Question

IQs are known to be normally distributed with mean 100 and standard deviation 15. In a random sample of 37 people, find the probability that the average IQ is between 96 and 103.

Answer 1

Solution :

Given that,

mean = = 100

standard deviation = = 15

= / n = 15 / 37 = 2.4660

= P[(96 - 100) /2.4660 < ( - ) / < (103 - 100) / 2.4660)]

= P(-1.62 < Z < 1.22)

= P(Z < 1.22) - P(Z < -1.62)

= 0.8888 - 0.0526

= 0.8362

Probability = 0.8362