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7 October, 04:37

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 69 customers. The dealer will be fined if the number of customers who report favorably is between 40 and 46. The dealership will be dissolved if fewer than 40 customers report favorably. It is known that 73% of the dealer's customers report favorably on satisfaction surveys. 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.)

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  1. 7 October, 04:59
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    Step-by-step explanation:

    The probability that the dealer will be fined is 0.0948

    To find p (a < = Z < = b) = F (b) - F (a)

    P (X < 20) = (20 - 30.5) / 3.4489

    = - 10.5/3.4489

    = - 3.0444

    = P (Z < - 3.0444) from standard normal table

    = 0.00117

    P (X < 26) = (26 - 30.5) / 3.4489

    = - 4.5/3.4489 = - 1.3048

    = P (Z < - 1.3048) From standard normal table

    = 0.09599

    P (20 < x < 26) = 0.09599 - 0.00117 = 0.0948

    The answer in this question is 0.0948
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