1. ## poisson

Customers arrive at photocopying machine at an average rate of 2 each 5 minutes. Assume that this arrivals are independent, with a constant arrival rate and hat thi sproblem follows a poisson distributionn, with X denoting the number of arrivals in a 5 minute period and mean lambda=2. Find the probab that more than 2 customers arrive in a 5 minute period.

This is no problem to solve. I just want to see if I got Poisson well. Like, 5 is my time base, right. So I know, that 2 persons come each 5 minutes. So people are computed on a base(subinterval) of 5 minutes. So if they ask me the prob that more than 2 customers arrive in a 1 minute period I should do

2: 5= x:1 x=2/5=0.4 and my lambda would be 0.4
so P=1 - P(0) - P(1) -P(2)

right?? Thank you so much

2. Originally Posted by 0123
Customers arrive at photocopying machine at an average rate of 2 each 5 minutes. Assume that this arrivals are independent, with a constant arrival rate and hat thi sproblem follows a poisson distributionn, with X denoting the number of arrivals in a 5 minute period and mean lambda=2. Find the probab that more than 2 customers arrive in a 5 minute period.
The probability that $k$ people arrive in that time interval is $P(k) = \frac{2^ke^{-2}}{k!}$. Here $k\geq 2$.
Thus,
$P(k\geq 2) = 1 - P(0\leq k \leq 1) = 1 - e^{-2} - 2e^{-2} \approx .59$

3. Originally Posted by ThePerfectHacker
The probability that $k$ people arrive in that time interval is $P(k) = \frac{2^ke^{-2}}{k!}$. Here $k\geq 2$.
Thus,
$P(k\geq 2) = 1 - P(0\leq k \leq 1) = 1 - e^{-2} - 2e^{-2} \approx .59$
well actually it's not. Because it asks for "more than 2" so you should exclude also P(X=2) and the result is 0. 3233.

But this is clear. I was wondering about the case in which the period to be examined changes.Can you help? Thanks