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Probability with the Poisson Distribution

Date: 10/16/98 at 08:40:18
From: Anonymous
Subject: Poisson Distribution Probability Questions

I was working on the following two Poisson Distribution Probability
problems. Do I have the correct values for lambda and a?  We are using
Stataquest software, so once we know lambda and a, when can solve the

Problem 1:
You are expected to receive 75 service requests every two days. These 
data were based on a 20-day sample.

I determined lambda to 75/2 = 37.5
a. What is the probability of receiving 25 requests on Thursday? 
   Here, a = 25.  
   ANSWER: = .75 %

b. What is the probability of receiving 90 requests combined for 
   Monday, Wednesday and Friday? So, a = 30, lamda = 112.5. 
   ANSWER: = 9.75%

c. What is the probability of receiving more than 85 requests on two
   consecutive days? So, a = 42.5 and lambda = 75. 
   ANSWER: = 4.44%

Problem 2:
Historically, two orders per hour for custom drape have been received.
Below is a sample taken over the past 2 months.

   Order per hour             0    1    2    3    4    5 or more
   Number of Occurrences     30   25    30   24   10   6

   lambda would be 2 for historically data
   lambda would be 1.82 for sample data

a. What is the probability of receiving 20 orders in an eight hour 
   shift? a = 2.5
   ANSWER: 21.79%

b. What is the probability of receiving no orders in the first three 
   hours of the shift? Here, lambda = 1.82*3 = 5.46, a = 0.  
   ANSWER: .43%

c. How does the sample compare to the expected probability for 
   receiving less the five orders per hour? ANSWER: The sample is less 
   likely to occur then the expected, since the sample lambda is 1.82  
   and the expected lambda is 2.00.


Date: 10/16/98 at 09:09:12
From: Doctor Statman
Subject: Re: Poisson Distribution Probability Questions

Dear Mark,

I know what lambda is in a Poisson context, and I think I understand 
what a is. If a is the value for which you wish to calculate the 
probability, for example, the probability of getting a = 3 calls in 
one day, then I think we are on the same page.

I think problem 1(a) is correct.

For part (b) of problem 1, I think your lambda of 112.5 is correct, 
but I think your a should be 90 (at least if I understand what a is).  
Over three days you would combine the three "daily" lambdas to get 
112.5, but you would also use the three day total of requests, which 
is 90.

For part (c) of problem 1, lambda is clearly 75, but now I am nervous 
about my understanding of a. I think a should be 86, 87, 88, ... all 
the way to infinity, because you want more than 85 requests. As in 
part b, you are using a new Poisson variable based on two days, so you 
are really looking for more than 85 requests, not 42.5.

For problem 2, part (a), I think your lambda should be 16 and your a 
should be 20, because you are looking for 20 in eight hours.

For part (b) I would use lambda of 6, but I can see why you might use
the estimate provided by the data. If I am given the true value of 2 
per hour, and if I have reason to believe it really is correct, I will 
use it. For part (b) I agree that a is zero.

In part (c) I think they want you to compute the probability of getting 
less than 5 (0, 1, 2, 3, or 4) in one hour, and then compare that to 
the observed percentage of times that fewer than 5 orders came in 
in one hour. I think it is more than just comparing the 1.82 to the 2.

I hope this helps. Good luck!

Statistically yours,

- Doctor Statman, The Math Forum   

Date: 10/16/98 at 11:33:51
From: Doctor Anthony
Subject: Re: Poisson Distribution Probability Questions

Problem 1(a):

Lambda for 1 day is 75/2 = 37.5 as you stated, so:

   P(25) = 37.5^25/25! e^(-37.5) = 0.00748

Problem 1(b):

If you are considering 3 days, lambda = 3 x 37.5 = 112.5, so:

   P(90) = 112.5^90/90! e^(-112.5) = 0.003747

Problem 1(c):

For a two-day period, lambda = 75, and so the mean is 75, 
variance = 75, and s.d. = 8.66.

Since we are looking for at least 85 requests, we need to find the 
probability that there are 86 requests, 87 requests, 88 requests, and 
so on, and add them up. Alternatively, since we know the probabilities 
must sum to 1, we could find the probabilities of 0 requests up to the 
probability of 85 requests, sum them, and subtract that result from 1. 
However, that would be a lot of extraneous work. Since the number of 
requests is large, we can use the Normal approximation to the Poisson:

       85.5 - 75   10.5
   z = --------- = ---- = 1.2125   and   A(z) = 0.8873
         8.66      8.66 

Thus, the tail area = 1 - 0.8873 =  0.1127. So the probability of 86 or 
more requests is 0.1127.

Problem 2(a):

If we accept your mean from the data of 1.82 per hour, then for an 
8-hour shift the mean will be 14.56, so you must find:

   P(20) = 14.56^20/20! e^(-14.56)  =  0.03579

If you decide to use the historical value for the mean, your lambda 
would be 2 * 8 = 16.

Problem 2(b):

The mean = 3 x 1.82 = 5.46, so:

   P(0) =  e^(-5.46)  =  0.004253

Again, if you use the historical data, your lambda is 6.

Problem 2(c):

If we take the mean as 2 per hour before taking the sample, then we 

  P(0) + P(1) + P(2) + P(3) + P(4)

  P(0) =     e^(-2)   =  0.135335
  P(1) = 2/1 x P(0)   =  0.270670
  P(2) = 2/2 x P(1)   =  0.270670
  P(3) = 2/3 x P(2)   =  0.180447
  P(4) = 2/4 x P(3)   =  0.090223
                Total =  0.9473455

Whereas the sample gave 100% less than 5 per hour.
- Doctor Anthony, The Math Forum   
Associated Topics:
High School Probability

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