Question

IQs are known to be normally distributed with mean 100 and standard deviation 15. In a random sample of 33 people, find the probability that the average IQ is between 96 and 104.

Answer 1

Given that ,

mean = = 100

standard deviation = = 15

n = 33

= 100

=  / n= 15 / 33=2.61

P(96<     < 104) = P[(96-100) /2.61 < ( - ) /   < (104-100) /2.61 )]

= P(-1.53 < Z <1.53 )

= P(Z <1.53) - P(Z < -1.53)

Using z table

=0.937 - 0.0630

probability= 0.8740

=