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The heights of young women aged $20$ to $29$ follow approximately the $\mathcal{N}(64, 2.7)$ distribution. Young men the same age have heights distributed as $\mathcal{N}(69.3, 2.8)$.

Height is measured in inches.

What percent of young women are taller than the tallest among the shortest $25\%$ of young men?

A young woman in that age group is randomly selected; independently, a young man in that age group is randomly selected. What is the probability that the young man is taller than the young woman by more than $5.3$ inches?

A young woman in that age group is randomly selected; independently, a young man in that age group is randomly selected. What is the probability that the young man is taller than the young woman by more than $5$ inches?

Edit:

I did attempt the first question, but I am not confident in my answer. I did the inverse probability for when $P(X \leq x)$ is $0.25$, and found that $x = 67.4114$ for the men. I think this means that $25\%$ of men are shorter than that height. So then I found the $P(X \leq 67.4114)$ for the women and subtracted that from $1$ to get $10.3208\%$ as the answer.

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    $\begingroup$ I am not sure where to even begin. I understand how to do the first question were it in regards to what percent of women are taller than the mean height of men. But aside from that I am lost. I am mostly looking for an idea of where to begin on these three questions, not the exact answer. $\endgroup$ – user64532 Aug 7 '15 at 13:46
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    $\begingroup$ I did attempt the first question, but I am not confident in my answer. I did the inverse probability for when P(X </= x) is 0.25, and found that x = 67.4114 for the men. I think this means that 25% of men are shorter than that height. So then I found the P(X </= 67.4114) for the women and subtracted that from 1 to get 10.3208% as the answer. $\endgroup$ – user64532 Aug 7 '15 at 13:49
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    $\begingroup$ That's the sort of information that is very useful to include in the posted question (you can still edit it in). I haven't checked the numbers yet but your method is correct. The other part of the question uses a different method; try looking at math.stackexchange.com/a/1015246/139123 (basically same problem, just different numbers). $\endgroup$ – David K Aug 7 '15 at 13:54
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(1) Your answer is correct. It appears you might be using software instead of normal tables to get so many decimal places of accuracy. [To use normal tables, you would have to 'standardize' (convert to standard normal distributions), then get something like four digits of accuracy.] In R software, this computation is as follows, without standardizing.

 qnorm(.25, 69.3, 2.8)
 ## 67.41143
 1 - pnorm(67.4114, 64, 2.7)
 ## 0.1032081

In the graph below, 25% of the probability under the blue curve (for men, at right) lies to the left of the dashed line at 67.4114, and 10.32% of the probability under the orange curve lies to the right of the same vertical line. (I recommend that you always try to draw sketches for such problems, especially as problems become more intricate than this one. Even very rough sketches can help catch gross computational or logical errors.)

enter image description here

(2) Let $X$ be the height of a randomly chosen woman and $Y$ be the height of a randomly chosen man. This part requires you to look at the distribution of the difference $D = Y - X$. Then $D$ is normally distributed with $$E(D) = \mu_D = \mu_M - \mu_W = 69.3 - 64 = 5.3$$ and $$V(D) = \sigma_D^2 = \sigma_M^2 + \sigma_W^2.$$ Notice that you subtract the means and add the $variances$. (So far, you have been dealing with standard deviations.) Then you want $P(D > 5.3).$ From what you have shown, I don't think you should have trouble from there on. (Make a sketch. Even without computations, the answer should be obvious.)

I don't know if you care for simulations, but here are results of a million simulated performances of this 2-person experiment. Simulated results are not perfectly accurate, but you can use them as a 'reality check' on your work.

 x = rnorm(10^6, 64, 2.7)
 y = rnorm(10^6, 69.3, 2.8)
 d = y - x
 mean(d);  sd(d);  mean(d > 5.3)
 ## 5.302365   # approx E(D)
 ## 3.889633   # approx SD(D)
 ## 0.50055    # approx P(D > 5.3)

(3) This part is very similar to part (2), but the result is not obvious, and you have a little computation to do. (My simulated answer is nearer to 0.53 than to 0.54.)

If this does not put you on the right track, or if you have unresolved questions, please leave a Comment.

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