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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?

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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 $Var[S]$.

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  • $\begingroup$ I have trouble for calculating E(Y^2)? Could you please help? $\endgroup$ Commented Oct 27, 2013 at 16:18
  • $\begingroup$ Do you mean $E_S[S^2|N]$? Where are you getting stuck? Remember that $E[X^2] = Var[X] + E[X]^2$ for any random variable $X$. $\endgroup$
    – Patrick
    Commented Oct 27, 2013 at 16:20
  • $\begingroup$ wait a minute...I think V(S) =V(10Y) = 10^2*V(Y) = 100V(Y) = 700 is right. Isn't it? $\endgroup$ Commented Oct 27, 2013 at 16:25
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    $\begingroup$ The question says it takes approximately 10 minutes to service each customer. To me that means it takes 10 minutes on average. I don't think we can assume it is always 10 minutes which leads to your answer. $\endgroup$
    – Patrick
    Commented Oct 27, 2013 at 16:30
  • $\begingroup$ What is your suggestion? $\endgroup$ Commented Oct 27, 2013 at 16:40

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