Assume the IQ scores are normally distributed with a mean of

Assume the IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability that a randomly selected person has an IQ score greater than 118.

Solution

Normal Distribution
Mean ( u ) =100
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 118) = (118-100)/15
= 18/15 = 1.2
= P ( Z >1.2) From Standard Normal Table
= 0.1151