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December 7th, 2017, 03:20 AM  #1 
Newbie Joined: Dec 2017 From: London Posts: 3 Thanks: 0  Poisson Method?
Hi! I have been given this task A restaurant based in a city centre monitors the rate at which customers are arriving and leaving. It finds during a particular lunchtime that customers arrive at a mean rate of 4.5 in 30 minutes, while they leave at a mean rate of 2.2 in 30 minutes. Giving your answers to 2 d.p. calculate the probability that a) exactly 6 customers arrive in one 30minute period. [3] b) at most 3 customers leave in one 30minute period. [2] c) between 2 and 5 customers inclusive leave in one 30minute period. [4] d) at most 12 customers arrive in a 1hour period. [2] e) between 5 and 7 customers inclusive leave in a 1hour period. [4] Can anyone advise me on how to work it out using the Poisson method? And then I think i will be able to handle the rest of it from there? I am just unsure on how to take the first step towards answering it. Thanks for any help it's greatly appreciated! 
December 7th, 2017, 03:06 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
just assume that arrivals and departures are distributed with a Poisson distribution of the appropriate rate. Read up on the distribution and see if you can figure it out. Shout back if not. 
December 11th, 2017, 07:54 AM  #3 
Newbie Joined: Dec 2017 From: London Posts: 3 Thanks: 0  Hi Romsek
So, I have been reading up on the Poisson Distribution. I have worked out that P(x;μ) = (e4.5) (μx) / x! so the answer to the first question is P(x=6) = 0.128 However, how would I work it out if the question asks at most three customers and between 2 and 5 customers? As I am not sure how I would change the calculation if I am working out at most or between numbers? Thanks for all your help! 
December 11th, 2017, 08:35 AM  #4  
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390  Quote:
2 to 5 customers is 2 through 5 customers. You can either use the CDF of the Poisson distribution, $P[k]$ and compute $P[3]$ in the first case and $P[5]P[1]$ in the second. It's probably easiest though to just sum up the individual pmf values, $p[k]$, as the CDF formula is a bit arcane. In the first case you would just have $\sum \limits_{k=0}^3~p[k]$ In the second case $\sum \limits_{k=2}^5~p[k]$  

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mean or averages, method, poisson, probability 
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