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

Answer:

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

Step-by-step 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:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

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

P(X > x)  = 6% = 0.06

$P( X > x) = P( z > \displaystyle\frac{x - 500}{50})=0.06$  

$= 1 - P( z \leq \displaystyle\frac{x - 500}{50})=0.06$  

$=P( z \leq \displaystyle\frac{x - 500}{50})=1-0.06=0.94$  

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

$P(z < 1.555) = 0.94$

$\displaystyle\frac{x - 500}{50} = 1.555\\x = 577.75$  

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%.