Question

Assume that women's heights have a distribution that is symmetric and unimodal, with a mean of 68 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 2 inches. a. What women's height corresponds to a Z-score of - 1.20? 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 his or her peers, who is taller? inches. a. The women's height that corresponds to a Z-score of - 1.20 is (Type an integer or a decimal. Do not round.) while is only standard deviations above the mean for b. because vis standard deviations above the mean for (Type integers or decimals rounded to two decimal places as needed.)

Answer 1

The women's height that corresponds to a Z-score of -1.20 is 65 inches. Comparing Evelyn Akhator and Draymond Green to their respective peers by their Z-scores, Evelyn (Z-score of 2.80) is relatively taller in the WNBA than Draymond (Z-score of 2.00) is in the NBA.

Step-by-step explanation:

The question deals with the calculation of specific heights corresponding to given Z-scores within normal distributions representing women's and men's heights.

a. To find the height that corresponds to a Z-score of -1.20 for women, we use the formula x = μ + (z × σ), where μ is the mean, z is the Z-score, and σ is the standard deviation. For women, μ = 68 inches and σ = 2.5 inches. Plugging in the values, we get x = 68 + (-1.20 × 2.5), which equals 65 inches.

b. To determine who is taller compared to their peers between Evelyn Akhator and Draymond Green, we calculate their respective Z-scores. For Evelyn, z = (75 - 68) / 2.5 = 2.80, and for Draymond, z = (79 - 75) / 2 = 2.00. Therefore, Evelyn is taller compared to her peers in the women's league than Draymond is to his peers in the men's league since 2.80 is greater than 2.00 standard deviations above the mean.