Combining Exponential Distributions for Concession Stand Wait Times

  • Thread starter Thread starter rhyno89
  • Start date Start date
  • Tags Tags
    Exponential
Click For Summary
SUMMARY

The discussion focuses on calculating the total wait time for customers at a concession stand, where each wait time follows an exponential distribution with a mean of 2 minutes. The total wait time, T, is the sum of four independent exponential random variables, leading to a mean of 8 minutes and a variance of 16. The participant initially attempted to combine probability density functions (pdfs) but shifted to using the properties of exponential distributions, ultimately deriving a pdf of 1/64e^(-x/64) for the total wait time.

PREREQUISITES
  • Understanding of exponential distributions and their properties
  • Knowledge of mean and variance calculations for independent random variables
  • Familiarity with probability density functions (pdfs)
  • Basic skills in statistical modeling and analysis
NEXT STEPS
  • Study the properties of the exponential distribution in detail
  • Learn about the moment-generating function (mgf) and its application in combining distributions
  • Explore the Central Limit Theorem and its implications for sums of random variables
  • Investigate real-world applications of exponential distributions in queuing theory
USEFUL FOR

Students studying probability and statistics, particularly those focusing on queuing theory and exponential distributions, as well as professionals in operations research and service management looking to optimize wait times.

rhyno89
Messages
18
Reaction score
0

Homework Statement


A concession stand serves customers with each customer starting as soon as the prior one finishes. The wait times are from an exponential distribution with mean = 2 minutes. Therefore the Total time is the summation of xi for i 1 to 4. Find the mean , variance and distribution of T.


Homework Equations





The Attempt at a Solution


I tried combining the pdfs to get the distribution and thought that I could work it from there. My first try was by comibining the mgf but was fairly certain that it would just make it more complicated so I shifted to trying to work with the exponential to the 4th since it was the same distribution. I got .0625e^-2x.

I thought that this was simple enough but this way they it could not have been another exponential distribution since the value for lambda was different throughout the equation.

As of now I was thinking that it is easier than it looks and the mean of this is simply (mu * n) which would give me 8 minutes.
The variance would then be 1/(.5^2) for each one and since they are independent would be 4*4=16

This seems not only way too easy but also way too high for a variance

Any tips to get me on the right track would be very appreaciated
 
Physics news on Phys.org
what i did was just combine the means since they are independent...resulting in a total mean of 8 minutes. This resulting in a lambda of 1/8 and therefore a variance of 1/64. Finally it would have a pdf of 1/64e^-x/64...

im pretty sure that's right but can anyone clarify?
 

Similar threads

Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
5
Views
5K
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
6
Views
2K
Replies
56
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
6K