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Scores on a test have a mean of 66 and Q3 of 81. The scores have a distribution that is approximately normal. Find the standard deviation. Round your answer to the nearest tenth.

  • statistics -

    Q3 = 75 percentile

    In a table in the back of your stat book labeled something like "areas under normal distribution," find the Z score equivalent. Put value in formula below.

    Z = (x - μ)/ SD = (81-66)/SD

    Solve for SD.

  • statistics -

    (81-66)/SD = 0.67(z score of 0.75)
    15/SD= 0.67
    15/0.67
    SD=22.3880

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