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?

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%

answer: 1,380

step-by-step explanation:

1: you turn %15 into a decimal by moving the decimal point two places to the left.

2: and then you multiply .15 by 1,200 , which will get you 180.00

3: from there you will add 180 to 1,200

4: which will get you 1,380

hope it *im kind of slow so idk if my answer is right*

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.