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Suppose men's heights follow a normal distribution $X \sim \mathcal{N}(\mu_1,\sigma_1^2)$ and women's heights follow a normal distribution $Y \sim \mathcal{N}(\mu_2,\sigma_2^2)$. How can I find the median height of the entire population of adults, supposing an equal number of men and women?

I think the solution is a height $h$ such that $P(X<h) + P(Y<h) = 1$ but I don't know how to find that $h$.

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  • $\begingroup$ Welcome to the site, @Jose. Do you know how to format formulae using $\LaTeX$? It looks like you started some formula, but did nit finish it. $\endgroup$ – StasK Oct 4 '14 at 20:28
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Your expression in the bottom is correct and can be proved using the law of total probability. Let's define two random variables $H$ and $G$ such that $H|G=1 \sim \mathcal{N}(\mu_1, \sigma_1^2)$ and likewise for $H|G=0$.

So, $Pr(H<h) = 0.5$ = $Pr(H<h|G=0)\cdot Pr(G=0) + Pr(H<h|G=1)\cdot Pr(G=1) = 0.5$. The important thing is that with a perfect 0.5 mixture, the solution to the median is quite easy, just calculate the $h$ that is symmetrically in the middle of the densities, with equal and opposite $\mathcal{Z}$-scores.. i.e. solve

$$\left( \frac{h - \mu_1}{ \sigma_1^2} \right) = -\left( \frac{h - \mu_2}{ \sigma_2^2} \right) $$

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