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 26 and 28. The dealership will be dissolved if fewer than 26 report favorably. It is known that 61% 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.) b.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.)

Explanation / Answer

Since there are only two out comes: report favorably, report unfavorably, this is a binomial distribution

P( report favorably) = p = 0.61

P(report unfavorably) = q = 1- 0.61 = 0.39

number of surveys = n = 40.

mean = np = 40*0.61 = 24.4

variance = npq = 40*0.61*0.39 = 9.516

sd = square root of variance = 3.085

Using normal distribution , applying continuity correction to z values to convert discrete probability distribution to continuous distribution, as normal distribution is a continuous distribution

1. Probability that the dealer will be fined = P ( 25.50<=x<=28.50) = P(x<28.50) - P(x<25.50) = 0.9081 - 0.6393 = 0.2688.

Probability that the dealer will be fined = 0.2688

2. Probability that the dealership will be dissolved = P(x < = 25) = applying continuity correction = P(x<25.50)

from normal distribution table , P(x<25.50) = 0.6393.

Probability that the dealership will be dissolved = 0.6393

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