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
1 answer:
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 and standard deviation , the z-score of a measure X is given by:
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 . The standard deviation is of . 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>
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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