Salaries for an large engineering company are normally distributed with a mean of $80,000 and a standard deviation of $8,450. An employee was having their annual appraisal and the manager indicated that the employee salary for next year has a Z-score of 1.24. Approximately how much will this employee be paid next year in salary? Salary=$[salary] Enter your answer to the nearest hundred of dollars, i.e 50,185 would be entered 50,200.

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Answer:

Salary: $90,500

Step-by-step explanation:

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

In this problem, we have that:

[tex]\mu = 80000, \sigma = 8450[/tex]

An employee was having their annual appraisal and the manager indicated that the employee salary for next year has a Z-score of 1.24. Approximately how much will this employee be paid next year in salary?

This is X when [tex]Z = 1.24[/tex]. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.24 = \frac{X - 80000}{8450}[/tex]

[tex]X - 80000 = 1.24*8450[/tex]

[tex]X = 90478[/tex]

Rouded to the nearest hundred of dollars:

Salary: $90,500

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