Math/Stat 352 Lecture 8 Sections 4.3 and 4.4 Commonly Used Distributions: Poisson, hypergeometric, geometric, and negative binomial. 1
The Poisson Distribution Poisson random variable counts the number of (rare) events that happen with intensity λ. λ = number of events that happen during a unit time (area) interval. Examples: 1. Number of earthquakes in Reno during 10 years, then λ could be number of quakes in 1 year; 2. Number of hits on a web site in 1 hour, then λ could be number of hits on that web site in 1 min. 2
Poisson R.V.: pmf, mean, and variance If X ~ Poisson(λ), the probability mass function of X is Mean: µ X = λ Variance: 2 σ X λ x e λ, for x = 0, 1, 2,... px ( ) = PX ( = x) = x! 0, otherwise = λ Note: X is a discrete random variable and λ must be a positive constant. Example: Number of hits on a web site in 1 min has a Poisson distribution with parameter λ=100. Find the probability that there are 90 hits on that website in 1 min. Soln: X ~ Pois(100), P(X=90)= e 111 111 99 = 0.0250389 (I used MINITAB) 99! 3
The Poisson Distribution approximation to Binomial One way to think of the Poisson distribution is as an approximation to the binomial distribution when n is large and p is small. It is the case the binomial mass function depends almost entirely on the mean np, and very little on the specific values of n and p. The approximation: X Bin(n, p), Y-Poiss(λ=np) P(X=k)= n k pk (1 p) n k e λ λ k k! =P(Y=k) We can therefore approximate the binomial probabilities with Poisson probabilities, where λ = np. 4
The Poisson approximation to Binomial - Example Suppose X~Bin(5000, 0.0004). Find (a) P(X=0), (b) P(X=3) using Binomial and Poisson probabilities. Solution: (a) Using Binomial probs: P(X=0)= 5000 0. 0000 0 (1 0. 0000) 5555 0 = 0. 111111111. 0 Using Poisson Approximation: λ=(5000)(0.0004)=2. P(X=0) e 2 2 0 0! =0.135335283. (b) P(X=3)= 5000 0. 0000 3 (1 0. 0000) 5555 3 = 0. 111111111. 3 Using Poisson Approximation: λ=(5000)(0.0004)=2. P(X=0) e 2 2 3 3! =0.180447044. 5
The Poisson approximation to Binomial pmf Poisson(1) Binomial(1000,.001] dpois(q, 1) 0.0 0.1 0.2 0.3 dbinom(q, 1000, 1/1000) 0.0 0.1 0.2 0.3 0 2 4 6 8 10 Number of succsses 0 2 4 6 8 10 Number of succsses
Example Particles are suspended in a liquid medium at a concentration of 6 particles per ml. A large volume of the suspension is thoroughly agitated, and then 3 ml are withdrawn. What is the probability that exactly 15 particles are withdrawn? Soln: X=# of particles withdrawn in 3mL, Average # particles in 3mL= (6particles/1mL)(3mL)=18particles X ~Pois(18). P(X=15) = e 11 11 11 11! =0.0786. 7
Example Typographical errors occur very rarely in well edited books. The probability that any given page contains at least one error is 0.005, and errors occur independently from one page to another. What is the probability that a book of 400 pages will contain exactly one page with errors? Soln: X= # of pages with errors in the book, X ~Bin(400, 0.005): n-large, p- small. Use Poisson approximation to the Binomial probability. Mean rate of pages with errors in a 400 page book is: 400(0.005)=2. Take Y ~Pois(2). Then P(X=1) P(Y=1) = e 2 2 1 1! =0.271. 8
Poisson process X(t) = number of events that occur in some time (area) t. As t 0, P( X(t)=1) λt, P(X(t) = 0) 1 - λt, and P(X(t) > 1) 0. For any disjoint time intervals (areas of space), the numbers of events occuring in those intervals are independent Poisson random variables. We call such a process a Poisson process with rate (parameter) λt. Example: Let electric pulses arrive at a counter at the average rate of 6 per minute, according to a Poisson process. Find the probability that at least 1 pulse is received on (a) 0.5 min (b) 2 min? Soln: (a) X=#pulses in 0.5 min, X~Pois(3). P(X=1) = e 3 3 1 (b) X=#pulses in 2 min, X~Pois(12). P(X=1) = e 11 11 1 11! 3! = 0.149361. = 0.0000737. 9
Hypergeometric Distribution Typical experiment: A box contains 10 balls: 4 red and 6 blue. Sample 3 balls out of the box. What is the probability that 1 of them is red? Consider a finite population containing two types of items: S and F. A simple random sample is drawn from the population, that is no replacement. Each item sampled constitutes a Bernoulli trial. However, as each item is selected, the probability of successes in the remaining population decreases or increases, depending on whether the sampled item was a success or a failure. Thus the trials are not independent, so the number of successes in the sample does not follow a binomial distribution. The number of successes in the sample has the hypergeometric distribution. 10
Hypergeometric pmf Assume a finite population contains N items, of which R are classified as successes and N R are classified as failures. Assume that n items are sampled from this population, and let X represent the number of successes in the sample. Then X has a hypergeometric distribution with parameters N, R, and n, which can be denoted X ~ H(N, R, n). The probability mass function of X is R N R x n x, max(0, R+ n N) x min( nr, ) px ( ) = PX ( = x) = N n 0, otherwise Example: For the balls experiment, X= # red balls in the sample of 3 balls has hypergeometric distribution with N=10, R=4, and n=3. 11
Mean and Variance of the Hypergeometric Distribution If X ~ H(N, R, n), then Mean of X: Variance of X: µ = X σ 2 X nr N R R N n = n 1 N N N 1 12
Example Of 50 buildings in an industrial park, 12 have electrical code violations. If 10 buildings are selected at random for inspection, what is the probability that exactly 3 of the 10 have code violations? What are the mean and variance of X? Soln: X=# of buildings in the sample that have code violations. X ~ H(50, 12, 10). EX= nr/n =(10)(12)/50= 2.4; On average between 2 and 3 houses in the sample will have code violations. VarX= σ 2 X R R N n = n 1 N N N 1 = 11 11 55 1 11 55 55 11 55 1 = 1.5 13
Geometric Distribution Typical experiment: Toss a coin until the first H show up. Count the number of tosses necessary to get the first H. Assume that a sequence of independent Bernoulli trials is conducted, each with the same probability of success, p. Let X represent the number of trials up to and including the first success. Then X is a discrete random variable, which is said to have the geometric distribution with parameter p. We write X ~ Geo(p). 14
Geometric R.V.: pmf, mean, and variance If X ~ Geo(p), then The pmf of X is µ = 1 p The mean of X is. 2 1 p The variance of X is σ =. X 2 p p p x px ( ) = PX ( = x) = 0, otherwise X x 1 (1 ), = 1,2,... 15
Example A test of weld strength involves loading welded joints until a fracture occurs. For a certain type of weld, 80% of the fractures occur in the weld itself, while the other 20% occur in the beam. A number of welds are tested. Let X be the number of tests up to and including the first test that results in a beam fracture. 1. What is the distribution of X? 2. Find P(X = 3). 3. What are the mean and variance of X? Soln: 1. Success=fracture in weld; X ~ Geo(0.8). 2. P(X=3)=(0.8)(0.2) 2 = 0.032. 3. EX= 1/0.8= 1.25. On average we need 1 to 2 tests to get fracture in the weld. VarX=(1-0.8)/0.8 2 = 0.3125. 16
Negative Binomial Distribution Typical experiment: Toss a coin until the r th H shows up. Count the number of tosses necessary to get the r th H. The negative binomial distribution is an extension of the geometric distribution. Let r be a positive integer. Assume that independent Bernoulli trials, each with success probability p, are conducted, and let X denote the number of trials up to and including the r th success. Then X has the negative binomial distribution with parameters r and p. We write X ~ NB(r,p). Note: If X ~ NB(r,p), then X = Y 1 + + Y n where Y 1,,Y n are independent random variables, each with Geo(p) distribution. 17
Negative Binomial R.V.:pmf, mean, and variance If X ~ NB(r,p), then The pmf of X is x 1 r x r p(1 p), x= rr, + 1,... px ( ) = PX ( = x) = r 1 0, otherwise r The mean of X is µ = X. p 2 r(1 p) The variance of X is σ = X 2. p 18
Beam fractures example cont. Find the mean and variance of X, where X represents the number of tests up to and including the third beam fracture. Soln: X ~ NB(3, 0.8). EX = r/p= 3/0.8 = 3.75. On average we need to check between 3 and 4 welds to get 3 rd fractured weld. 2 r(1 p) VaxX = σ = = 3(1-0.8)/0.8 2 = 0.9375. X 2 p Multinomial Trials read in the text book, please. 19