The heights of adult men in America are normally distributed, with a mean of 69.1 inches and a standard deviation of 2.68 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.8 inches and a standard deviation of 2.55 inches. a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? \( z= \) b) If a woman is 5 feet 11 inches tall, what is her \( z \)-score (to two decimal places)? \( z= \) c) Who is relatively taller? The 5 foot 11 inch American woman The 6 foot 3 inch American man

To find the z-scores, let's tackle the calculations! For the man who is 6 feet 3 inches tall (which is 75 inches), we can calculate the z-score using the formula: \[ z = \frac{(X - \mu)}{\sigma} \] Where \( X \) is the height, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Plugging in the numbers for men: \[ z = \frac{(75 - 69.1)}{2.68} \approx 2.19 \] So, \( z \approx 2.19 \). Now for the woman who is 5 feet 11 inches tall (which is 71 inches): Using the same formula but now using the parameters for women, we get: \[ z = \frac{(71 - 64.8)}{2.55} \approx 2.43 \] So, \( z \approx 2.43 \). Comparing the two z-scores, the woman has a higher z-score, indicating that she is relatively taller compared to her gender's average height. The 5 foot 11 inch American woman is relatively taller! In conclusion, when you convert heights into z-scores, you can see who's standing out in their respective gender groups! It's a fun way to visualize how individuals compare to their peers.