Question

GIVEN: It's been reported that the mean height for an American female is 64 inches with a standard deviation of 3 inches, and the interquartile range is 1.33 standard deviations.

Answer 1

Answer:

the percentage of women greater than 72 inches is 0.38%

Explanation:

It's been reported that the mean height for an American female is 64 inches with a standard deviation of 3 inches, and the interquartile range is 1.33 standard deviations. Calculate the percentage of women greater than 72 inches.

Since mean height equals the median height and the interquartile range = 1.3  standard deviations, this is a normally distributed problem. Given that:

mean (m) = 64 inches and standard deviation (s) = 3 inches.

To calculate the percentage of women greater than 72 inches we need to calculate the z score at 72 inches. Z score shows the relationship between a group of data and the mean.

Z score (Z) $= \frac{x-m}{s}$

substituting values:

$Z=\frac{72-64}{3} =\frac{8}{3} = 2.67$

P( X > 72) = P(Z > 2.67) = 1 - P(Z < 2.67) = 1 - 0.9962 = 0.0038

Therefore, the percentage of women greater than 72 inches is 0.38%