Question

The heights of the men age 18 and over in HANES5 averaged 69 inches; the SD was 3 inches. Suppose the histogram of their heights follows the normal curve. Such a man that is 6 feet tall is at the percentile of this height distribution. (Enter the nearest whole number.) What is the 25th percentile, to the nearest inch

Answer 1

a) 84th percentile.

b) The 25th percentile is of 67 inches.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

The heights of the men age 18 and over in HANES5 averaged 69 inches; the SD was 3 inches.

This means that
`\mu = 69, \sigma = 3`

Such a man that is 6 feet tall is at the percentile of this height distribution.

6 feet = 6*12 inches = 72 inches. So this percentile is the p-value of Z when X = 72. So

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

`Z = (72 - 69)/(3)`

`Z = 1`

`Z = 1` has a p-value of 0.84, so 84th percentile.

What is the 25th percentile, to the nearest inch

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

`-0.675 = (X - 69)/(3)`

`X - 69 = -0.675*3`

`X = 67`

The 25th percentile is of 67 inches.