Poisson distribution problem help

[gfields]

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?

 

[micromass]

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?

 

[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??
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]

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:

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

 

[CellCoree]

Sure, I would be happy to help with your Poisson distribution problem. Let's break down the problem into smaller parts to make it easier to understand and solve.

First, we need to understand what a Poisson distribution is. It is a probability distribution that is used to model the number of events that occur in a specific time period, given the average rate of occurrence. In this case, the time period is 10 minutes and the average rate of occurrence is 2 calls every 10 minutes.

Next, we need to determine the probability that in a 10 minute period, more calls will arrive than the system can handle. To do this, we need to find the probability of having 6 or more calls in a 10 minute period, as 5 is the maximum number of calls the system can handle. This can be calculated using the Poisson distribution formula:

P(x ≥ 6) = 1 - P(x ≤ 5)

= 1 - e^(-2) (2^0/0! + 2^1/1! + 2^2/2! + 2^3/3! + 2^4/4! + 2^5/5!)

= 1 - 0.406

= 0.594

Therefore, the probability that in a 10 minute period, more calls will arrive than the system can handle is 0.594 or 59.4%.

I hope this helps you understand and solve the problem. Let me know if you have any further questions.