Question

A supervisor has determined that the average salary of the employees in his department is $40,000 with a standard deviation of $15,000. A sample of 25 of the employees’ salaries was selected at random. Assuming that the distribution of the salaries is normal, what is the probability that the average for this sample is between $36,000 and $42,000?

Answer 1

The z-score for 36,000 is (36,000 - 40,000) / (15,000 / √25) = -1.333

The z-score for 42,000 is (42,000 - 40,000) / (15,000 / √25) = 0.6667

P(36,000 < x < 42,000) = P(-1.3333 < z < 0.6667)

P(z < 0.6667) - P(z < -1.3333)

0.7475 - 0.0913

0.6562