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4 January, 20:34

For normally distributed data, what proportion of observations have z-scores satisfying the following conditions:

0.62 < z < 2.92

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  1. 4 January, 23:38
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    From standard tables for normal distribution,

    P (z<=0.62) = 0.7324

    P (z<=2.92) = 0.9983

    Therefore

    P (0.62 < z <2.92) = 0.9983 - 0.7324 = 0.2659

    The required proportion is 0.2659 or 26.6%

    Answer: 0.266 or 26.6%
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