The tensile strength of a particular metal alloy is normalit

The tensile strength of a particular metal alloy is normality distributed with fi = 43 and a = 4.5.What is the probability that tensile strength is at most 40

Solution

Mean ( u ) =43
Standard Deviation ( sd )=4.5
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P(X > 40) = (40-43)/4.5
= -3/4.5 = -0.6667
= P ( Z >-0.667) From Standard Normal Table
= 0.7475                  
P(X < = 40) = (1 - P(X > 40)
= 1 - 0.7475 = 0.2525