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 40 customers. The dealer will be fined if fewer than 26 customers report favorably and the dealership will be dissolved if fewer than 22 report favorably. It is known that 70% of the dealer\'s customers report favorably on satisfaction surveys.

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a.

What is the probability that the dealer will be fined?

b.

What is the probability that the dealership will be dissolved?

Solution

Normal Approximation to Binomial Distribution
Mean ( np ) =40 * 0.7 = 28
Standard Deviation ( npq )= 40*0.7*0.3 = 2.8983
Normal Distribution = Z= X- u / sd                   
P(X < 26) = (26-28)/2.8983
= -2/2.8983= -0.6901
= P ( Z <-0.6901) From Standard NOrmal Table
= 0.2451                  
P(X < 22) = (22-28)/2.8983
= -6/2.8983= -2.0702
= P ( Z <-2.0702) From Standard NOrmal Table
= 0.0192