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.

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

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

Answer 1

Solution :

Given that ,

mean = = 115

standard deviation = = 12

a)

P(x 95) = 1 - P(x   95)

= 1 - P[(x - ) / (95 - 115) / 12]

= 1 -  P(z -1.67)   

= 1 - 0.0475

= 0.9525

= 0.9525 * 600 = 571.5

Answer = 571.5

b)

Using standard normal table ,

P(Z > z) = 30%

1 - P(Z < z) = 0.3

P(Z < z) = 1 - 0.3

P(Z < 0.52) = 0.7

z = 0.52

Using z-score formula,

x = z * +

x = 0.52 * 12 + 115 = 121.24

The minimum IQ of the top 30 applicants is: 121.24