Question

The IQs of 600 applicants to a certain college are approximately normally distributed with a mean of 115 and a standard deviation of 12. Note that IQs are recorded to the nearest integers.

1) If the college requires an IQ of at least 95, how many of these students will be rejected?

2) What is the minimum IQ of the top 30 applicants?

Answer 1

Answer:

a)

Given,

Mean = 115

Standard deviation = 12

sample n = 600

P(X < 95) = P(X < 94.5) [continuity correction]

= P((x-mu)/s < (94.5 - 115)/12)

= P(z < - 1.71)

= 0.0436329 [since from z table]

= 0.0436

Expected value = np

= 600*0.0436

= 26

Rejected students = 26

b)

To give the minimum IQ of the top 30 applicants

P(Z > z) = 30%

P(Z < z) = 1 - 0.30

= 0.70

since from the standard normal distribution table

z = 0.52

let us consider,

x = z*s + u

substitute values

= 0.52*12 + 115

= 6.24 + 115

= 121.24

x = 121.24

So minimum IQ of the top 30 applicants = 121.24