A large manufacturing firm tests job applicants. Test scores are normally distributed with a mean of 500 and a standard deviation of 50. Management is considering placing a new hire in an upper-level management position if the person scores in the upper sixth percent of the distribution. What is the lowest score a new hire must earn to qualify for a responsible position?
A)50
B)625
C)460
D)578

Question

Quizplus · Accepted Answer

To find the score corresponding to the upper sixth percent of the distribution, we need to find the z-score that separates the top 6% of the scores from the rest.

Using a z-table, we find that the z-score at the 94th percentile (100% - 6%) is approximately 1.56.

We can use the formula z = (x - μ) / σ to find the corresponding raw score. Plugging in the given values, we get:

1.56 = (x - 500) / 50

Solving for x, we get:

x = 1.56 * 50 + 500 = 578

Therefore, the lowest score a new hire must earn to qualify for an upper-level management position is 578.

Note that choices A, B, and C are all outside the possible range of scores, since they are either too low or too high.

A Large Manufacturing Firm Tests Job Applicants