Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of 7 per hour. If it takes approximately 10 minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period.
If we take $Y=y$ as the number of costumer served in a given hour, then the expected value $E(Y) = 7$, the variance will also be $V(Y) = 7$.
If we take $T$ to denote the total service time, the we have $T = 10Y$. Then the expected total time service per hour would be $10E(Y) = 70$.
I am having trouble finding the variance. I know that we have
$$V(T) = E(T^2) - (E(T))^2$$ $$V(10Y) = E(100Y^2)-(E(10Y))^2$$ $$V(10Y) = 100E(Y^2)-4900$$
This is where I am stuck. Is there any formula for $E(Y^2)$ provided that we know the value of $E(Y)$?
I looked at the formula for variance, and thought I could derive $V(Y10)=10^2V(Y)$, is this correct?
A basic fact, very important: If $k$ is a constant, the variance of $kX$ is $k^2$ times the variance of $X$. So in our case the variance is $700$.
For a proof, let $W=kX$. Then $\text{Var}(W)=E(W^2)-(E(W))^2$.
But $E(W^2)=E((kX)^2)=E(k^2X^2)=k^2E(X^2)$ and $(E(W))^2=(E(kX))^2=(kE(X))^2=k^2(E(X))^2$.