Question

Salaries for various positions can vary significantly, depending on whether or not thecompany is in the public or private sector. The U.S. Department of Labor posted the2007 mean salary for human resource managers employed by the federal governmentas $76,503. Assume the annual salaries for this type of job are normally distributedand have a standard deviation of $8850.a)What is the probability that a randomly selectedhuman resource manager received over $100,000in 2007?

Answer 1

Answer:

0.0039 = 0.39% probability that a randomly selectedhuman resource manager received over $100,000 in 2007.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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

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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 76503, standard deviation of 8850.

This means that $\mu = 76503, \sigma = 8850$

a)What is the probability that a randomly selected human resource manager received over $100,000 in 2007?

This is 1 subtracted by the pvalue of Z when X = 100000. So

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

$Z = \frac{100000 - 76503}{8850}$

$Z = 2.66$

$Z = 2.66$ has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.0039 = 0.39% probability that a randomly selectedhuman resource manager received over $100,000 in 2007.