Question

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 military 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?

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 join this branch of the military because they are too short or too tall?

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

B. Yes, because the percentage of women who meet the height requirement is fairly large.

C. No, because the percentage of women who meet the height requirement is fairly small.

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 requirements, 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.)

Answer #1

a)

probability
=P(58<X<80)=P((58-63.4)/2.2)<Z<(80-63.4)/2.2)=P(-2.45<Z<7.55)=1-0.0071=0.9929~
99.29 % |

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

b)

for 1th percentile critical value of z= | -2.33 | |

therefore corresponding value=mean+z*std deviation= | 58.27 |

for 98th percentile critical value of z= | 2.05 | |

therefore corresponding value=mean+z*std deviation= | 67.91 |

new height requirement are"

**at least 58.3 in and at most 67.9 in**

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