A car manufacturer is concerned about poor customer satisfac

A car manufacturer is concerned about poor customer satisfaction at one of its dealerships. The management decides to evaluate the satisfaction surveys of its next 64 customers. The dealer will be fined if the number of customers who report favorably is between 27 and 48. The dealership will be dissolved if fewer than 27 report favorably. It is known that 64% of the dealer’s customers report favorably on satisfaction surveys.

a.

What is the probability that the dealer will be fined? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

What is the probability that the dealership will be dissolved? (Round intermediate calculations to 4 decimal places, “z” value to 2 decimal places, and final answer to 4 decimal places.)

Solution

Normal Distribution
Mean ( np ) =40.96
Standard Deviation ( npq )= 64*0.64*0.36 = 3.84
Normal Distribution = Z= X- u / sd
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 27) = (27-40.96)/3.84
= -13.96/3.84 = -3.6354
= P ( Z <-3.6354) From Standard Normal Table
= 0.00014
P(X < 48) = (48-40.96)/3.84
= 7.04/3.84 = 1.8333
= P ( Z <1.8333) From Standard Normal Table
= 0.96662
P(27 < X < 48) = 0.96662-0.00014 = 0.9665                  
b)                  
P(X < 27) = (27-40.96)/3.84
= -13.96/3.84= -3.6354
= P ( Z <-3.6354) From Standard NOrmal Table
= 0.0001