Question

The Mentoring Group gives an exam to make sure those who apply are qualified as teachers. The mean score is 88 with a known standard deviation of 10. It is decided that they will accept for interviews those applicants above the 40^th percentile. What is the cutoff score for acceptance?

Group of answer choices

79.3

88.4

85.5

91.3

Answer 1

The mean score is 88 with a known standard deviation of 10. It is decided that they will accept applicants whose scores are above 40th percentile.

So, we have to find the 40th percentile of the distribution.

Let, X be the random variable denoting the score.

So, X~Normal(88,100).

So, (X-88)/10~Normal(0,1).

We have to find p, such that P(X<p)=0.4.

So,

P((X-88)/10<(p-88)/10)=0.4

ie. P(Z<(p-88)/10)=0.4

Where, Z is the standard normal variate.

ie. phi((p-88)/10)=0.4

Where, phi is the distribution function of the standard normal variate.

Comparing with the standard normal table, we find that

(p-88)/10=-0.25 (approximately)

So, p-88=-2.5

So, p=88-2.5

So, p=85.5.

So, the cutoff score for acceptance is (c) 85.5.