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12 December, 21:18

The scores of a high school entrance exam are approximately normally distributed with a given mean m = 82.4 and standard deviation = 3.3. What percentage of the scores are between 75.8 and 89?

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  1. 13 December, 00:57
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    Mean, m=82.4

    standard deviation, sigma=3.3

    Need proportion of sample between 75.8 and 89.

    first calculate Z-scores of 75.8 and 89

    Z (75.8) = (75.8-82.4) / 3.3=-2

    Z (89) = (89-82.4) / 3.3=2

    So

    P (75.8
    =P (-2
    =P (Z<2) - P (Z<-2)

    =0.9772-0.0227

    =0.9545
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