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12 July, 18:14

Suppose the mean income of firms in the industry for a year is 75 million dollars with a standard deviation of 17 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 110 million dollars? Round your answer to four decimal places.

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  1. 12 July, 21:41
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    Answer and Explanation:

    Given:

    μ = 75 million

    SD = 17 million

    Probability (x) raw data = 110 million

    Computation:

    = Probability (x) < 110 million

    = Probability [ (x-μ) / SD] < [ (110 - 75) / 17]

    [ (x-μ) / SD] = Z

    = Probability [z] < [ (35) / 17]

    = Probability [z] < [2.05882353]

    Using z calculator:

    P-value from Z-Table:

    Z score = 0.98024

    Therefore, probability is 0.98024
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