# Finding the pdf for a random variable $Z_{1}$ [closed]

I'm having some problem getting started with this problem:

Suppose the heights of women are normally distributed with mean $\mu_{1}$ and variance $\sigma_{1}^{2}$, and the heights of men are normally distributed with mean $\mu_{2}$ and variance $\sigma_{2}^{2}$. Also assume that the percentage of women is $\theta \in (0, 1)$. Let $Z_{1}$ be the height of a randomly chosen person. Find the pdf for $Z_{1}$.

Thanks for any help!

## closed as off-topic by heropup, Claude Leibovici, Watson, N. F. Taussig, user26857Feb 16 '16 at 11:01

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Let $f_{Z_m}$ be the p.d.f. for the first normal and $f_{Z_w}$ be the p.d.f. for the second one.
Then $p(Z_1<c)=p(Z_m<c\mid\text{man})\cdot p(\text{man})+p(Z_w<c\mid\text{woman})\cdot p(\text{woman})=(1-\theta)F_{Z_m}(c)+\theta F_{Z_w}(c)$
this equals $(1-\theta)\int_{-\infty}^c f_{Z_m}(x)dx+\theta\int_{-\infty}^c f_{Z_w}(x)dx$
Taking derivative w.r.t. $c$ gives
$(1-\theta)f_{Z_m}(x)+\theta f_{Z_w}(x)$.