Question

The heights of adult men in America are normally distributed, with a mean of 69.8 inches and a standard deviation of 2.67 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.54 inches. What percentage of men are SHORTER than 67 inches

Answer 1

14.69% of men are SHORTER than 67 inches

Explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The heights of adult men in America are normally distributed, with a mean of 69.8 inches and a standard deviation of 2.67 inches.

This means that
`\mu = 69.8, \sigma = 2.67`

What percentage of men are SHORTER than 67 inches

This is the pvalue of Z when X = 67. So

`Z = (X - \mu)/(\sigma)`

`Z = (67 - 69.8)/(2.67)`

`Z = -1.05`

`Z = -1.05` has a pvalue of 0.1469

14.69% of men are SHORTER than 67 inches