Question

8.) The U.S. Army requires women’s heights to be between 57 in and 79 in.

a.) Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join the Army because they are too short or too tall?

b.) If the U.S. Army changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?

Note: Women’s heights are normally distributed with a mean of 62.6 in and standard deviation of 2.4 in

10.) Assume SAT score are normally distributed with a mean ? = 1508 and a standard deviation of ? = 315.a.) If 1 SAT score is randomly selected, find the probability that it is between 1530 and 1575
b.) If 25 SAT scores are randomly selected, find the probability that they have a mean between 1530 and 1575

c.) Why can the central limit theorem be used in part (b), even though the sample size does not exceed 30?

11.) The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 71 in. Heights of men are normally distributed with a mean of 69 in and a standard deviation of 2.8 in.

a.) If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending. b.) if half of the 200 passengers are men, find the probability that the mean height of the 100 men is less than 71 in

c.) When considering the comfort and safety of passengers, which result is more relevant: The probability from part (a) or the probability from part (b)?

d.) When considering the comfort and safety of passengers, why are women ignored in this case?

Answer 1

#8.
a)
P(57 < X < 79)
= P((57 - 62.6)/2.4 < z < (79 - 62.6)/2.4)
= P(-2.3333 < z < 6.8333)
= 0.9902

i.e. 99.02%

b)
lower limit = 62.6 - 2.33*2.4 = 57.008
upper limit = 62.6 + 2.0537*2.4 = 67.5289

i.e. 57.01 in to 67.53 in

#10.
a)
P(1530 < X < 1575)
= P((1530 - 1508)/315 < z (1575 - 1508)/315)
= P(0.0698 < z < 0.2127)
= 0.0564

b)
P(1530 < X < 1575)
= P((1530 - 1508)/(315/sqrt(25)) < z (1575 - 1508)/(315/sqrt(25)))
= P(0.3492 < z < 1.0635)
= 0.2197

c)
population is normally distributed and population standard deviation is known