Math CentralQuandaries & Queries


Question from lisa, a student:

An employer gives a pre-employment evaluation to a large group of applicants. The scores are normally distributed with a mean of 154 and a standard deviation of 21. The employer wants to interview only those applicants who score in the top 15%. What should the cut off score be for the interviews?

Hi Lisa.

The scores on the text X are distributed as the normal distribution with μ = 154 and σ = 21. You need to transform X to the standard normal random variable Z using Z = (X - μ)/σ = (X - 154)/21. You want to interview only those who scored in the top 15% so you need the cut off score s to satisfy

Pr(Z > s ) = 0.15


Pr((X - 154)/21 > s ) = 0.15

Use the standard normal table to find s and then solve (X - 154)/21 = s for X.


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