Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $0.0498$.

share|improve this question
    
You need some assumption about the relative duration of the calls, such as all of them being the same length. –  T.. Nov 18 '10 at 2:53

1 Answer 1

up vote 2 down vote accepted

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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