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13 June, 01:46

The Arc Electronic Company had an income of 90 million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 75 million dollars with a standard deviation of 11 million dollars. If incomes for this industry are distributed normally, what is the probability that a randomly selected firm will earn more than Arc did last year? Round your answer to four decimal places.

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  1. 13 June, 05:32
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    0.0869

    Step-by-step explanation:

    The arc electronic company had an income of 90 million dollars last year.

    Mean (μ) = 75 million dollars

    Standard deviation (σ) = 11 million dollars

    Probability that the randomly selected will earn more than arc did last year = Pr (x>90)

    Using normal distribution,

    Z = (x - μ) / σ

    Z = (90 - 75) / 11

    Z = 15/11

    Z = 1.36

    From the normal distribution table, 1.36 = 0.4131

    Φ (z) = 0.4131

    Recall that when Z is positive, Pr (x>a) = 0.5 - Φ (z)

    = 0.5 - 0.4131

    = 0.0869
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