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 40 customers. The dealer will be fined if the number of customers who report favorably is between 22 and 26. 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. Use Table 1 a. What is the probability that the dealer will be fined? (Round your intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.) Probability b. What is the probability that the dealership will be dissolved? (Round your intermediate calculations to 4 decimal places, "z" value to 2 decimal places, and final answer to 4 decimal places.) Probability

Explanation / Answer

The mean of the sample is .70
standard deviation of the sample is square root(PQ/n) where P=0.70, Q=0.30 and n =40
standard deviation of the sample is square root(0.70*0.30/40) = 0.0724
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mean(p) = 0.73 and std dev(p) = 0.005
22/40 = 0.55
26/40 = 0.65

a) Probability(P) ( 22 < X < 26 ) = P ( X < 26) - P ( X < 22 )
z-value for 26 is (0.65 - 0.70) / 0.0724 = -0.69 and
P ( X < 26 ) = 0.2451
z-value for 22 is (0.55 - 0.70) / 0.0724 = -2.07 and
P ( X < 22 ) = 0.0192
therefore
Probability(P) ( 22 < X < 26 ) = 0.2541 - 0.0192 = 0.2349
b) P ( X < 40 ) = 0.2349

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