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StClowers
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PLEASE HELP!!! WILL GIVE MEDALS FAST!!! A survey found that women’s heights are normally distributed with mean 63.8 in and a standard deviation 2.3 in. A branch of the military requires women’s heights to be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?
The percentage of women who meet the height requirement is __%.
For the new height requirements, this branch of the military requires women’s heights to be at least __ in and at most __ in.
 5 months ago
 5 months ago
StClowers Group Title
PLEASE HELP!!! WILL GIVE MEDALS FAST!!! A survey found that women’s heights are normally distributed with mean 63.8 in and a standard deviation 2.3 in. A branch of the military requires women’s heights to be between 58 in and 80 in. a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements? The percentage of women who meet the height requirement is __%. For the new height requirements, this branch of the military requires women’s heights to be at least __ in and at most __ in.
 5 months ago
 5 months ago

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coureges Group TitleBest ResponseYou've already chosen the best response.1
I'm not going to use calculus now, using a simple standard deviation cumulative distribution function (that should be "2nd" "vars" and the second option) and entering normcdg(58,80,63.8,2.3) returns the value .99416, %99.416, of women fall between 58 and 80 inches tall.
 5 months ago

coureges Group TitleBest ResponseYou've already chosen the best response.1
Furthermore, using an inverse normal distribution for .01 (%1 shortest) and .98 (2% tallest, since invnorm is always left tail) I get 58.449 and 68.523 inches for the second part.
 5 months ago

StClowers Group TitleBest ResponseYou've already chosen the best response.0
ok so then that would be 58 and 69, correct? How did you get that?
 5 months ago

StClowers Group TitleBest ResponseYou've already chosen the best response.0
@coureges
 5 months ago

coureges Group TitleBest ResponseYou've already chosen the best response.1
invnorm(.01,63.8,2.3)=58.4493 invnorn(.98,63.8,2.3)=68.5236 invnorm is the third option on the above stated menu.
 5 months ago

StClowers Group TitleBest ResponseYou've already chosen the best response.0
ok great. Thanks!!
 5 months ago
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