A car manufacturer is concerned about poor customer satisfaction at one of its d

Question

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 50 customers. The dealer will be fined if the number of customers who report favorably is between 39 and 43. The dealership will be dissolved if fewer than 39 report favorably. It is known that 72% of the dealer’s customers report favorably on satisfaction surveys. Use Table 1.

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.)

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 50 customers. The dealer will be fined if the number of customers who report favorably is between 39 and 43. The dealership will be dissolved if fewer than 39 report favorably. It is known that 72% of the dealer’s customers report favorably on satisfaction surveys. Use Table 1.

Explanation / Answer

This is a problem using the Sampling Distribution of the Proportion
The mean of the sample is .73
standard deviation of the sample is square root(PQ/n) where P=0.73, Q=0.27 and n =69
standard deviation of the sample is square root(0.73*0.27/69) = 0.053446438

mean(p) = 0.73 and std dev(p) = 0.05
40/69 = 0.579710145 approx 0.58
46/69 = 0.666666667 approx 0.67

a) Probability(P) ( 40 < X < 46 ) = P ( X < 46) - P ( X < 40 )
z-value for 46 is (0.67 - 0.73) / 0.05 = -1.2 and
P ( X < 46 ) = 0.1151
z-value for 40 is (0.58 - 0.73) / 0.05 = -3 and
P ( X < 40 ) = 0.0013
therefore
Probability(P) ( 40 < X < 46 ) = 0.1151 - 0.0013 = 0.1138
b) P ( X < 40 ) = 0.0013

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