The heights of women aged 20 to 29 are approximately normal with mean 64 inches and standard deviation 2.7 inches. men the same age have mean height 69.3 inches with standard deviation 2.8 inches. what are the z-scores for a woman 6 feet tall and a man 6 feet tall? what information do the z-scores give that the actual heights do not?

To find the z-scores, find the difference in the number of inches for each value from the mean and then divide that by the standard deviation. For women at 72 inches (6'), this would be (72-64) / 2.7, or a z-score of +2.963. For a male at 72 inches, this would be (72-69.3)/2.8, or +0.964. This shows that the woman at 6 feet tall is more rare in the distribution than a man at 6 feet tall.

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