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 68 customers. The dealer will be fined if the number of customers who report favorably is between 40 and 44. The dealership will be dissolved if fewer than 40 customers report favorably. It is known that 70% of the dealer's customers report favorably on satisfaction surveys. Use Table 1. a. What is the probability that the dealer will be fined? (Round "z" value to 2 decimal places, and final answer to 4 decimal places.) b. What is the probability that the dealership will be dissolved? (Round "z" value to 2 decimal places, and final answer to 4 decimal places.)

Explanation / Answer

This is a problem using the sampling distribution of the proportion

The mean of the sample is 0.70 (70% report the favourably)

P = 0.70, Q= 0.30

standard deviation of the sample is (PQ/n) where P = 0.70, Q = 0.30 and n = 68

standard deviation of the sample is (0.70*0.30/68) = 0.003 = 0.054

Mean(p) = 0.70 and standard deviation = 0.054

40/68 = 0.5882

44/68 = 0.6470

a) probability (P) (40 < X < 44) = P (X<44) - P(X<40)

Z value for 44 is ( 0.647 - 0.70)/ 0.054 = -0.053/0.054 = - 0.98

and P(X <44) = 0.1635 (From area under the standard normal curve)

Z value for 40 is ( 0.5882 - 0.70) / 0.054 = -2.07 and

P (X < 40) = 0.0192 (From area under the standard normal curve)

Therefore, probability (P) (40 < X < 44) = 0.1635 - 0.0192 = 0.1443

b) P(X<40) = 0.0192

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