371chapter2f2011

371chapter2f2011 - Chapter 2 Bernoulli Trials 2.1 The...

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Unformatted text preview: Chapter 2 Bernoulli Trials 2.1 The Binomial Distribution In Chapter 1 we learned about i.i.d. trials. In this chapter, we study a very important special case of these, namely Bernoulli trials (BT). If each trial has exactly two possible outcomes, then we have BT. Because this is so important, I will be a bit redundant and explicitly present the assumptions of BT. The Assumptions of Bernoulli Trials. There are three: 1. Each trial results in one of two possible outcomes, denoted success ( S ) or failure ( F ). 2. The probability of S remains constant from trial-to-trial and is denoted by p . Write q = 1- p for the constant probability of F . 3. The trials are independent. When we are doing arithmetic, it will be convenient to represent S by the number 1 and F by the number 0. One reason that BT are so important, is that if we have BT, we can calculate probabilities of a great many events. Our Frst tool for calculation, of course, is the multiplication rule that we learned in Chapter 1. or example, suppose that we have n = 5 BT with p = 0 . 70 . The probability that the BT yield four successes followed by a failure is: P ( SSSSF ) = ppppq = (0 . 70) 4 (0 . 30) = 0 . 0720 . Our next tool is extremely powerful and very useful in science. It is the binomial probability distribution . Suppose that we plan to perform/observe n BT. Let X denote the total number of successes in the n trials. The probability distribution of X is given by the following equation. P ( X = x ) = n ! x !( n- x )! p x q n- x , for x = 0 , 1 , . . ., n. (2.1) 19 To use this formula, recall that n ! is read n-factorial and is computed as follows. 1! = 1; 2! = 2(1) = 2; 3! = 3(2)(1) = 6 , 4! = 4(3)(2)(1) = 24; and so on. By special deFnition, 0! = 1 . I will do an extended example to illustrate the use of Equation 2.1. Suppose that n = 5 and p = 0 . 60 . I will obtain the probability distribution for X . P ( X = 0) = 5! 0!5! (0 . 60) (0 . 40) 5 = 1(1)(0 . 0102) = 0 . 0102 . P ( X = 1) = 5! 1!4! (0 . 60) 1 (0 . 40) 4 = 5(0 . 60)(0 . 0256) = 0 . 0768 . P ( X = 2) = 5! 2!3! (0 . 60) 2 (0 . 40) 3 = 10(0 . 36)(0 . 064) = 0 . 2304 . P ( X = 3) = 5! 3!2! (0 . 60) 3 (0 . 40) 2 = 10(0 . 216)(0 . 16) = 0 . 3456 . P ( X = 4) = 5! 4!1! (0 . 60) 4 (0 . 40) 1 = 5(0 . 1296)(0 . 40) = 0 . 2592 . P ( X = 5) = 5! 5!0! (0 . 60) 5 (0 . 40) = 1(0 . 0778)(1) = 0 . 0778 . You should check the above computations to make sure you are comfortable using Equation 2.1. Here are some guidelines for this class. If n 8 , you should be able to evaluate Equation 2.1 by hand as I have done above for n = 5 . or n 9 , I recommend using a statistical software package on a computer or the website I describe later. or example, the probability distribution for X for n = 25 and p = 0 . 50 is presented in Table 2.1....
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371chapter2f2011 - Chapter 2 Bernoulli Trials 2.1 The...

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