Poisson Distribution

Transcription

1 Page 1 of 5 Poisson Distribution ** An rv is said to have a Poisson distribution if the pmf of X is p(; l) = e -l l, =,1,2, for some l >! otherwise E(X) = V(X) = l s.d. s = λ (pg. 128) Let us review the criteria and basic assumptions for utilizing the Poisson distribution: 1. The value of λ is frequently a rate per unit time or area. (pg. 128) 2. λ must be greater than zero for all possible values. (pg. 128) Now we can answer the questions! 3-79). Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter λ = 8t. a). What is the probability that eactly 5 small aircraft arrive during a 1-hour period? At least 5? At least 1? X = the number of small aircraft that arrive during time t Given t = 1, we can solve for λ: λ = 8t = (8)(1) = 8 small aircraft

2 Page 2 of 5 To find the probability of eactly 5 small aircraft arriving given that λ = 8 we can use the definition of a Poisson distribution. P(X = 5) = e -8 (8) 5 =.91 5! 4 P (X 5) = 1 P(X 4) = 1 - e -8 8 = = 1-F(5) **F(5) can be found on Figure 2 (Poisson CDF). = 1 - e -8 (8 /! /1! /2! /3! /4! 8 5 /5!) = 1.1 =.9 **There is a very useful tool in our book to make these large summations really fast and painless! In the appendi tables there are several cumulative properties. On page 73, Table A.2 displays the cumulative Poisson probabilities. To use this table we need to know λ and. Let s use the table to solve this net probability. 9 P(X 1) = 1 P(X 9) = e -8 8 = 1-F(9) **(Figure 2). = In this case, λ = 8 and = 9. We read 9 down and 8 across to find.717 = = Figure 1- Poisson PMF with l = 8

3 Page 3 of Figure 2- Poisson CDF with l = 8 b). What are the epected value and standard deviation of the number of small aircraft that arrive during a 9-minute period? ** When using Poisson distribution, E(X) = V(X) = l s = λ (pg. 13) We are given that t = 1.5 hours so now we can solve for λ. λ = 8t = E(X) λ = (8)(1.5) = 12 arrivals are epected over 1.5 hrs. The standard deviation (σ) is the square root of the epected value Thus, σ = λ σ = 12 = arrivals over 1.5 hours.

4 Page 4 of 5 c). What is the probability that at least 2 small aircraft arrive during a 2-½ hour period? That at most 1 arrive during this period? Given that t = 2.5 we can solve for λ. λ = 8t λ = (8)(2.5) = 2 Now that we know λ we can solve the probabilities. P (X 2) = e -λ λ X=2! **Although it would be possible to solve the problem this way, there is a much easier way to approach it. 19 P (X 2) = 1 P (X 19) = 1 e -λ λ = 1-F(19) =! = 1 - e -2 (2 /!+2 1 /1!+2 2 /2! /19! ) = 1.47 (from table) =.53 1 P (X 1) = e -2 2 =! =.11 (from table) Figure 3- Poisson PMF for l = 2

5 Page 5 of 5 CDF for 3-79 Part c F() Figure 4- Poisson CDF for l = 2

6 Page 1 of ). Suppose that trees are distributed in a forest according to a twodimensional Poisson process with parameter α, the epected number of trees per acre, equal to 8. a). What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? T(a) = No. of trees in an acre Since the epected number per acre is 8 trees, the epected number per quarteracre is 2 trees. P(a)~Poisson( = αa) The parameter = α = (8 trees per acre) (1/4 acre) = 2 trees Now that we know the parameter, we can solve for the P(X 16). 16 P(T(a=.25) 16) = e -2 2 =! = F(16) =.221 (from table) b). If the forest covers 85, acres, what is the epected number of trees in the forest? If one acre has an epected value of 8 trees, then the entire forest of 85, acres has an epected value of 6,8, trees. E(entire forest)= E(T(a=85,)) = (α)(total acreage) = (8)(85,) =6,8, trees

7 Page 2 of 2 c). Suppose you select a point in the forest and construct a circle of radius.1 mile. Let X = the number of trees within that circular region. What is the pmf of X? (hint: 1 square mile = 64 acres.) To solve this problem we need to pull out a few tricks. First thing to recall is that the area of a circle is πr 2. Area = πr 2 = (π)(.1mile) 2 =.314 miles 2 The units of miles are not useful to us because all the previous information has been in acres. Now we can use the hint that was given to convert the miles to acres..314 miles 2 64 acres = 2.1 acres 1 mile 2 ** So, now we can finally show the pmf of T. T has a Poisson distribution with parameter 2.1 OR p ( T ; 2.1) T = Number of Trees per.1 Mile Radius Figure 5- Poisson Distribution with parameter 2.1