In: Statistics and Probability
IQ scores are scaled to have a mean of 100 and a standard deviation of 15. They are also supposed to follow a normal distribution. Suppose a random sample of 81 people are given IQ tests. Calculate the probability that the sample mean IQ is greater than 102.
Solution :
Given that ,
mean = = 100
standard deviation = = 15
n = 81
= 100
= / n = 15/ 81=1.6667
P( >102 ) =1 - P( < 102)
=1- P[( - ) / < (102-100) /1.6667 ]
= 1-P(z <1.20 )
Using z table
= 1- 0.8849
probability=0.1151