The law of total variance states:
$$
\operatorname{Var}(Y)=E[\operatorname{Var}(Y \mid X)]+\operatorname{Var}(E[Y \mid X])
$$
Let's apply this with $Y=S$ and $X=N$.
Step 1: Find $E[S \mid N]$
Given $N=n, S$ is a sum of $n$ i.i.d. random variables.
$$
E[S \mid N=n]=E\left[\sum_{i=1}^n X_i\right]=\sum_{i=1}^n E\left[X_i\right]=n \mu
$$
So, $E[S \mid N]=N \mu$.
Step 2: Find $\operatorname{Var}(S \mid N)$
Given $N=n, S$ is a sum of $n$ i.i.d. random variables.
$$
\operatorname{Var}(S \mid N=n)=\operatorname{Var}\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n \operatorname{Var}\left(X_i\right)=n \sigma^2
$$
So, $\operatorname{Var}(S \mid N)=N \sigma^2$.
$$
\operatorname{Var}(S)=E[\operatorname{Var}(S \mid N)]+\operatorname{Var}(E[S \mid N])
$$
- First term: $E[\operatorname{Var}(S \mid N)]=E\left[N \sigma^2\right]=\sigma^2 E[N]=\sigma^2 \nu$
- Second term: $\operatorname{Var}(E[S \mid N])=\operatorname{Var}(N \mu)=\mu^2 \operatorname{Var}(N)=\mu^2 \tau^2$
Therefore:
$$
\operatorname{Var}(S)=\nu \sigma^2+\mu^2 \tau^2
$$
Final Answer: (C) $\nu \sigma^2+\mu^2 \tau^2$
