A company asserts that 81 of the customers who purchase its

A company asserts that 81% of the customers who purchase its special lawn mower will require no repairs during the first 2 years of ownership. Your personal study has shown that only 68 of the 100 in your sample lasted the 2 years without incurring repair expenses. What is the probability of your sample outcome or less if the actual repair expense-free percentage is 81%? (Give your answer correct to four decimal places.)

Solution

Proportion ( P ) =0.68
Standard Deviation ( sd )= Sqrt (P*Q /n) = Sqrt(0.68*0.32/100)=0.0466
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
                  
P(X > 0.81) = (0.81-0.68)/0.0466
= 0.13/0.0466 = 2.7897
= P ( Z >2.79) From Standard Normal Table
= 0.0026                  
P(X <= 0.81) = 1 - P(X>0.81)
= 0.9974