1. ## Poussin distraction

Poisson distribution

Hi guys this question has mind blanked me

A call centre receives an average of 10 calls per hour. What is the probability that the next call will occur at least 5 minutes from now?

Do I need to find the means as 10/12 =.8333 and ten do the poussin distribution for 0 1 2 3 4 and 5

Any help will be greatly appreciated

2. ## Re: Poussin distraction

Assuming the calls follow a Poisson process as indicated. Then the occurence time of next call follows an exponential distribution.

More specifically, if the increment of a (homogeneous) Poisson process within a time interval of length follows a Poisson distribution with mean , then the inter-arrival time follows an exponential distribution with mean (with the same time unit as defined)

3. ## Re: Poussin distraction

Why did you 'delete' your old post when this is the *exact* same question?

You correctly have that the distribution of the number of calls in the next 5 minutes will follow a Poisson(10/12) distribution. If the next calls occurs 'at least 5 minutes from now' that means that there must be 0 calls in the next 5 minutes.

BGM's answer is why I mentioned in the previous thread that there is more than one way to do this.

4. ## Re: Poussin distraction

Oh Dason's comment recall my memory. I just overlook the problem as it does not require the knowledge about Poisson process - simply knowing there is no arrival in the time interval is enough.

5. ## Re: Poussin distraction

Originally Posted by BGM
Oh Dason's comment recall my memory. I just overlook the problem as it does not require the knowledge about Poisson process - simply knowing there is no arrival in the time interval is enough.
I like these types of problems because there typically are multiple ways to do them. I find that my students hate these types of problems for that same exact reason.

6. ## Re: Poussin distraction

I asked my lecturer and he said we definitely need to use the poisson distribution! Should I used .83333 and apply this for time intervals 0 1 2 3 4 and 5, because when I do so it gives me a probability nearly at 1, does that senright

7. ## Re: Poussin distraction

Originally Posted by Dason
You correctly have that the distribution of the number of calls in the next 5 minutes will follow a Poisson(10/12) distribution. If the next calls occurs 'at least 5 minutes from now' that means that there must be 0 calls in the next 5 minutes.
Notice the last line. Notice how it doesn't mention any other values you would need to calculate the probability for other than 0.

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