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I have the following problem I'm trying to solve:

I know that the quantity of complains in a call center is a Poisson variable with $\lambda=18 $ costumers/hour, and that the probability of being able to solve a complain is $0.35$. They ask me about the probability of having more than $50$ complaints without solution in $8$ hours.

I tried to solve it by obtaining a new Poisson process with $\lambda=0.35\times18$, which describes the quantity of complains unsolved in certain time.

So, to solve the problem I would have to find $F_x (x)$ and do this:

\begin{align} P(\text{more than 50 complaints in 8 hours}) &= 1 - P (\text{less than 50 complaints in 8 hours}) \\&= 1 - F_x(x) \end{align}

My problem is that I need to sum $50$ terms of this Poisson distribution in order to solve the problem (to find the $F_x(x)$, which is exhausting, and I quite doubt this is the objective of the problem).

I think the problem has more to do with using a Normal distribution to get some sort of approximation, but I don't know how many calls there will be (I can't find how many terms I should solve, if I think that the variables I'm adding are each one of the calls, that I can think as being Bernoulli Variables, each one with a probability of $0.35$ of being solved).

I'd appreciate any help. Thank you very much.

Maazul
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FDrico
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1 Answers1

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The number of unresolved complaints in $1$ hour does indeed (under our assumptions) have Poisson distribution with parameter $(0.35)(18)$. This assumes essentially that a complaint is either resolved instantly or remains unresolved, which is not quite reasonable.

Let $X$ be the number of unresolved complaints in $8$ hours. This then also has Poisson distribution, with parameter $\lambda=(8)(0.35)(18)=50.4$.

The normal distribution with mean $\lambda$ and variance $\lambda$ is a reasonably good approximation to the Poisson with the large parameter $\lambda=50.4$. So now we have a standard normal calculation to do.

André Nicolas
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  • I didn't know you could do that (multiply $\lambda$ by $8$, just like that. I was thinking something way less straightforward: If i have $8Pois(\lambda= 0.3518)>50$, I could say that i need: $Pois(\lambda=0.35*18)>50/8=6.25)$ and then i wouldn't be able to use a Normal approximation because 6.25 is a value too low. All of this because if i assume that i can what you did, wouldn't i be assuming that the Poisson distribution "has no memory" and therefore i'm just adding two identally distributed and independent variables up? Thank you for your time! – FDrico May 20 '13 at 12:54
  • The mean of the Poisson is $50.4$, awfully close to $50$, particularly considering the fact that the standard deviation is $\sqrt{50.4}$, big. The probability of $\lt 50$ is the probability of $\le 49$, Which is roughly the probability that a standard normal is $\le \frac{49-50.4}{\sqrt{50.4}}$. This is a somewhat crude approximation, maybe one should use a continuity correction. Anyway, the answer will be under $0.5$, but not much under. – André Nicolas May 20 '13 at 13:02
  • And yes, if we have a Poisson, number of accidents in an hour is Poisson parameter $\lambda$, then the number of accidents in $t$ hours has Poisson distribution parameter %\lambda t$. Important fact. – André Nicolas May 20 '13 at 13:05
  • Thank you very much for your answer! – FDrico May 20 '13 at 13:15
  • You are welcome. What you will find most useful in the long run is that the content of the last comment. A closely related important fact is that the sum of independent Poisson, with possibly different parameters, is Poisson. – André Nicolas May 20 '13 at 13:24