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27 July, 03:40

Assume that blood pressure readings are normally distributed with a mean of 120 and a standard deviation of 8. If 100 people are randomly selected, find the probability that their mean blood pressure will be greater than 122.

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  1. 27 July, 04:09
    0
    0.0062

    Step-by-step explanation:

    Let X be the random variable blood pressure. The information we are given are

    mean blood pressure=μx=120

    standard deviation blood pressure=σx=8

    we have to find P (mean bp>122) = P (xbar>122).

    P (Xbar>122) = P (xbar-μxbar/σxbar>122-μxbar/σxbar) = P (Z>122-μxbar/σxbar)

    Now, we have to find μxbar and σxbar. We know that

    μxbar=μx

    σxbar=σx/sqrt (n).

    So, μxbar=μx=120

    σxbar=σx/sqrt (n) = 8/sqrt (100) = 0.8

    P (Xbar>122) = P (Z>122-120/0.8) = P (Z>2.5) = P (0
    From normal distribution table P (0
    P (Xbar>122) = 0.5-0.4938=0.0062

    There is 0.62% chance that the mean blood pressure of 100 people will be greater than 122
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