LaneHW 6Engr. 323Problem 3-79Page 1 of 5

Poisson Distribution

** An rv is said to have a Poisson distribution if the pmf of X is

p(x; λ) = e-λλx , x = 0,1,2,… for some λ >0 x!

0 otherwise

E(X) = V(X) = λ s.d. σx = λ (pg. 128)

Let us review the criteria and basic assumptions for utilizing the Poissondistribution:

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 x values. (pg. 128)

Now we can answer the questions!

3-79). Suppose small aircraft arrive at a certain airport according to aPoisson process with rate α = 8 per hour, so that the number of arrivalsduring a time period of t hours is a Poisson rv with parameter λ = 8t.

a). What is the probability that exactly 5 small aircraft arrive during a1-hour period? At least 5? At least 10?

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

LaneHW 6Engr. 323Problem 3-79Page 2 of 5

To find the probability of exactly 5 small aircraft arriving given that λ = 8 we canuse the definition of a Poisson distribution.

P(X = 5) = e-8(8)5 = 0.091 5!

4

P (X ≥ 5) = 1 – P(X ≤ 4) = 1 - ∑ e-88x

x=0 x = 1-F(5) **F(5) can be found on Figure 2 (Poisson CDF).

= 1 - e-8(80/0! + 81/1! + 82/2! + 83/3! + 84/4! 85/5!)

= 1 – 0.1 = 0.900

**There is a very useful tool in our book to make these large summations really

fast and painless! In the appendix tables there are several cumulative properties.

On page 703, Table A.2 displays the cumulative Poisson probabilities. To use this

table we need to know λ and x. Let’s use the table to solve this next probability. 9

P(X ≥ 10) = 1 – P(X ≤ 9) = ∑ e-88x = 1-F(9) **(Figure 2). x=0 x

In this case, λ = 8 and x = 9. We read 9 down and 8 across to find 0.717 = 1- 0.717 = 0.283

Figure 1- Poisson PMF with λ = 8

0

0.05

0.1

0.15

0 1 2 3 4 5 6 7 8 9 10x

LaneHW 6Engr. 323Problem 3-79Page 3 of 5

Figure 2- Poisson CDF with λ = 8

b). What are the expected value and standard deviation of the number ofsmall aircraft that arrive during a 90-minute period?

** When using Poisson distribution,

E(X) = V(X) = λ σ = λ (pg. 130)

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 expected over 1.5 hrs.

The standard deviation (σ) is the square root of the expected value

Thus, σ = λ

σ = 12 = 3.464 arrivals over 1.5 hours.

00 . 10 . 20 . 30 . 40 . 50 . 60 . 70 . 80 . 9

0 1 2 3 4 5 6 7 8 9 1 0x

LaneHW 6Engr. 323Problem 3-79Page 4 of 5

c). What is the probability that at least 20 small aircraft arrive duringa 2-½ hour period? That at most 10 arrive during this period?

Given that t = 2.5 we can solve for λ.

λ = 8tλ = (8)(2.5) = 20

Now that we know λ we can solve the probabilities. ∞

P (X ≥ 20) = ∑ e-λλx

X=20 x!

**Although it would be possible to solve the problem this way, there is amuch easier way to approach it. 19

P (X ≥ 20) = 1 – P (X ≤ 19) = 1 – ∑ e-λλx = 1-F(19) x=0 x!

= 1 - e-20(200/0!+201/1!+202/2!+… +2019/19! )

= 1 – 0.470 (from table) = 0.53

10

P (X ≤ 10) = ∑ e-2020x = 0.011 (from table) x=0 x!

Figure 3- Poisson PMF for λ = 20

0

0.002

0.004

0.006

0.008

0 1 2 3 4 5 6 7 8 9 10x

LaneHW 6Engr. 323Problem 3-79Page 5 of 5

Figure 4- Poisson CDF for λ = 20

C D F fo r 3 - 7 9 P a rt c

00 . 0 0 5

0 .010 . 0 1 5

0 2 4 6 8 1 0

x

F(x)

LaneHW 6Engr. 323Problem 3-79Page 1 of 2

3-85). Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter α, the expected number of treesper acre, equal to 80.

a). What is the probability that in a certain quarter-acre plot, therewill be at most 16 trees?

T(a) = No. of trees in an acre

Since the expected number per acre is 80 trees, the expected number per quarter-acre is 20 trees.

P(a)~Poisson(x = αa)

The parameter = α = (80 trees per acre) (1/4 acre) = 20 trees

Now that we know the parameter, we can solve for the P(X ≤ 16).

16

P(T(a=0.25) ≤ 16) = ∑ e-2020x = F(16) = 0.221 (from table) x=0 x!

b). If the forest covers 85,000 acres, what is the expected number oftrees in the forest?

If one acre has an expected value of 80 trees, then the entire forest of85,000 acres has an expected value of 6,800,000 trees.

E(entire forest)= E(T(a=85,000)) = (α)(total acreage) = (80)(85,000)

=6,800,000 trees

LaneHW 6Engr. 323Problem 3-79Page 2 of 2

c). Suppose you select a point in the forest and construct a circle ofradius 0.1 mile. Let X = the number of trees within that circularregion. What is the pmf of X? (hint: 1 square mile = 640 acres.)

To solve this problem we need to pull out a few tricks. First thing to recall is thatthe area of a circle is πr2.

Area = πr2 = (π)(0.1mile)2 = .0314 miles2

The units of miles are not useful to us because all the previous information hasbeen in acres. Now we can use the hint that was given to convert the miles toacres.

.0314 miles2 × 640 acres = 20.1 acres1 mile2

** So, now we can finally show the pmf of T.

T has a Poisson distribution with parameter 20.1 OR p ( T ; 20.1)

Figure 5- Poisson Distribution with parameter 20.1

00.010.020.030.040.050.060.070.080.09

0.1

0 20 40 60 80 100T = Number of Trees per 0.1 Mile Radius