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Harlamova29_29 [7]
3 years ago
6

A survey found that​ women's heights are normally distributed with mean 63.4 in and standard deviation 2.2 in. A branch of the m

ilitary 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?
Mathematics
1 answer:
Ede4ka [16]3 years ago
8 0

Answer:

Step-by-step explanation:

Given that a survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.2 in.

Let X be the random variable representing women heights.

P(58<X<80)

Z for 58 = -2.454

Z for 80 = 7.545

P(-2.454<Z<7.545) = 0.4945+0.5=0.9945

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\boxed{\sf d. \ \frac{a}{b}  =  \frac{10}{7}}

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