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 May 30th, 2012, 08:47 AM #1 Newbie   Joined: May 2012 Posts: 4 Thanks: 0 Poisson distribution question Hello. I am self studying statistics and have become stuck on a question in the chapter on the Poisson distribution. Here it is: A car hire firm has two cars which it hires out day by day. The number of demands for a car on each day is distributed as a Poisson distribution with a mean of 1.5. And the question is: What proportion of demands has to be refused? Although it seems very simple I'm struggling to understand what the question is asking for. Any help you can give me would be gratefully received! Thanks.
 May 30th, 2012, 02:38 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Re: Poisson distribution question The simplest way is to calculate the probability that the demand will be met, that is the number of requests is 0, 1, or 2, which is e^(-x) (1 + x + x^2/2), where x = 1.5. Once you have this, subtract from 1 to get the answer you want.
 May 30th, 2012, 11:58 PM #3 Newbie   Joined: May 2012 Posts: 4 Thanks: 0 Re: Poisson distribution question Thanks for your reply mathman. The answer I arrived at was 0.1913, however the textbook gives it as 18.6%. I think the question wants you to give the proportion of all possible demands that have to be refused. I am confused as I have learnt that a Poisson model assumes there can be an unlimited number of events (demands) in the given time interval. Am I way off the mark here? The textbook is only an introduction to statistics so I don't believe it can be too difficult.
 May 31st, 2012, 12:20 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Poisson distribution question I found an online Poisson distribution calculator, and the value it returns for this problem is: $P$$X>2$$\approx0.191153169461941$ Observe that: $1-\frac{1+1.5+\frac{1.5^2}{2}}{e^{1.5}}\approx0.1911 5316946194183$
 May 31st, 2012, 03:07 PM #5 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Re: Poisson distribution question My guess as to what is being asked. Let x be the Poisson mean and let Q = e^(-x). The average number of requests = Q?nx^n/n!, where sum from 0 to ?. The average number of turndowns = Q?(n-2)x^n/n!, where sum is from 3 to ?. I did a quick estimate of the ratio and it seems to be about the same as the book answer.
 June 1st, 2012, 12:01 AM #6 Newbie   Joined: May 2012 Posts: 4 Thanks: 0 Re: Poisson distribution question That's great, thank you very much for your help. Could you please show how you derived those sums and also how to evaluate them? It's beyond my level I'm afraid but I'm interested to know. Thanks again for your time.
June 1st, 2012, 02:56 PM   #7
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Re: Poisson distribution question

Quote:
 Originally Posted by MVCastlequeen That's great, thank you very much for your help. Could you please show how you derived those sums and also how to evaluate them? It's beyond my level I'm afraid but I'm interested to know. Thanks again for your time.
The probability that there will be n requests is Qx^n/n!, so multiply by n to get the weighted number of requests. For the turn downs, since 2 can be satisfied, then n-2 is the number of turndowns for n > 2, again weighted by the probability that there will be n requests.

 June 3rd, 2012, 01:28 PM #8 Newbie   Joined: May 2012 Posts: 4 Thanks: 0 Re: Poisson distribution question Thank you.

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