Assume that adults have IQ scores that are normally distribu

Assume that adults have IQ scores that are normally distributed with a mean of mu = 105 and a standard deviation sigma = 20. Find the probability that a randomly selected adult has an IQ between 88 and 122. The probability that a randomly selected adult has an IQ between 88 and 122 is .

Solution

Normal Distribution
Mean ( u ) =105
Standard Deviation ( sd )=20
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 88) = (88-105)/20
= -17/20 = -0.85
= P ( Z <-0.85) From Standard Normal Table
= 0.19766
P(X < 122) = (122-105)/20
= 17/20 = 0.85
= P ( Z <0.85) From Standard Normal Table
= 0.80234
P(88 < X < 122) = 0.80234-0.19766 = 0.6047