Chapter 3.4-3.6

Chapter 3.4-3.6 - Statistics 511: Statistical Methods Dr....

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Unformatted text preview: Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2010 Lecture 6: The Binomial, Hypergeometric, Negative Binomial and Poisson Distributions Devore: Section 3.4-3.6 Sept, 2010 Page 1 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2010 Binomial Experiment An experiment for which the following four conditions are satisfied is called a binomial experiment 1. The experiment consists of a sequence of n trials, where n is fixed in advance of the experiment. 2. The trials are identical, and each trial can result in one of the same two possible outcomes, which are denoted by success (S) or failure (F). 3. The trials are independent 4. The probability of success is constant from trial to trial and is denoted by p. Sept, 2010 Page 2 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2010 Suppose each trial of an experiment can result in S or F, but the sampling is without replacement from a population of size N. If the sample size n is at most 5% of the population size, the experiment can be analyzed as though it were exactly a binomial experiment. Given a binomial experiment consisting of n trials, the binomial random variable X associated with this experiment is defined as X = the number of Ss among n trials Sept, 2010 Page 3 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2010 Because the pmf of a binomial rv X depends on the two parameters n and p, we denote the pmf by b(x;n,p). The binomial pmf is b ( x ; n,p ) = ( n x ) p x (1- p ) n- x x = 0 , 1 , 2 ,...,n otherwise Sept, 2010 Page 4 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2010 Expected value and variance of a binomial RV Let X 1 ,...,X n be mutually independent Bernoulli random variables, each with success probability p. Then, Y = n X i =1 X i is a binomial random variable with pmf b ( x ; n,p ) . The expected value is E Y = E n X i =1 X i ! = n X i =1 E X i = np Sept, 2010 Page 5 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2010 The variance is V ar Y = var n X i =1 X i ! = n X i =1 V ar X i = n X i =1 p (1- p ) = np (1- p ) Sept, 2010 Page 6 Statistics 511: Statistical Methods Dr. Levine Purdue University Fall 2010 Example A card is drawn from a standard 52-card deck. If drawing a club is considered a success, find the probability of 1. exactly one success in 4 draws (with replacement) 2. no successes in 5 draws (with replacement)....
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Chapter 3.4-3.6 - Statistics 511: Statistical Methods Dr....

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