In the above problem, I understand that the Poisson random variable determines the sample size corresponding to Y. But I'm not sure about how to approach the problem.
If the problem didn't involve the poisson random variable, I could find the moment-generating function of the sum of i.i.d normally distributed random variables and find the first two moments to find the variance. But how does this problem change with the introduction of the poisson random variable? Any help would be great!
Guide:$$Y=\frac{\sum_{i=1}^N X_i}{N}$$
By law of total variance,
$$\operatorname{Var}(Y)=\mathbb{E}(\operatorname{Var}(Y|N))+\operatorname{Var}(\mathbb{E}(Y|N))$$
$$\operatorname{Var}(\mathbb{E}(Y|N))=\operatorname{Var}(\mu)=0$$
Hence you just have to evaluate: $$\mathbb{E}(\operatorname{Var}(Y|N))$$
Edit:
When $N=0$, $Y=0, \operatorname{Var}(Y|N=0)=0$
\begin{align} \mathbb{E}[\operatorname{Var}(Y|N)]&=\sigma^2 \sum_{n=1}^\infty \frac{1}{n} \frac{\exp(-\lambda)\lambda^n}{n!} \end{align}