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.

Suppose a customer service rep gets on average exactly 2 calls per minute. What are the odds of getting 1 call in the next minute?

This problem calls (no pun intended) for the Possion distribution : $$ P\{X = k\} = e^{-2} \frac{2^k}{k!}$$

Here's where I'm stuck wondering. The probability of there being one call $$P\{X = 1\} = e^{-2} \frac{2^1}{1!} \approx .2707 $$ is the same as the probability of getting two calls: $$P\{X = 2\} = e^{-2} \frac{2^2}{2!} \approx .2707 $$

I understand that the calculations show they are the same (obviously), but conceptually, I don't see how they can be the same. And also, why should one have the same probability as the average, but not three (which has a probability of $.1804$). Unless, am I calculating it incorrectly?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

The distribution can't be symmetric, since the probability for arbitrarily high numbers of calls is non-zero, whereas the probability for negative numbers of calls is zero; thus the probabilities for $3$ and $4$ must be less than the probabilities for $1$ and $0$.

There's no contradiction in the fact that $1$ has the same probability as $2$, and in fact it's true for any integer average number of calls $n$ that $n$ calls and $n-1$ calls have the same probability, since $\mathrm e^{-n}n^n/n!=\mathrm e^{-n}n^{n-1}/(n-1)!$. In general the average and maximum of an asymmetric distribution don't coincide, so there's no reason to expect the maximum to be exactly at $n$.

share|improve this answer
    
Thanks a lot, that was really helpful. –  Imray Dec 12 '12 at 21:34
    
@Imray: You're welcome! –  joriki Dec 12 '12 at 21:35

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.