Adults have IQ scores that are normally distibuted with a mean of 100 and a standard deviation of 15. Find the probability that a randomly...

IQ x scores are normally distributed with mean , m = 100
and standard devition , s =15.

Therefore z = (x-m)/s is the
standard normal variate.

Therefore when IQ , x = 131.5, z =
(131.5-100)/15 = 2.1.

Therefore , the probability that x
> 131.5 is as good as P(z > 2.1)
.

P(z> 2.1) = 1- P(z<= 2.1), where P(z
< = 2.1) could be obtained from the tables

P(z
> = 2.1) = 1 - 0.98214 = 0.01786