1
$\begingroup$

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!

$\endgroup$

closed as off-topic by heropup, Claude Leibovici, Watson, N. F. Taussig, user26857 Feb 16 '16 at 11:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Claude Leibovici, Watson, N. F. Taussig, user26857
If this question can be reworded to fit the rules in the help center, please edit the question.

4
$\begingroup$

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

$(1-\theta)f_{Z_m}(x)+\theta f_{Z_w}(x)$.

$\endgroup$

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