Question

According to the Bureau of Labor Statistics, the average weekly pay for a U.S. production worker was $441.84 in 1998 (The World Almanac, 2000). Assume that the available data indicate that wages are normally distributed with a standard deviation of 120. How much does a production worker have to make to be in the top 30% of wage earners? Give your answer to 2 decimal places

Answer 1

Using the normal distribution, it is found that a production worker has to make $542.64 a week to be in the top 30% of wage earners.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

In this problem:

• The mean is of $\mu = 441.84$.

• The standard deviation is of $\sigma = 120$.

• The lower bound of the top 30% is the 70th percentile, which is X when Z has a p-value of 0.7, so X when Z = 0.84.

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

$0.84 = \frac{X - 441.84}{120}$

$X - 441.84 = 0.84(120)$

$X = 542.64$

A production worker has to make $542.64 a week to be in the top 30% of wage earners.

You can learn more about the normal distribution at brainly.com/question/24663213