Question

27. The average hourly wage of workers at a fast food restaurant is $7.25/hr
with a standard deviation of $0.50. Assume that the distribution is normally distributed. If a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $8.00?*

Answer 1

Answer:

0.0668 = 6.68% probability that the worker earns more than $8.00

Step-by-step explanation:

When the distribution is normal, we use 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.

The average hourly wage of workers at a fast food restaurant is $7.25/hr with a standard deviation of $0.50.

This means that $\mu = 7.25, \sigma = 0.5$

If a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $8.00?

This is 1 subtracted by the pvalue of Z when X = 8. So

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

$Z = \frac{8 - 7.25}{0.5}$

$Z = 1.5$

$Z = 1.5$ has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% probability that the worker earns more than $8.00