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 $h$.

  • $\begingroup$ 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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.