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