Assume that military aircraft use ejection seats designed for men weighing between 141.2 lb and 202 lb. If women's weights are normally distributed with a mean of 172.6 lb and a standard deviation of 45.9 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications? The percentage of women that have weights between those limits is 49.21 %. (Round to two decimal places as needed.) Are many women excluded with those specifications? A. Yes, the percentage of women who are excluded, which is the complement of the probability found previously, shows that about half of women are excluded. B. Yes, the percentage of women who are excluded, which is equal to the probability found previously, shows that about half of women are excluded. C. No, the percentage of women who are excluded, which is the complement of the probability found previously, shows that very few women are excluded. D. No, the percentage of women who are excluded, which is equal to the probability found previously, shows that very few women are excluded. 9:18 1 ME = 172.6, TF=45.9 F→ Female; =P/ 141-2-172.6 X-ME < 202-172-6 :45-9 (या P(-0.6841 < z <0.6405) P(Z
Assume that military aircraft use ejection seats designed for men weighing between 141.2 lb and 202 lb. If women's weights are normally distributed with a mean of 172.6 lb and a standard deviation of 45.9 lb, what percentage of women have weights that are within those limits? Are many women excluded with those specifications? The percentage of women that have weights between those limits is 49.21 %. (Round to two decimal places as needed.) Are many women excluded with those specifications? A. Yes, the percentage of women who are excluded, which is the complement of the probability found previously, shows that about half of women are excluded. B. Yes, the percentage of women who are excluded, which is equal to the probability found previously, shows that about half of women are excluded. C. No, the percentage of women who are excluded, which is the complement of the probability found previously, shows that very few women are excluded. D. No, the percentage of women who are excluded, which is equal to the probability found previously, shows that very few women are excluded. 9:18 1 ME = 172.6, TF=45.9 F→ Female; =P/ 141-2-172.6 X-ME < 202-172-6 :45-9 (या P(-0.6841 < z <0.6405) P(Z
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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