Calculating Poisson Distribution for Telephone Calls in College Switchboard

Click For Summary

Homework Help Overview

The problem involves calculating the probability of waiting longer than 1.5 minutes for the second telephone call in a Poisson process, where calls arrive at an average rate of three calls every four minutes.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants express confusion regarding the application of the Poisson distribution and its relation to the problem, with some questioning the relevance of the distribution to the current chapter's content.

Discussion Status

Some participants are exploring the relationship between the Poisson distribution and other statistical distributions, while others are attempting to clarify the meaning of the waiting time and the implications of the given rate of calls. Guidance has been offered regarding the expected number of calls in a given time frame.

Contextual Notes

There is mention of other statistical distributions covered in the chapter, which may lead to uncertainty about the applicability of the Poisson distribution in this context.

XodoX
Messages
195
Reaction score
0

Homework Statement


Telephone calls enter a college switchboard according to a Poisson process on the average of three calls every 4 minutes (i.e., at a rate of λ=0.75 per minute). Let W denote the waiting time in minutes until the second call. Compute P(W>1.5 minutes).


Homework Equations





The Attempt at a Solution



I don't get it. No idea how to do it. I guess 1.5 means here that at the most 1 event can occur here.
 
Physics news on Phys.org
XodoX said:

Homework Statement


Telephone calls enter a college switchboard according to a Poisson process on the average of three calls every 4 minutes (i.e., at a rate of λ=0.75 per minute). Let W denote the waiting time in minutes until the second call. Compute P(W>1.5 minutes).


Homework Equations





The Attempt at a Solution



I don't get it. No idea how to do it. I guess 1.5 means here that at the most 1 event can occur here.

You DO know how to do it. Your guess is correct: do you see why?

RGV
 
No, I don't. The poisson distribution is not in this chapter. It's Weibull, Gompertz, extreme value, gamma, chi-square, and logonormal. I don't know which one of those it is.
 
XodoX said:
No, I don't. The poisson distribution is not in this chapter. It's Weibull, Gompertz, extreme value, gamma, chi-square, and logonormal. I don't know which one of those it is.

Well, it's related to the Gamma.

However, let me ask you: what does the Poisson distribution represent? Never mind if it is not in that chapter; it is either in another chapter or else in another book or else in thousands of web pages. So, you have a Poisson distribution with m = 1.5*0.75 = 9/8 = 1.125; that would be the expected number of calls to occur in a 1.5 minute period. You can find the probability distribution of the number of calls in a 1.5-minute period by using the Poisson distribution formula for mean m. Now ask: if you need to wait > 1.5 min for the second call, how many have arrived before time 1.5? What is the probability of that?

RGV
 
Last edited:

Similar threads

Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
56
Views
6K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K