At a gas station, the number of customers arriving during a certain hour is a random variable $N$ with distribution $$P(N = n) = \frac{e^{-30}30^n}{n!}, n=0,1,...$$ Compute the expected value and variance of the total sales $Z_n$ during that hour if the expected value and variance of a single sale are 3.67 and 2.24 respectively.
I found the expected value to be the expected value of $N$ times the expected value of a single sale. $E(N)$ is $$\sum_{n=0}^{\infty}\frac{ne^{-30}30^n}{n!} = e^{-30}30\sum_{n=1}^{\infty}\frac{30^{n-1}}{(n-1)!} = 30$$ So the expected value of the total sales $Z_n$ is 30(3.67) = 110.1. This is correct. My trouble is with the variance. It is stated that the variance of $Z_n$ is $nb^2$, where $b$ is the variance of $Y$. I thought that $n=30$ and $b=2.24$, but this would yield $150.5$. The correct answer should be $470.5$. I know how to go about finding the variance of the number of customers (I got 30), but I don't know if that's needed here.
EDIT:
I found $Var(N)$ (for number of customers) by using a trick from the book where you first find $E(N(N-1))$ then add $E(N)$ to get $E(N^2)$: $$E(N(N-1)) = \sum_{n=0}^\infty \frac{n(n-1)e^{-30}30^n}{n!} = 30^2e^{-30}\sum_{n=2}^\infty \frac{30^{n-2}}{(n-2)!} = 900$$ Then $E(N^2) = 900 + 30 = 930$.
Finally $Var(N) = 930 - 30^2 = 30$.
If the single sales are each $X_i$, then the total sales are $Z=\sum_{i=1}^N X_i$.
Note $E[N] = \operatorname{Var}[N] = 30$ (show this!).
The expected value of total sales is $$E[Z]=E[N]E[X] = (30)(3.67)=110.1$$ and the variance of total sales is $$\operatorname{Var}[Z]=E[N]\operatorname{Var}[X] + \operatorname{Var}[N](E[X])^2$$ $$=(30)(2.24) + (30)(3.67)^2=471.267$$
The variable $Z$ is a random sum of random variables. That is, the number $N$ of terms is random, and each of the terms $X_i$ is identically distributed. Presumably, the number of terms and each of the terms all are independent.