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Poisson distribution |
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| Jun20-11, 08:44 AM | #1 |
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Poisson distribution
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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| Jun20-11, 08:48 AM | #2 |
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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? |
| Jun20-11, 07:14 PM | #3 |
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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. |
| Jun20-11, 07:19 PM | #4 |
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Poisson distribution[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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