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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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).

share|cite|improve this answer
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

Your Answer


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.