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 between 88 and 112,

Solution

Normal Distribution
Mean ( u ) =100
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 88) = (88-100)/15
= -12/15 = -0.8
= P ( Z <-0.8) From Standard Normal Table
= 0.21186
P(X < 112) = (112-100)/15
= 12/15 = 0.8
= P ( Z <0.8) From Standard Normal Table
= 0.78814
P(88 < X < 112) = 0.78814-0.21186 = 0.5763