Poisson distribution


by gfields
Tags: distribution, poisson
gfields
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#1
Jun20-11, 08:44 AM
P: 2
I need help with aPoisson distribution problem please. Question is: company capable of handling 5 calls every 10 min on new system. Prior to new system, company analysts determined incoming calls to the system are Poisson distributed w/ a mean equal to 2 every 10 min. what is the probability that in a 10 min period more calls will arrive than the system can handle?
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micromass
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#2
Jun20-11, 08:48 AM
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Hi gfields!

What have you tried?

Obviously, we're going to work with a random variable X such that X is Poisson([itex]\lambda[/itex]) distributed. What do you think [itex]\lambda[/itex] is in this case?
gfields
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#3
Jun20-11, 07:14 PM
P: 2
I am taking this class online....no instructor...no interaction. I need some guidance on how to think this through. The book is helpful but still lacks clarity.

I understand that the mean before the old system was 2 calls for 10 minutes.

I need to define the segment unit to do the problem... 10 minutes??
The mean is already defined for me. 2
Defining the segment size would be next. I need some discussion here to understand exactly what to use.The event of interest would be P>5 correct?

Once this information is calculated.....the Poisson table can be used to find the probability and a comment on the adequacy of the new system can be made.

micromass
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Jun20-11, 07:19 PM
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Poisson distribution


Quote Quote by gfields View Post
I am taking this class online....no instructor...no interaction. I need some guidance on how to think this through. The book is helpful but still lacks clarity.

I understand that the mean before the old system was 2 calls for 10 minutes.

I need to define the segment unit to do the problem... 10 minutes??
Good!

The mean is already defined for me. 2
Also good.

Defining the segment size would be next. I need some discussion here to understand exactly what to use.The event of interest would be P>5 correct?
Indeed, you'll need to calculate [itex]P\{X>5\}[/itex]. You can use tables to calculate this, but you can also do it by hand easily:

[itex]P\{X>5\}=1-P\{X\leq 5\}=1-e^{-2}(1+2+\frac{2^2}{2!}+\frac{2^3}{3!}+\frac{2^4}{4!}+\frac{2^5}{5!})[/itex]


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