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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 30-minute period.  b) at most 3 customers leave in one 30-minute period.  c) between 2 and 5 customers inclusive leave in one 30-minute period.  d) at most 12 customers arrive in a 1-hour period.  e) between 5 and 7 customers inclusive leave in a 1-hour period.  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. Thanks from DecoMate 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) = (e-4.5) (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
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Quote:
 Originally Posted by DecoMate So, I have been reading up on the Poisson Distribution. I have worked out that P(x;μ) = (e-4.5) (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!
at most 3 customers is 0 through 3 customers
2 to 5 customers is 2 through 5 customers.

You can either use the CDF of the Poisson distribution, $P[k]$ and compute

$P$ in the first case and $P-P$ 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 $\sum \limits_{k=2}^5~p[k]$ Tags mean or averages, method, poisson, probability Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post tttim Advanced Statistics 0 December 13th, 2015 06:24 AM mathsforumuser Advanced Statistics 1 June 19th, 2014 09:21 AM shin777 Algebra 3 February 3rd, 2014 09:29 AM Dodi Calculus 0 February 10th, 2012 10:13 AM Joolz Advanced Statistics 1 October 24th, 2009 08:43 PM

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