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the heights of adult men are normally distributed with a mean of 68 inches and a standard deviation of 3 inches

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The fraction of women above a man's average height is approximately Option E: 1%.

How to find the percentage from the z-score?

To find the fraction of women above a man's average height, we need to calculate the probability that a woman's height is greater than the average height of men.

Using the z-score formula:

z = (x - μ) / σ

where:

x = raw score = 68 inches

μ = mean = 63 inches

σ = standard deviation = 2 inches

Calculating the z-score:

z = (68 - 63) / 2

z = 2.5

Using a standard normal distribution table or calculator, we can find the probability corresponding to a z-score of 2.5.

Looking up the probability for z > 2.5, we find that it is approximately 0.0062097, which is equivalent to 0.62%.

Therefore, the fraction of women above a man's average height is approximately 0.62%, which is closest to answer choice E. 1%.

Complete question is:

Suppose the average man's height is 68 inches, with a standard deviation of 3 inches, and is normally distributed. Also suppose that the average woman's height is 63 inches, with a standard deviation of 2 inches, and is normally distributed. What fraction of women are above a MAN'S average height? (Remember to pick the closest answer.) A. 10% B. 5% C. 3% D. 2% E. 1%

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