Question

A survey found that women's heights are normally distributed with mean 63.7 in and standard deviation 23 in A branch of the military requires women's heights to be between 50 in and on 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 shorter too? b. If this branch of the military changed the height requirements so that wil women are eligible except the shortest and the test 2%, what are the new he requirements? Cick to view of the table. Click to view of the table . The percentage of women who meet the height requirements (Hound to two decimal places as needed.) Are many women being denied the opportunity to join this branch of the military because they are too short or tool OA No, because only a small percentage of women are not allowed to join this branch of the military because of her high OB Yes, because we percentage of women are not allowed to join this branch of the day because of their height OC Yes, because the percentage of women who meet the height requirement is tasty large OD Ns, became the percentage of women who meet the height requirements witymall . For the new requirements, a branch of eitwy requires women's health and not Round to one decimaces needed)

Answer 1

Answer:

Given,

Mean = 63.7 , Standard deviation = 2.3

a)

P(58 < X < 80) = P((58 - 63.7)/2.3 < (x-u)/s < (80 - 63.7)/2.3)

= P(-2.48 < z < 7.09)

= P(z < 7.09) - P(z < -2.48)

= 1 - 0.0065691 [since from z table]

= 0.9934

= 99.34%

b)

P(Z < z) = 1% = 0.01

P(z < - 2.33) = 0.01

since from standard normal table

z = - 2.33

(x - 63.7)/2.3 = - 2.33

x = 58.3

Now,

P(Z > z) = 2% = 0.02

1 - P(Z < z) = 0.02  

P(Z < z) = 1 - 0.02 = 0.98

P(Z < 2.05 ) = 0.98

since from standard normal table

z = 2.05

(x - 63.7)/2.3 = 2.05

x = 68.415

Here we can say that, the new height is 58.3 in & taller height is 68.415 in.