Suppose the heights (in inches) of adult males in the United States are normally distributed with a mean of 72 inches and a standard deviation of 2 inches. Find the percent of men who are no more than 68 inches tall. The percent of men who are 68 inches tall or below is

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Hinton

Mathematics

13 October, 05:33

Suppose the heights (in inches) of adult males in the United States are normally distributed with a mean of 72 inches and a standard deviation of 2 inches. Find the percent of men who are no more than 68 inches tall. The percent of men who are 68 inches tall or below is

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Carmen Lutz · Answer 1

Note that 72-68 = 4, and that 4 is 2 std. dev. from the mean.

68 would be 2 std. dev. below the mean. Alternatively, we could say that the z-score corresponding to a raw score of 68 is - 2.

Using either a calculator with built-in stat. functions, or a table of z-scores, we can find the area under the curve to the left of 68. That area is 0.023.

We could also use the Empirical Rule here. That Rule states that 95% of data whose distribution is normal lie within 2 std. dev. of the mean.

This means that (95%/2), or 47.5% of the data lies between 68 and the mean. To the left of the mean is 0.5 of the total area under the curve. So the percentage of males whose heights are no more than 68 inches is 50% less 45.5%, or 2.5% (answer, just slightly different from the previous result).