chap5S2 - Chapter5.4 Tostart, X B x=1 success x=0 failure X...

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Chapter 5.4 Binomial Distribution A special application of a discrete probability distribution is the binomial distribution. To start, introduce the random variable that takes two outcomes: B X x = 1 “success” x = 0 “failure” The probability distribution function of is: B X for 0 < p < 1 (the probability of success) p ) X ( P B = = 1 p 1 ) X ( P B = = 0 This is known as the Bernoulli distribution. The mean and variance are calculated as: p p p ) x ( P x ) X ( E x B X B = + = = = μ ) (1 0 1 ) p 1 ( p p p p p p p ) x ( P x ) X ( E ) X ( Var 2 2 2 x 2 2 X 2 B B B = = + = = μ = ) (1 (0)(0) (1)(1) Chapter 5 17
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Now consider that a random experiment with the outcome of success or failure is repeated n times. Each trial produces success or failure with probabilities p and (1 p ) respectively. Assume independence so that the result of one trial does not influence the result of any other trial. Let the random variable X be the number of successes in n trials. The probability distribution function of X is defined as: ) ( P ) x ( P trials t independen n in successes x = for x = 0, 1, 2, . . . , n This is known as the binomial distribution . Chapter 5 18
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A calculation formula for the probabilities can be obtained as follows. In n independent trials, the probability of x successes and ( n x ) failures is: x n x ) p 1 ( p The number of combinations of x successes in n trials is: ! ) x n ( ! x ! n C n x = Therefore, the probability distribution function for the binomial distribution is: x n x ) p 1 ( p ! ) x n ( ! x ! n ) x ( P = for x = 0, 1, 2, . . . , n Chapter 5 19
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Note: a calculation rule is 1 ! 0 = Examples of application of the binomial distribution are: n ) p 1 ( ) ( P ) 0 X ( P = = = trials n in successes no 1 n ) p 1 ( p n ) ( P ) 1 X ( P = = = trials n in success one The cumulative probability function of the binomial distribution may have useful application and is calculated as: for x = 0, 1, 2, . . . , n ) x X ( P ) x ( F = For example, ) 1 X ( P ) 0 X ( P ) ( P ) 1 X ( P ) 1 ( F = + = = = = trials n in success one most at Chapter 5 20
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Example: Exercise 5.39, page 156. A company installs new heating furnaces. For any installation, the probability of a return visit for a repair is 0.15. Six installations are made in a given week. Assume independence of outcomes for these installations. Let the random variable X be the number of return visits. X follows a binomial distribution with: and 0.15 = p 0.85 = p 1 Find the probability that a return visit will be needed in more than one of the installations. ] [ 1 )] 1 X ( P ) 0 X ( P [ 1 ) 1 X ( P 1 ) 1 X ( P 5 6 0.85) (6)(0.15)( (0.85) + = = + = = = > At this point, it is clear that the numerical calculations can be tedious.
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chap5S2 - Chapter5.4 Tostart, X B x=1 success x=0 failure X...

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