0
$\begingroup$

Studying the probability of waiting time between two orders where the number of calls in a given period of time follows a Poisson distribution, a seller got the following probability density function for the waiting time:

$$f(W) = 0.2e^{(−0.2W)}$$

where the waiting time $W$ is measured in minutes. In this case, obtain the probability of getting at least two calls in $5$ minutes.

I wonder what are the relationships between the exponential distribution and Poisson distribution in this example. I mean, what should the parameter lambda of the Poisson distribution be?

I thought, because the given time interval is $5$ (minutes) and the parameter of the exponential distribution above is $0.2$, the lambda should be $5*0.2=1$?

Is this correct?

$\endgroup$
0
$\begingroup$

The Poisson process can be interpreted as the process that counts the total number of events that have occurred when the waiting times between events are i.i.d. exponential.

Hence, if you find the holding/waiting times between some random events are i.i.d. and distributed exponentially with parameter $\lambda$, then the total number of events will be distributed as a Poisson distribution with parameter $\lambda t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.