Question

A large manufacturing firm tests job applicants who recently graduated from college. The 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 6 percent of the distribution. What is the lowest score a college graduate must earn to qualify for a responsible position?

Answer 1

The lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.

Explanation:

We are given the following information in the question:

Mean, μ = 500

Standard Deviation, σ = 50

We are given that the distribution of test score is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

We have to find the value of x such that the probability is 0.06.

P(X > x) = 6% = 0.06

`P( X > x) = P( z > \displaystyle(x - 500)/(50))=0.06`

`= 1 - P( z \leq \displaystyle(x - 500)/(50))=0.06`

`=P( z \leq \displaystyle(x - 500)/(50))=1-0.06=0.94`

Calculation the value from standard normal z table, we have,

`P(z < 1.555) = 0.94`

`\displaystyle(x - 500)/(50) = 1.555\\x = 577.75`

Hence, the lowest score a college graduate must be 577.75 or greater to qualify for a responsible position and lie in the upper 6%.