Variance of a Sum of $n$ Variables A random number of dice are rolled. If the number of dice rolled has the Poisson (12) distribution, find the standard deviation of the total number of spots showing.
I thought I could use the following:
$$\sigma^2=\frac{1}{n}\sum^6_{k=1}[x(k)-E(X)]^2$$
$$= 2\cdot [(1-42)^2+(2-42)^2+(3-42)^2+(4-42)^2+(5-42)^2+(6-42)^2]$$
But this approach is magnitudes off. Is there a rule that because the number of dice rolled has the Poisson distribution $\sigma^2=\frac{1}{n}\sum^6_{k=1}[x(k)]^2 $?
 A: Let $Y$ be the total number of spots showing and $N\sim\mathrm{Pois}(\lambda)$. Then $Y= \sum_{i=1}^N X_i$ where the $X_i$ are i.i.d. with uniform distribution on $\{1,2,3,4,5,6\}$. We have $\mathbb E[X_1]=\frac72$ and so for each $n\geqslant0$,
$$
\mathbb E[Y\mid N=n] = \mathbb E\left[\sum_{i=1}^N X_i\mid N=n \right] =\mathbb E\left[\sum_{i=1}^n X_i\right] = \frac72n,
$$
so $\mathbb E[Y\mid N]=\frac72N$. Moreover, $$\mathbb E[X_1^2] = \sum_{i=1}^6 i^2\mathbb P(X_1=i) = \frac16\sum_{i=1}^6 i^2 = \frac{91}6   $$
so
$$
\mathrm{Var}(X_1) = \mathbb E[X_1^2] - \mathbb E[X_1]^2 = \frac{91}6 - \left(\frac72\right)^2 = \frac{35}{12}.
$$
For each $n\geqslant 0$ we have
\begin{align}
\mathrm{Var}(Y\mid N=n) &= \mathrm{Var}\left(\sum_{i=1}^N X_i\mid N=n \right)\\
&= \mathrm{Var}\left(\sum_{i=1}^n X_i \right)\\
&=\sum_{i=1}^n\mathrm{Var}(X_i)\\
&= n\mathrm{Var}(X_1)\\
&= \frac{35}{12}n
\end{align}
so $\mathrm{Var}(Y\mid N) = \frac{35}{12}N$. It follows from the law of total variance that
\begin{align}
\mathrm{Var}(Y) &= \mathbb E[\mathrm{Var}(Y\mid N)] + \mathrm{Var}(\mathbb E[Y\mid N])\\
&= \mathbb E\left[\frac{35}{12}N\right] + \mathrm{Var}\left(\frac72 N\right)
\end{align}
Now, $\mathrm{Var}(N)=\mathbb E[N]=\lambda$, and so
$$
\mathrm{Var}(Y) = \frac{35}{12}\lambda + \frac{49}4\lambda = \frac{91}6\lambda.
$$
The standard deviation is just the square root of the variance, so
$$
\sigma(Y) = \sqrt{\frac{91}6\lambda}.
$$
In the case where $\lambda=12$, we have
$$
\sigma(Y) = \sqrt{\frac{91}6\cdot12} \approx 13.49074.
$$
