# Poisson Dist.

• Mar 27th 2010, 02:57 PM
Bushy
Poisson Dist.
The number of people arriving for treatment at an emergency room can be modelled by a Poisson process with a rate parameter of five per hour. What is the probability that more than 12 people arrive during a two hour period?

I said $>12$ people arriving in 2 hours would be the same as finding $>6$ people in one hour using the same parameter $\lambda = 5$

Using $P(X) = \frac{e^{-\lambda}\lambda^x}{x!}$

$P(X>6) = 1- P(X\leq 6) = 1-0.762 = .238$

but this was not correct. I think the problem here is my logic.

(Wondering)
• Mar 27th 2010, 03:31 PM
mr fantastic
Quote:

Originally Posted by Bushy
The number of people arriving for treatment at an emergency room can be modelled by a Poisson process with a rate parameter of five per hour. What is the probability that more than 12 people arrive during a two hour period?

I said $>12$ people arriving in 2 hours would be the same as finding $>6$ people in one hour using the same parameter $\lambda = 5$ Mr F says: This is not correct.

Using $P(X) = \frac{e^{-\lambda}\lambda^x}{x!}$

$P(X>6) = 1- P(X\leq 6) = 1-0.762 = .238$

but this was not correct. I think the problem here is my logic.

(Wondering)

Calculate Pr(Y > 12) where Y ~ Poisson(mean = 10).
• Mar 27th 2010, 03:31 PM
Random Variable
I'm fairly certain that those aren't equivalent statements.

Let Y be the number of people who arrive during any two hour period.

Then Y follows a Poisson distribution with a mean of 10.

and $P(Y>12) = 1-P(Y \le 12) = 1 - \sum_{j=0}^{12} \frac{e^{-10} 10^{j}}{j!}$