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14 January, 00:04

In the United States, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. If height is normally distributed, what percent of the women in this country are between 63 and 70.5 inches tall?

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  1. 14 January, 02:17
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    Z = (x - mean) / SD

    The z-score for x = 63 is: z = (63 - 65.5) / 2.5 = - 1

    The z-score for x = 70.5 is: z = (70.5 - 65.5) / 2.5 = 2

    Therefore we need the probability that - 1 < z < 2. This is equivalent to the probability that z < - 1, subtracted from the probability that z < 2. From z-tables, we find that this is equal to 0.9772 - 0.1587 = 0.8185. This means that 81.85% of women's heights fall within this range.
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