Understanding the Poisson distribution

Let $P(x,u) = \dfrac{e^{-u} u^x}{x!}$ be a random variable. I understand the $u$ is the mean average of success, and $x$ is the random variable. So, how come when I assign $x=x$ $P(x,u)$ is significantly lower then $1$? If I sell two cars per day on average, my chances of selling two cars tomorrow should be pretty good right? Instead I got $P(2,2) = 0.2706$.

Would anyone care to explain? Thanks!

P.S. I was looking for a guide on how to write mathematical symbols here, but I couldn't find anything. Any links provided would be helpful.

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1 Answer

The expected value is an average value. Some days you sell fewer than two cars, other days you sell 2 or more cars. On average, you sell 2 cars per day. But this does not imply that you sell exactly two cars per day most of the time. Here, there is a small, but significant, probability that you sell a large number of cars on a given day, which brings the average up.

As you found, around 27% of days, you sell exactly two cars. You can calculate that around 64% of days, you sell 0 or 1 cars. That leaves 9% of days where you sell more than 2 cars; and that "drives" the average number of cars sold per day up (to 2).

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To expand on this: to sell on average 2.64 cars per day is not to sell exactly 2.64 cars during a day. In fact, one never sells exactly 2.64 cars during a given day. –  Did Dec 2 '11 at 9:56
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