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
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I am not sure how this response has been obtained? Can you help me? Can you go step-by-step, essentially as though you are doing an introduction, please? Because, I feel so lost. Someone tried to explain it in the second picture, but I cannot interpret it, as I need rationale to go with answers, not merely the answer.
In advance, I sincerely appreciate your efforts.

Transcribed Image Text: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.

Transcribed Image Text: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<O:6405) -P(ZE =O.684))
%3D
6.7391-O•2470
0.4921
49.21%0
as
100-49.2| = 50.79 %,
excluded with those specificationy
so, optiom A|.
blease Rate!!
any doubts please ask ! thank you :))
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