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22 October, 22:57

In a recently administered iq test, the scores were distributed normally, with mean 100 and standard deviation 15. what proportion of the test takers scored between 70 and 130?

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  1. 22 October, 23:20
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    To solve for proportion we make use of the z statistic. The procedure is to solve for the value of the z score and then locate for the proportion using the standard distribution tables. The formula for z score is:

    z = (X - μ) / σ

    where X is the sample value, μ is the mean value and σ is the standard deviation

    when X = 70

    z1 = (70 - 100) / 15 = - 2

    Using the standard distribution tables, proportion is P1 = 0.0228

    when X = 130

    z2 = (130 - 100) / 15 = 2

    Using the standard distribution tables, proportion is P2 = 0.9772

    Therefore the proportion between X of 70 and 130 is:

    P (70
    P (70
    P (70
    Therefore 0.9544 or 95.44% of the test takers scored between 70 and 130.
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