Poisson Distribution (Poisson Random Variable)

Transcription

1 Poisson Distribution (Poisson Random Variable) Practical applications for Poisson random variables include 1. Number of phone calls per hour (day or week, etc.) received at an exchange or a call center 2. Number of cars arriving at a tunnel (or intersection or toll booth) per hour (or day, etc.) 3. Number of flaws per unit length in a cable or optical fiber 4. Number of visitors per unit time at a Webserver 5. Number of potholes per mile on California roads 6. Number of earthquakes per year (or decade, etc.) and so on. The Poisson distribution deals with number of random events within an interval (e.g., time interval, length interval, area interval, etc.). In fact, the Poisson process is a counting process in that it expresses the number random events that occur within an interval. An event arrival (think of it as a subinterval) is assumed to occur in a time much, much less than the interval of interest over which events are counted. Therefore, events do not overlap with each other. This allows us to set the following conditions upon a Poisson process: 1. The probability of an event occurring within the interval is taken to be constant proportional to the length of the interval. 2. Intervals are independent with respect to the probability of an event occurring. In other words, the probability of an event occurring in one segment is the same for all other segments. 1 P a g e

2 Introduction of the Poisson Distribution Let the random variable X denote the number of occurrences within the entire interval. Parameter is the mean (or average) number of occurrences over the interval. The Poisson distribution is given by k e P( X = k) =, for k = 1,2,3, k! EXAMPLE: The average number of potholes in Santa Rosa is 1.2 potholes per 1000 feet. You drive for two miles on an errand what is the probability of encountering 10 potholes in your two-mile drive in Santa Rosa? Note: One mile is 5,280 feet, therefore, a two-mile drive is 10,560 feet in length. To obtain the parameter we compute it by 1.2 flaws 10,560 feet = = potholes/mile 1,000 feet 2 miles e (6.3360) P(10 potholes in 2 miles) = = ! The mean value of potholes/mile = = potholes/mile Next, what is the probability of encountering 3 or more potholes in one mile of travel? 2 P a g e

3 Average number of potholes in one mile = = potholes/mile. Let X = the number of potholes encountered. P( X 3) = 1 P( X = 0) + P( X = 1) + P( X = 2) Poisson Probability e (6.336) e (6.336) e (6.336) = ! 1! 2! = = 1 (0.0486) = The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be sold tomorrow? Solution: This is a Poisson experiment in which we know the following: = 2; since 2 homes are sold per day, on average. k = 3; since we want to find the likelihood that 3 homes will be sold tomorrow. e = We plug these values into the Poisson formula as follows: k 2 3 e e 2 ( ) 8 P( X = k) = P( k, ) = = = = k! 3! 6 Thus, the probability of selling 3 homes tomorrow is P a g e

4 Cumulative Poisson Probability A cumulative Poisson probability refers to the probability that the Poisson random variable is greater than some specified lower limit and less than some specified upper limit. Cumulative Poisson Example Suppose the average number of lions seen on a 1-day safari is 5. What is the probability that tourists will see fewer than four lions on the next 1-day safari? Solution: This is a Poisson experiment in which we know the following: = 5; since 5 lions are seen per safari, on average. k = 0, 1, 2, or 3; since we want to find the likelihood that tourists will see fewer than 4 lions; that is, we want the probability that they will see 0, 1, 2, or 3 lions. e = To solve this problem, we need to find the probability that tourists will see 0, 1, 2, or 3 lions. Thus, we need to calculate the sum of four probabilities: P(0, 5) + P(1, 5) + P(2, 5) + P(3, 5). To compute this sum, we use the Poisson formula: P( X 3,5) = P(0,5) + P(1,5) + P(2,5) + P(3,5) e (5) e (5) e (5) e (5) PX ( 3,5) = ! 1! 2! 3! PX ( 3,5) = = Thus, the probability of seeing at no more than 3 lions is These numbers can be taken from the below table. 4 P a g e

5 Poisson table of probabilities: x e P( X = x) =, for x = 1,2,3, x! 5 P a g e

6 6 P a g e

7 We observe that the distributions are (i) unimodal (ii) exhibit positive skew (that decreases a λ increases) (iii) centered roughly on λ (iv) the variance (spread) increases as λ increases Examples of Poisson distributions: Poisson probability distributions as function of k (=n) for several values of = (= 1, 2, 4, 8). 7 P a g e