I have to solve the following problem to solve, and I'm not sure I'm following the right track.
Let $X$, the number of accidents in a factory in a week, be given by a r.v. with mean $\mu$ and variance $\sigma^2$. Let $N$ the number of people involved in the different accidents be independent and have mean $\xi$ and variance $\tau^2$. Give the mean and the variance of the number of people involved in accidents in a week.
So I'm computing $Z$, number of people involved in accidents during the week. In order to do so, I'm using the fact that $E(Z)=E(E(Z \mid X))$, since $Z\geq 0$ and it's discrete. So I understand:
$$ E(Z \mid X) = X \xi $$ because the number of people people involved in the different accidents is independent. So $E(Z) = \mu \xi$.
For the variance:
$$\mathrm{Var}(Z) = E(\mathrm{Var}(Z \mid X)) + \mathrm{Var}(E(Z \mid X)) $$
So I evaluated $\mathrm{Var}(Z \mid X) = X \tau^2$ because $\mathrm{Var}(Z \mid X) = \mathrm{Var}(\sum_{i=1}^{X} N_i) = \sum_{i=1}^{X} \mathrm{Var}(N_i)$ due to independence (and I denote $N_i$ as the number of involved people in the $i$-th accident). Thus:
$$\mathrm{Var}(Z) = E(\tau^2 X) + \mathrm{Var}(X \xi) $$
$$\mathrm{Var}(Z) = \mu \tau^4 + \xi^2 \sigma^2 $$
Is it right? Or have I used the wrong properties? Thanks!