In a particular IQ test with a mean of 100 and standard devi

In a particular IQ test with a mean of 100 and standard deviation of 10 is standardized to normal model. Aproximatley what percentage of scores fall either above an IQ of 110 or below and IQ of 80?

Solution

Mean ( u ) =100
Standard Deviation ( sd )=10
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P( X > a or X < b ) = P ( X > a ) + P( X < b)
P(X < 80) = (80-100)/10
= -20/10= -2
= P ( Z <-2) From Standard Normal Table
= 0.0228
P(X > 110) = (110-100)/10
= 10/10 = 1
= P ( Z >1) From Standard Normal Table
= 0.1587
P( X < 80 OR X > 110) = 0.0228+0.1587 = 0.181405