Poisson Distribution and Median

The number of automobile accidents at the corner of Wall and Street is assumed to have Poisson distribution with a mean of five per week. Let A denote the number of automobile accidents that will occur next week. Find

(a) $\Pr[A<3]$

(b) The median of A

(c) $\sigma_A$

Since this is a Poisson distribution, the probability function is:

$\Pr(A=k)=\cfrac{\lambda^k}{k!} e^{-\lambda}\tag{1}$ where $\lambda=5$ because the rate of occurrence is 5 accidents per week. Therefore:

(a) $\Pr[A<3]=e^{-5}(5^0 + 5 ^1 + 5^2/2!)=0.247$

(c) $\sigma_A=\sqrt{\lambda}=\sqrt{5}$

Although I wasn't sure how to calculate (b). Equation (1) is valid for $k>0$ so how do you find the median for a probability distribution which can take an infinite number of k values?

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The Median is the value for which at least half are greater than or equal to and at least half are less than or equal to. So you want to find the value such that $\Pr[A<n]<1/2$ and $\Pr[A\le n] \ge 1/2$. This works out to 5.

I think there may be an arithmetic error in your answer to part a (I get about half that)

Part c is almost certainly due to a theorem and not a definition.

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Thanks @deinst. So for a Poisson Distribution, the median always occurs at the expected value. Is this correct? –  user1527227 Jul 29 '13 at 20:43
The median occurs at the point where half of the probability is above or below. With the Poisson distribution the median is always an integer, but the expected value need not be. –  deinst Jul 30 '13 at 11:35