member at random, what is the probability the crew member earns: a. Between $20.50 and $24.00 per hour

Answer 1

Answer:

0.3413 = 34.13% probability that the crew member earns between $20.50 and $24.00 per hour

Step-by-step explanation:

Problems of normal distributions 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.

A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary was $20.50, with a standard deviation of $3.50.

This means that $\mu = 20.5, \sigma = 3.5$

a. Between $20.50 and $24.00 per hour

This is the pvalue of Z when X = 24 subtracted by the pvalue of Z when X = 20.5. So

X = 24

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

$Z = \frac{24 - 20.5}{3.5}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 20.5

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

$Z = \frac{20.5 - 20.5}{3.5}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.8413 - 0.5 = 0.3413

0.3413 = 34.13% probability that the crew member earns between $20.50 and $24.00 per hour