# Calculating variance of a sum

The number of students per day has the distribution N ∼ Poisson(10). The students of CSUEB withdraw money from a cash machine according to the following probability function (X):

X | 50 | 100 | 200

P(X = x) | 0.3| 0.5 | 0.2

Let $$T_N = X_1 +X_2 +···+X_N$$ be the total amount of money withdrawn in a day, where each $$X_i$$ has the probability function in the table, and $$X_1,X_2,...$$ are independent of each other and of N. Here $$T_N$$ is a randomly stopped sum, stopped by the random number of N customers.

Find $$Var(T_N)$$.

Attempt:

I've found E[X] = 105 and Var[X] = 2725

My attempt at solving this is:

$$Var[T_N | N] = Var[X_1] + \cdot\cdot\cdot+Var[X_N] = 2725N = 27250$$

I thought this was pretty straight forward but I'm not too confident. Is it correct to make the assumption that $$Var[T_N] = 2725N$$?

Update:

If I use the property

$$V[X] = V[E[X|Y]]+E[V[X|Y]$$

and

$$E[T_N|N] = 105N$$, $$V[T_N|N] = 2725N$$, $$V[N] = 10$$, $$E[N] = 10$$

then,

$$V[T_N] = V[105N] + E[2725N] = 105^2N + 2725N = 137500$$

The variability in $$T_N$$ arises not only due to $$X_i$$ bit also due to random number of summands $$N$$.
\begin{align} E(X) &= E(E(X\mid Y)) \\V(X) &= V(E(X\mid Y)) + E(V(X\mid Y)) \end{align}
Note: What you have calculated is in fact $$V(T_N\mid N)$$.