Variance of Product of Ind. Variables Whats wrong with my approach to answer the following question?
The number of customers arriving to a fast food restaurant between 7 am and 9 am has the Poisson distribution with mean 40. Suppose that each customer spends in the restaurant a random amount, uniformly distributed between $8$ and $20$ (the amounts spent by different customers are independent, and also independent of the number of the arriving customers). Let X be the total amount spent by the customers between 7 am and 9 am. Find EX and Var(X).
Let $Y$ be the number of customers in the fast food restaurant between  $7$ am and $9$ am. It is given that $f_Y = \frac{40^ye^{-40}}{y!}$ with $R_Y = \{0,1,\dots\}.$
Let $Z$ be the amount spent by a costumer between $7$ am and $9$ am. It is given that $f_Z = \frac{1}{12}$ with $R_Z = (8,20)$. $X$ is given by $X = YZ$. Since $Y$ and $Z$ are independent, 
\begin{align*}
    E(X) = E(YZ) &= E(Y)E(Z)\\
    &= (40)(14)\\
    &= 560
\end{align*}
and 
\begin{align*}
    Var(YZ) &= Var(Y)Var(Z) + Var(Y)E(Z)^2 + Var(Z)E(Y)^2\\
    &=(40)(12) +(40)(14)^2+ (12)(40)^2\\
    &= 27520
\end{align*}
 A: Wait... $Z$ is the amount of time spent by which customer? 
No.   If $Z_k$ is the amount of time spent by the $k$eth customer, what you seek is the expectation and variance for the random variable  $$X:=\sum_{k=1}^Y Z_k$$
That is the sum of $Y$ independent random variables, which is not the same thing as $YZ_1$, the sum of $Y$ occurrences of the same random variable.

Given $Y\sim\mathcal{Pois}(40)$, we know $\mathsf E(Y)=\mathsf{Var}(Y)=40$
Given for any $k$ that  $Z_k\sim\mathcal U(8;20)$, we know $\mathsf E(Z_k)= 14$ and $\mathsf {Var}(Z_k)=12$
We are also given that $Y, Z_1, Z_2,\ldots$ are mutually independent.
Then indeed it is that :
$$\begin{split}\mathsf E(X) &= \mathsf E(\mathsf E(\sum_{k=1}^Y Z_k\mid Y))\\[1ex]&=\mathsf E(Y)\mathsf E(Z_1)\end{split}$$
However, it is instead that: $$\begin{split}\mathsf{Var}(X) &= \mathsf E(\mathsf {Var}(\sum_{k=1}^Y Z_k\mid Y))+\mathsf{Var}(\mathsf E(\sum_{k=1}^YZ_k\mid Y))\\[1ex] &= \mathsf E(Y~\mathsf {Var}(Z_1))+\mathsf{Var}(Y~\mathsf E(Z_1))\\[1ex]& = \mathsf E(Y)~\mathsf{Var}(Z_1)+\mathsf E(Z_1)^2~\mathsf{Var}(Y)\\\end{split}$$
