meansnormal distributionstatisticsvariance
It is often said that the height of men and that of women follow normal distribution with different means and variances. As graphs suggest, it appears true.
Then, what is the whole distribution of both men and women? Is it normal distribution or other distribution?
(I'm asking a distribution the height of randomly chosen people follow, including both men and women.)
I thought it is related to the reproductive property of the normal distribution, but it appears slightly different.
(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.)
(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.
I would break this question up into two pieces. First, what height are half of the women taller than? Since the mean of the women distribution is $165$ and the median of a normal distribution is its mean, the answer is 165.
The second half of the question is: What proportion of men are less than the height we just calculated. i.e. What proportion of men are less than 165 cm. That is just asking what is the probability that a normal distribution with mean 178 and standard deviation 8 is less than or equal to 165. So you need to calculate:
$$P(X \leq 165)$$ where $X$ is the normal distribution for male heights.
Best Answer
That would be an example of a mixture distribution. And the mixture of two normals is not necessarily normal (in fact it is almost never normal). Also, the idea that men and women's heights follow normal distributions is obviously an idealization. At the very least, any normal distribution can produce negative values; and that is nonsense for heights.
Anyway, even if you model the heights of men and women as normal variables say $M, W \in \mathbb{R}$ distributed as $\sim N(\mu_M, \sigma_M^2), N(\mu_W, \sigma_W^2)$ the distribution of people in general, say $P \in \mathbb{R}$, would be a mixture of $M, W$ for e.g. if men and women are equally numerous, $P$ would be $M$ with probability $\frac{1}{2}$ while $W$ with the probability $1 - \frac{1}{2} = \frac{1}{2}$ too. This means the density of $P$ is $$ f_P(x) = \frac{1}{2} f_M(x) + \frac{1}{2}f_W(x) = \frac{1}{2}\frac{1}{\sigma_W\sqrt{2\pi}}e^{-\frac{1}{2\sigma_W^2}(x - \mu_W)^2} + \frac{1}{2} \frac{1}{\sigma_M\sqrt{2\pi}}e^{-\frac{1}{2\sigma_M^2}(x - \mu_M)^2} $$ which can't be expressed as a normal density unless very special conditions are met for e.g. the heights of men and women have same mean $\mu_M = \mu_W$ and deviation $\sigma_M = \sigma_W$.