# [20]Normal probability density

Normal probability density function

The score on a test have a mean of 50 and a standard deviation of 5. Assume the variable is normally distributed.

(a)If a personnel manager wishes to select from the top 10% of applicants who take the test, find the cutoff score.

(b)What is the probability that a randomly selected applicant will score between 40 and 45.

Standard Normal Distribution Table:

http://i.imgur.com/QYB4asD.png

### 1 個解答

• 7 年 前
最佳解答

(a) Let X ~ N(50,5^2), Z ~ N(0,1)

Let P(Z < k) = 0.9, where k is a constant

Since P(Z < 1.28) = 0.8997 and P(Z < 1.29) = 0.9015,

1.28 < k < 1.29

By linear extrapolation,

(P(Z < k) - P(Z < 1.28)/(k-1.28) = (P(Z < 1.29) - P(Z < 1.28)/(1.29-1.28)

(0.9-0.8997)/(k-1.28) = (0.9015-0.8997)/0.01

k-1.28 = 0.0003/0.18 = 1/600

k = 1.28+1/600

Therefore P(Z < 1.28+1/600) = 0.9

P(X < (1.28+1/600)*5+50) = 0.9

Cutoff score = (1.28+1/600)*5+50 = 56.4 (corr. to 3 sig.fig)

(b) The required probability

= P(40 < X < 45)

= P((40-50)/5 < Z < (45-50)/5)

= P(-2 < Z < -1)

= P(Z < -1) - P(Z < -2)

= 1 - P(Z < 1) - (1 - P(Z < 2))

= P(Z < 2) - P(Z < 1)

= 0.9772 - 0.8413

= 0.1359

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