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24 October, 01:00

The mean score of a competency test is 65, with a standard deviation of 4. use the empirical rule to find the percentage of scores between 53 and 77. (assume the data set has a bell-shaped distribution.)

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  1. 24 October, 01:06
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    To solve this problem, we use the t statistic. The z score is calculated as:

    z = (x - u) / s

    where x is the sample value = 53 and 77, u is the mean score = 65 while s is the standard deviation = 4

    when x = 53

    z = (53 - 65) / 4 = - 3

    From the standard distribution tables, the p value at z = - 3 is:

    P (z=-3) = 0.0013

    when x = 77

    z = (77 - 65) / 4 = 3

    From the standard distribution tables, the p value at z = 3 is:

    P (z=3) = 0.9987

    The percentage P between the two would be the difference:

    P (53 < x < 77) = 0.9987 - 0.0013 = 0.9974

    Answer:

    0.9974 or 99.74%
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