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a Secure https:/wwwmathxl.com/student/player Testaspx?testld-iss1760708centenwinsyes APPLIED STATISTICS SPRING 2017 MMW 8 00-9:30 Quiz: Quiz 11 Submit Quiz This Question: 1 pt 6 of 10 (0 complete) This Quiz: 10 pts possible A survey Question Help between found that women's heights are normally distributed with mean 639 n and standard deviation 24 in A branch of the m tary requires women's heights to be 58 in and 80 in a. Find the percentage of women meeting the height requirement Are many women being dened the opportunity to join this branch of the military because they are too short or b. If this branch of the military changes the height requrements so that women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements? all glisk to view page 1 of the table. Qick to iew pago 2of be table. a. The percentage of women who meet the height requirement is (Round to two decimal places as needed) Are many women being denied the opportunity to jon this branch of the military because they aretoo short or too tar? o A. No, because the percentage of women who meet the height requirementis fairly smal O B. No, because only a small percentage of women are not allowed this branch of the military because of their height O c. Yes, because the percentage of women who meet the height requirement is fary O D. Yes because a large percentage of women are not allowed to join this branch of the military because of their height. b. For the new height requirement this branch of the military requires women's heights to be at least in and at most in (Round to one decimal place as needed) Cack to select your answer(s).

Explanation / Answer

Mean = 63.9 in

Standard deviation = 2.4 in

a) P(58 < X < 80) = P(X < 80) - P(X < 58)

= P(Z < (80-63.9)/2.4) - P(Z < (58-63.9)/2.4)

= P(Z < 6.71) - P(Z < -2.46)

= 1 - 0.0069

= 0.9931

B. No, because only small percentage of women are not allowed to join this branch of military because of their height

b. For shortest 1%,

P(X < A) = P(Z <(A - 63.9)/2.4) = 0.01

(A - 63.9)/2.4 = -2.33

A = 58.3 in

For tallest 2%,

P(X < B) = P(Z < (B - 63.9)/2.4) = 0.98

(B - 63.9)/2.4 = 2.05

68.8 in

The new height requirements are, at least 58.3 in and at most 68.8 in

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