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18 January, 09:53

In a recent survey of college professors, it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $95.25 and a standard deviation of $27.32. What is the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.50?

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  1. 18 January, 10:07
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    0.0918

    Step-by-step explanation:

    We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

    The mean and standard deviation of average spending of sample size 25 are

    μxbar=μ=95.25

    σxbar=σ/√n=27.32/√25=27.32/5=5.464.

    So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

    The z-score associated with average spending $102.5

    Z=[Xbar-μxbar]/σxbar

    Z=[102.5-95.25]/5.464

    Z=7.25/5.464

    Z=1.3269=1.33

    We have to find P (Xbar>102.5).

    P (Xbar>102.5) = P (Z>1.33)

    P (Xbar>102.5) = P (0
    P (Xbar>102.5) = 0.5-0.4082

    P (Xbar>102.5) = 0.0918.

    Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
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