1. 1 5.6 Poisson Distribution and the Poisson Process Some experiments result in counting the numbers of particular events occur in given times or on given physical objects, such as : a line segment, an area, a volume, or perhaps a piece of a metal. For example, we may want to count the number of flaws in 100 feet of a wire, number of white cells in a drop of blood. number of misprint on a page of a book. Each count can be looked upon as a random variable associated with an approximate Poisson process. The experiments are called Poisson experiments.

2. 2 Properties of Poisson Process The number of outcomes occurring in one time interval or specified region is independent of the number that occurs in any other disjoint time interval or region of space. The probability that a single outcome will occur during a very short time interval or in a small region is proportional to the length of the time interval or the size of the region and does not depend on the number of outcomes occurring outside this time interval or region.

3. 3 The probability that more than one outcome will occur in such a short time interval or fall in such a small region is negligible. Let X be the number of outcomes occurring during a Poisson experiment, then X is called a Poisson random variable. The probability distribution of X is called the Poisson distribution. The possible values for X are 0, 1, 2, 3, ….

4. 4 Poisson Distribution Let X be a Poisson random variable, representing the number of outcomes occurring in a given time interval or specified region denoted by t. Then p(x; t)= P(X = x) =, x = 0, 1, 2, …., Where is the average number of outcomes per unit time or per unit region. (rate of the occurrence of the outcomes) Table A.2 contains Poisson probability sum P(X  r) = =

5. 5 Example 5.19, page 137 During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4. What is the probability that 6 particles enter the counter in a given millisecond. Solution: = 4, t = 1, t = 4, x = 6. p(6;4)= =P(X  6) – P(X  5) = 0.1042 Example

6. 6 Example 5.20, page 137 Ten is the average number of oil tankers arriving each day at a certain port city. The facilities at the port can handle at most 15 tankers per day. What is the probability that on a given day tankers have to be turned away? Solution: Let X be the number of tankers arriving each day. = 10, t = 1, t = 10. Find P(X > 15). P(X > 15) = 1 – P(X  15) = 1 – P(15; 10) = 1 – = 0.0487.

7. 7 Theorem 5.5 The mean and variance of the Poisson distribution p(x; t) both have value t.  = t, = t Proof: (See Appendix A26)

8. 8 Example The white blood-cell count of a healthy individual can average as low as 6000 per cubic millimeter of blood. To detect a white-cell deficiency, a 0.001 cubic millimeter of blood is taken and the number of white cells X is found. (a) How many white cells are expected in a healthy individual? = 6000, t = 0.001, t = 6.  = t = 6 In a healthy individual, we would expect, on average, to see six white cells per 0.001 cubic millimeter drop of blood.

9. 9 (b) What is the probability of finding at most one white cell? P(X  1) = 0.0174 The chance of observing none or one white cell per 0.001 cubic millimeter drop of blood of a healthy individual is only 0.0174. In other words, in 1000 healthy individuals, only about 17 of them would provide the blood drops that contain none or one white cell. Hence, if either none or one white cell found in a blood drop of an individual, we would think that there is good evidence that this individual is not healthy, or there is a white-cell deficiency in his or her blood, although there is chance for error.

10. 10 Theorem 5.6 Let X be a binomial random variable with probability distribution b(x; n, p). If when n  , p  0 and  = np remains constant, then b(x; n, p)  p(x;  ) as n   That is to say: if,then Proof: (λ>0)

11. 11 Application: if X ~B(n, p) ， when n >10 ， p < 0.1 ， we have Example: The newborns’ chromosome( 染色体 ) abnormality rate is 1% according to the experience. Let X be the number of chromosome abnormality among 100 newborns. Please use the principles of Binominal and Poisson distribution to calculate the probability distribution of random variable X. n=100, X=0,1,2,…,100, p=0.01, λ= np=1.

12. 12 X P(X) BinominalPoisson 012345678012345678 0.3660 0.3697 0.1849 0.0610 0.0149 0.0029 0.0005 0.0001 0.0000 0.3679 0.1839 0.0613 0.0153 0.0031 0.0005 0.0001 0.0000 1.0000 Example

13. 13 Example 2 A computer hardware company manufactures a particular type of microchip. There is a 0.1% chance that any given microchip of this type is defective, independent of the other microchips produced. Determine the probability that there are at least 2 defective microchips in a shipment of 1000. Solution: Let X be the number of defectives in a shipment of 1000. Then X has the distribution b(x; 1000, 0.001).

14. 14 Solution: since n is large, p is small, an approximate model for X is the Poisson model with

15. 15 Example 3 The number of visits per minute to a particular Website providing news and information can be modeled using a Poisson distribution with mean 5. The Website can only handle 20 visits per minute and will crash if this number of visits is exceeded. Determine the probability that the site crashes in the next minute. Solution: Let X be the number of visits to the site in the next minute. Then X has the distribution p(x; 5).

16. 16 Example 4 Let X be the number of some animal’s eggs which are laid randomly, and the probability is given by. Suppose that the probability of every egg’s hatching a little animal is p and each egg can hatch out a little animal independently. If Y stands for the number of the animal’s offspring, what is the cumulative distribution of Y?