The number of dropped connections per call follows a Poisson distribution. From four calls, the number of dropped connections is $2,\,0,\,3,\,1$.
Obtain the maximum likelihood estimate that the next two calls will be completed without any accidental drops.

I know the maximum likelihood estimate of $\lambda$ is $1.5$. I think I am supposed to use this formula: $$f(x|\lambda) = \frac{\lambda^x e^{-\lambda}}{x!}$$

However, I do not know what $x$ should be. The answer should be $0.0498$.

  • $\begingroup$ You need some assumption about the relative duration of the calls, such as all of them being the same length. $\endgroup$ – T.. Nov 18 '10 at 2:53

x is the number of dropped connections, right? So $x=0$ for no dropped connections.

  • $\begingroup$ Then that would be e^-1.5 = 0.223. However, the answer should be 0.0498 $\endgroup$ – Raptrex Nov 18 '10 at 1:40
  • $\begingroup$ Squared (since there are two calls). $\endgroup$ – Yuval Filmus Nov 18 '10 at 1:41
  • $\begingroup$ Your right, thank you $\endgroup$ – Raptrex Nov 18 '10 at 1:49

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.