# Thread: Poisson distribution question.

1. ## Poisson distribution question.

The question is about a mechanic.

To summarise, he gets 7 calls per hour. The question wants to know what the probability is of him getting $\displaystyle <5$ callouts in one hour.
I understand the formula for poisson, which is:
$\displaystyle \frac{e^{-\lambda}\times\lambda^r}{r!}$ where$\displaystyle \lambda$is the mean. But I substitute $\displaystyle \lambda$=7, r=5 in and get the wrong answer. I then substiture $\displaystyle \lambda = \frac{7}{60}$ and r=5 but this is still incorrect.

The answer is 0.173 to 3 d.p.

I have the poisson cumulative frequency tables to save you some working.

Please help?

Edit: I now understand what $\displaystyle \lambda$ represents, and know how to get the correct answer from my cumulative frequency table. I just don't understand what the r! represents?

I don't expect a huge, half a page formula, because I wouldn't want to waste any one's time, but how would I get the answer without the Poisson cumulative frequency diagram?

2. P( R < 5 ) = P(0) + P(1) + P(2) + P(3) + P(4)
= 0.1729

R is just the number of events occurring, its usually denoted X. In your CDF table of poisson probabilities, look up Lamda = 7, X (or R) = 4

3. Oh, that obvious. Thanks, your simple explanation helped greatly.