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 $?
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 $$ \sigma(Y) = \sqrt{\frac{91}6\cdot12} \approx 13.49074. $$