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30 March, 09:41

A sample of final exam scores is normally distributed with a mean equal to 23 and a variance equal to 16. Part (a) What percentage of scores are between 19 and 27? (Round your answer to two decimal places.)

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  1. 30 March, 10:38
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    Let X = score of final exam.

    X~normal (23, 16)

    (a)

    percentage of score between 19 and 27

    = P (19 < X < 27)

    = P ((19-23) / sqrt (16) < Z < (27 - 23) / sqrt (16))

    = P (-1 < Z < 1)

    = 1 - P (Z = 1)

    = 1 - P (Z > = 1) - P (Z > = 1)

    = 1 - 2*P (Z > = 1)

    = 1 - 2 (0.1587)

    = 0.6826

    = 68.26%

    According to Normal Distribution Table

    P (Z>=1) = 1 - P (Z<1) = 1-0.8413

    So the final percentage is 68.26%
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