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?

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ChiKesselman ChiKesselman · Answer 1 · 2019-11-26T10:54:58+01:00

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:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

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

P(X > x) = 6% = 0.06

[tex]P( X > x) = P( z > \displaystyle\frac{x - 500}{50})=0.06[/tex]

[tex]= 1 - P( z \leq \displaystyle\frac{x - 500}{50})=0.06 [/tex]

[tex]=P( z \leq \displaystyle\frac{x - 500}{50})=1-0.06=0.94 [/tex]

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

[tex]P(z < 1.555) = 0.94[/tex]

[tex]\displaystyle\frac{x - 500}{50} = 1.555\\x = 577.75[/tex]

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%.