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Sagot :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 441.84[/tex].
- The standard deviation is of [tex]\sigma = 120[/tex].
- 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.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 441.84}{120}[/tex]
[tex]X - 441.84 = 0.84(120)[/tex]
[tex]X = 542.64[/tex]
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 https://brainly.com/question/24663213
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