# Poisson distribution

by gfields
Tags: distribution, poisson
 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?
 Mentor P: 18,019 Hi gfields! What have you tried? Obviously, we're going to work with a random variable X such that X is Poisson($\lambda$) distributed. What do you think $\lambda$ is in this case?
 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.
Mentor
P: 18,019
Poisson distribution

 Quote by gfields 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 $P\{X>5\}$. You can use tables to calculate this, but you can also do it by hand easily:

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

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