Variance of the sum of a Poisson-distributed random number of (normally distributed) random variables Suppose I have a fruit stand where I sell oranges. Every day I have an average of $\lambda$ people (Poisson distributed) each buying a normally distributed weight quantity of oranges with mean $\mu$ and standard deviation $\sigma$.
What's the mean and variance of the number of pounds of oranges I sell per day? 
OK, the mean is trivial: $\lambda \cdot \mu$. But how do I calculate the variance? 
Strangely enough I've come up with an answer through trial and error that seems to fit really well with my Monte Carlo simulations, but I can't convince myself that it's actually the correct answer in theory. If someone can point me to the "right" answer, I'll post the one I got empirically for comparison.
 A: Here's the answer I got through trial and error, which seems to come close to Monte Carlo simulated estimates in every scenario I have tried.
$\text{Var} = λ(μ^2+σ^2)$
(Edit: Yes, this agrees with Glen_b's updated answer, much to my delight!)
A: This is a compound Poisson distribution
The law of total variance gets you the answer.
Note that this answer doesn't rely on the normality of the $X$'s at all; it applies to any distribution with mean $\mu$ and variance $\sigma^2$.
In the following, $N$ is the Poisson r.v., $X$'s are individual components, $Y$ is the sum of components, $Y|N$ is the sum of components given a specific count of terms, $N$:
$$\text {Var}_{Y}(Y)=E_{N}\left[\text {Var}_{{Y|N}}(Y)\right]+\text {Var}_{N}\left[E_{{Y|N}}(Y)\right]$$
Substituting in:
$$=\text {E}_{N}\left[N\text {Var}_{X}(X)\right]+\text {Var}_{N}\left[N\text {E}_{X}(X)\right]\,$$
$$=\text {E}_{N}\left[N\sigma^2\right]+\text {Var}_{N}\left[N \mu\right]\,$$
$$=\sigma^2\text {E}_{N}\left[N\right]+\mu^2\text {Var}_{N}\left[N \right]\,$$
$$=\sigma^2\lambda+\mu^2\lambda$$
$$=(\sigma^2+\mu^2)\lambda$$
(... which is also $\lambda E(X^2)$)
A: By the law of the product of variances, if two random variables are independent, the variance of their product $\ D(X*Y)$ = $\ E(X)^2*D(X)+E(Y)^2*D(Y)+D(X)*D(Y)$
which is $\ E(X^2)*E(Y^2)-E(X)^2*E(Y)^2$. 
So, if the distributions are Poisson and Normal, I'd guess it's $\lambda*σ^2$.
