Customers arrive at a checkout counter in a department store according to a Poisson distribution at an average of seven per hour. If it takes approximately ten minutes to serve each customer, find the mean and variance of the total service time for customers arriving during a 1-hour period. (Assume that a sufficient number of servers are available so that no customer must wait for service.) Is it likely that the total service time will exceed $2.5$ hours?
Let $S$ be the total service time $= 10Y$.
What I have:
$$E(S) = 10E(Y) = 70$$
$$V(S) = 100V(Y) = 700$$
$$P(S>150) = P(Y>15) = 1-P(Y\le 15) = 1-.998 = .002$$
So the event is very unlikely to occur.
Is my answer right?
Best Answer
We can write $$S = \sum_{n=1}^{N}X_i$$
where $N$ is the number of customers that arrive within one hour and $X_i$ is the total service time for customer $i$.
The law of total expectation states: $E_X[X] = E_Y[E_X[X|Y]]$ and so for this case it amounts to: $$E_S[S] = E_N[E_S[S|N]] = E_N[E_X[N\times X]] = E_N[N E_X[X]]=E[N]E[X]$$.
In other words, your calculation of $E[S]$ is correct. Now to get the variance we start with:
$$Var[S] = E[S^2]-E[S]^2 = E_N[E_S[S^2|N]]-\{E_N[E_S[S|N]]\}^2$$
$$=E[Var[S|N]+E[S|N]^2]+\{E_N[E_S[S|N]]\}^2$$
$$=E[Var[S|N]]+Var[E[S|N]] = E[N]Var[X]+E[X]^2Var[N]$$
$$=E[N]E[X^2]$$
Where the last equality follows from the equality of a Possion random variable's mean and variance.
Finally, to calculate the probability, invoke the central limit theorem with your newly calculated $E[S]$ and $Var[S]$.