Question

hoights are nomally distributed with moan 63 7 in and standard deviation 2.3 in A b of women meeting the height requairoment Are many womon being denied the oppotunity to joun tas branch of the maltary branch of the miltary requres women's hHights to be betwoen 56 in and 80 a Find the porcentage b. If this trant, ot the metary changes the heght ' hey are too short or too tat? t reparoments so thiat alt women are oligiblo except tho shortest 15 and the tallest %, whal are the new henghe reparements? women are eli a. The percentage of wornon who moet tho hour' roquernert is r % (Round to two docimal places as needed ) Are many womon being densed tho opportunity lo jon thes branch of tho mildary bocause they are too short or too tal? only a small percontage of women are not alowed to yoin thes branch of the malary because of thoir heigh ов. Yes, because a large percentage of women are not allowed toponths branch ot the matary because of ther hort ° C. Yes, because the percentage of women who meet the height requ"emont tuny large O D. No, because the percentage of women who meet the height requaroment s farty smal b. For the new height requrements, thrs branch of the miltary requres women's heights to be af least n and at most Round to one decimal place as needed ) O Type here to search HD 1080

Answer 1

Sol:

P(58<x<80)

to be found

X~N(63.7,2.3)

use pnorm in R to get the probability

multiply with 100 to get the percentage

R code is

pnorm(80,mean=63.7,sd=2.3)-pnorm(58,mean=63.7,sd=2.3)

= 0.9933988

= 0.9933988*100

=99.33%

ANSWER:99.33%

mark option b

yes because large percentage of women are not allowed to join the branch of the military because of their heigh

Solutionb:

shorrtest 1%

convert percentile to z as

qnorm(0.01)=-2.326348

z=x-mean/sd

-2.326348=x-63.7/2.3

x=-2.326348*2.3+63.7

x=58.3494

x=58.3

and

qnorm(0.02)

-2.053749=x-63.7/2.3

x=-2.053749*2.3+63.7

x= 58.97638

x=59.0

ANSWER:atleast 58.3 and atmost 59