The heights of young women aged 20 to 29 follow approximately the N (64, 2.7) distribution. Young men the same age have heights distributed as N (69.3, 2.8). What percent of young men are shorter than the shortest among the tallest 25% of young women?

Answer: 10.703%

Step-by-step explanation:

Let minimum height of the tallest 25% of young women be M.

Let Q be the random variable which denotes the height of young women.

Therefore, Q - N (64,2.70)

Now, P (Q˂M) = 1-0.25

i. e. P[ (Q-64) / 2.7 ˂ (M-64) / 2.7] = 0.75

I. e. ф-1 [ (M-64) / 2.7] = 0.75 i. e. (M-64) / 2.7 = ф-1 (0.75) = 0.675 i. e. M = 65.8198 inches

Let R be the random variable denoting the height of young men

Therefore, R - N (69.3, 2.8)

i. e. (R-69.3) / 2.8 - N (0,1)

therefore the probability required = P (R ˂65.8198) = P[ (R-69.3) / 2.8 ˂ (65.8198 - 69.3) / 2.8]

this gives P[ (R-69.3) / 2.8 ˂] = ф (-1.2429) = 0.107033

From this, the percentage of young men shorter than the shortest amongst the tallest 25% of young women is 10.703%