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Assume that women's heights have a distribution that is symmetric and unimodal, with a mean of 66 inches, and the standard deviation is 2.5 inches. Assume that men's heights have a distribution that is symmetric and unimodal, with mean of 75 inches and a standard deviation of 3 inches. a. What women's height corresponds to a z-score of -1.90 ? b. Professional basketball player Evelyn Akhator is 75 inches tall and plays in the WNBA (women's league). Professional basketball player Draymond Green is 79 inches tall and plays in the NBA (men's league). Compared to, respectively, the populations of men and women, who is taller? a. The women's height that corresponds to a z-score of - 1.90 is 61.25 inches. (Type an integer or a decimal. Do not round.) b. Evelyn Akhator, because she is 1.33 standard deviations above the mean for women while Draymond Green is only 3.60 standard deviations above the mean for men. (Type integers or decimals rounded to two decimal places as needed.) ber.

Ask by Luna Lee. in the United States
Feb 05,2025

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**a.** 61.25 inches **b.** Evelyn Akhator is taller relative to the female population because she is **3.60** standard deviations above the mean for women, while Draymond Green is only **1.33** standard deviations above the mean for men.

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The calculation for a z-score involves the formula \( Z = \frac{(X - \mu)}{\sigma} \). To find the height related to a z-score of -1.90 for women, you would rearrange to \( X = Z \cdot \sigma + \mu \). Substituting the values, \( X = -1.90 \cdot 2.5 + 66 \), results in a height of 61.25 inches. When comparing heights to their respective populations, the z-scores reveal a lot. For Evelyn, her z-score is calculated as \( Z = \frac{(75 - 66)}{2.5} = 3.60 \), which indicates she's significantly taller than the average woman. For Draymond, \( Z = \frac{(79 - 75)}{3} = 1.33 \), showing he's also taller than the average man, but not as far above the mean as Evelyn is among women!

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