Widgets and boxes problem: expectation and variance. Why is this wrong? I'm taking the MITx: 6.041x Introduction to Probability - The Science of Uncertainty class to sharpen my probability skills. In one of the problems, the solution I came up with diverged from the correct solution, and I'm not sure why my solution isn't valid. I'd appreciate any insights. The problem is:
Widgets and crates. Widgets are stored in boxes, and then all boxes are assembled in a crate. Let $X$ be the number of widgets in any particular box, and $N$ be the number of boxes in a crate. Assume that $X$ and $N$ are independent integer-valued random variables, with expected value equal to 10, and variance equal to 16. Evaluate the expected value and variance of $T$, where $T$ is the total number of widgets in a crate.
The solution can be found here. My (incorrect) solution is below.
Finding $E[T]$
I started off thinking that the $T$, the total number of widgets, should be $X \cdot N$. Thus:
$E[T] = E[X \cdot N]$
$= E[X] \cdot E[N]$ (independence)
$= 10 \cdot 10 = 100$
Finding $var(T)$
I use the equation $var(T) = E[(T - E[T])^2] = E[T^2] - E[T]^2$. Thus:
$var(T) = var(X \cdot N) = E[X^2 \cdot N^2] - E[X \cdot N]^2$
We know $E[X \cdot N]^2 = 100^2$ from earlier.
I believe I can do $E[X^2 \cdot N^2] = E[X^2] \cdot E[N^2]$ due to independence. 
$X^2 = var(X) + E[X]^2 = 16 + 10^2 = 116$. This came from a rearrangement of the variance equation above. $N^2$ should be the same.
Thus $E[X^2] \cdot E[N^2] = E[116] \cdot E[116] = 116^2$
Finally $var(T) = 116^2 - 100^2$ ... but this is not the case. What am I missing?
 A: Because $X$ is the number of widgets in a particular box, whereas in your calculations you're treating it as the number of widgets in every box.  The number of widgets in a crate is actually $\sum_{i=1}^{N} X_i$, where $X_i$ is the number of widgets in box $i$.  The expected number should be
\begin{align}
\text{E} \left ( \sum_{i=1}^{N} X_i \right ) &= \text{E} \left [ \text{E} \left ( \sum_{i=1}^{N} X_i \mid N \right ) \right ] \\ 
&= \text{E} \left [ N \text{E} \left ( X_1 \right ) \right ] \\
&= \text{E}(X_1) \text{E}(N) \\
&= 100
\end{align}
which is the same as your answer but your argument is not correct.  For the variance we again use conditioning
\begin{align}
\text{Var} \left ( \sum_{i=1}^{N} X_i \right ) &= \text{Var} \left [ \text{E} \left ( \sum_{i=1}^{N} X_i \mid N \right ) \right ] + \text{E} \left [ \text{Var} \left ( \sum_{i=1}^{N} X_i \mid N \right ) \right ] \\
&= \text{Var}[N \text{E}(X_1)] + \text{E}[N \text{Var}(X_1)] \\
&= \text{E}(X_1)^2 \text{Var}(N) + \text{Var}(X_1) \text{E}(N) \\
&= 100 \cdot 16 + 16 \cdot 10 \\
&= 1760.
\end{align}
