Question

Heights are generally normally distributed. Men have a mean of 69.5 inches and standard deviation 2.4 inches. Women have a mean of 63.8 inches and standard deviation 2.6 inches. The US Air Force has a height requirement for their pilots to be between 64 inches and 77 inches.

*Make sure you are rounding z-scores properly to two
places.*

Part A: Find the two z-scores for women who meet this height
requirement z = (*smaller value)* and z
= (*larger value)*

Part B: Using the z-scores from part A, find the proportion of women who meet the height requirement. Give this answer as a decimal.

Part C: Find the two z-scores for men who meet this height
requirement z = (*smaller value)* and z
= (*larger value)*

Part D: Using the z-scores from part C, find the proportion of men who meet the height requirement. Give this answer as a decimal.

Part E: If the height requirement were changed to exclude only
the shortest 1% of women and tallest 1% of men, what are the new
heights? The short value would be in inches and the tall
value would be in inches. *Please give these two
values rounded properly to one decimal place.*

Answer #1

Z = (X - mean) / sd

a) Z (smaller value) = (64 - 63.8) / 2.6 = 0.08

Z (larger value) = (77 - 63.8) / 2.6 = 5.08

b) P(0.08 < Z < 5.08) = P(Z < 5.08) - P(Z < 0.08) = 1 - 0.5319 = 0.4681

c) Z (smaller value) = (64 - 69.5) / 2.4 = -2.29

Z (larger value) = (77 - 69.5) / 2.4 = 3.13

d) P(-2.29 < Z < 3.13) = P(Z < 3.13) - P(Z < -2.29) = 0.9991 - 0.011 = 0.9881

e) Z for shortest 1% = -z_{0.01} = -2.33

New height (smaller value) = 63.8 - 2.33 * 2.6 = 57.742 = 57.7

Z for tallest 1% = z_{0.01} = 2.33

New height (larger value) = 69.5 + 2.33 * 2.4 = 75.1

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