Question

6.3.21 Assigned Media A survey found that women's heights we normally distributed with mean 63.6 in and standard deviation 22 in Abranch of the military requires women's heights to be between 58 nad 80 in a. Find the percentage of women meeting the height requirement. Are many women beng denied the opportunity to join this branch of the military because they are too short or too tall? b. If this branch of the miltary changes the height requirements so that all women are eligible except the shortest 15 and the latest 2%, what are the new height requrements ? Click to view page 1 of the table. Click to view page 2 of the table a. The percentage of women who meet the height requirement (Round to two decimal places as needed)

Answer 1

of Ans: X is noomally distributed random variable represent the height Women in inches. Given that. mean l=63.6 standaard deviation r = 2.2 Therefore, XN(63-6, 2.4) It is given that the branch of the military requires height to be between 58in & so in women's

a) Now to find the peocentage women meeting the height requirement je. To find the probability that height of the women is in between 58 and 80 it. Pl 582x480) - P(58 x <Bs) = P 생뼹 8"생 58-63.6 2. 2. < 86 - 626 22 | 쁨 = (-2.55 X 1.45) Where w Norl) 2 SS 0 45

= P(586x9 80) = 1- P(Z >17.45) -PC25-2.55) =1-0-0.0054 P[58< X<80) = 0.9946 - 99.46 of women Thus, the percentage is gg. 46% meeting height requirement The poobability of women being denied the oppostunity is J-0.9945 =8-0054 i-e-547. Hence, it can not be said that there are many women who denied the opportunity because they are too fall of too small

b) In this situation the branch of the military changes the height requirement so that all women are eligible except the shortes it and the tallest at Nors to find out the new height requirement Given that height of shortes is tri Here XN N(63.6, 2.2 2 )

Then calculate the new height using following probability PCX5L) = 11 *P(x-ll 14)=0:1 Plzs 3:6 ) = 5 L-63-6 2.2 20.1 wher 228 o Z=KEINNCO,1) LGST from the standaod normal table 2-score value conresponding to probability 0-1 - 1.28 (negative values appers because We have consider left side area) L-63.6 =-1.28 2.2 L-83.6 =(1-28)(2.2) 1- 63-6 - 1.2812-2) L=60.784 L=60.8

The height of the largest 24 is given by P(x>,U) = 0.2 7P 40-450-2 e(z 30-62.00 where ZX-40N (0,0) from the standerod noomal table Z-score value conoponding to poobability 0-2 is 0.8416 -0.2 V-63.6 2-2 6 * U-63-6 = 0.8416 2.2 U = 0.8416(2.2) +63.6 U=65.45 The new height requirement are women's height to be between 60.8 in to 65.45 in.