Question

A queue with a single server receives 50 requests per second on average. The average time for the server to address a request is 10 milliseconds. What is the probability that there are exactly k requests in this system?

Answer 1

Answer:

$P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}$

Step-by-step explanation:

This is a typical example where the Poisson distribution is a good choice to model the situation.

In this case we have an interval of time of 50 milliseconds as average time for the server to address one request and 50 requests per second.  

By cross-multiplying we determine the expected value of requests every 50 milliseconds.  

We know 1 second = 1,000 milliseconds

50 requests __________ 1000 milliseconds

 x requests __________ 50 milliseconds

50/x = 1000/50 ===> x = 2.5  

and the expected value is 2.5 requests per interval of 50 milliseconds.

According to the Poisson distribution, the probability of k events in 50 milliseconds equals

$\bf P(X=k) = \displaystyle\frac{(2.5)^ke^{-2.5}}{k!}$