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20 July, 19:53

The heights of adult men in america are normally distributed, with a mean of 69.1 inches and standard deviation of 2.65 inches. what percentage of adult males in america are under 5 feet 7 inches tall? round your answer as a percentage to the nearest whole number.

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  1. 20 July, 21:32
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    The solution for this problem is:It's given that the heights are normally distributed.

    5 feet 7 inches = (5*12 + 7) inches = (60+7) inches = 67 inches.

    The z-score is (67 - 69.1) / (2.65) = - 1.136. The probability is 0.127978 or 12.8%, making probability of a height over 67 inches
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