Question

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 1

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%