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Molodets [167]
2 years ago
8

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
Mathematics
1 answer:
Oksanka [162]2 years ago
3 0

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.

<h3>Normal Probability Distribution</h3>

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 <u>X when Z = 0.84.</u>

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

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