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7 July, 10:42

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

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  1. 7 July, 13:38
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    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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