In the United States, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. If height is normally distributed, what percent of the women in this country are between 63 and 70.5 inches tall?

Question

Mandy

Mathematics

11 March, 07:21

In the United States, the mean average height of adult women is approximately 65.5 inches, with a standard deviation of 2.5 inches. If height is normally distributed, what percent of the women in this country are between 63 and 70.5 inches tall?

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Duncan · Answer 1

Z = (x - mean) / SD

The z-score for x = 63 is: z = (63 - 65.5) / 2.5 = - 1

The z-score for x = 70.5 is: z = (70.5 - 65.5) / 2.5 = 2

Therefore we need the probability that - 1 < z < 2. This is equivalent to the probability that z < - 1, subtracted from the probability that z < 2. From z-tables, we find that this is equal to 0.9772 - 0.1587 = 0.8185. This means that 81.85% of women's heights fall within this range.