Question

# Heights are generally normally distributed. Men have a mean of 69.5 inches and standard deviation 2.4...

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.

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% = -z0.01 = -2.33

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

Z for tallest 1% = z0.01 = 2.33

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