Lecture 7S

# Lecture 7S - Statistics 371-001 Spring 2009 Lecture#7 10...

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Statistics 371-001, Spring 2009 Lecture #7 10 Feb 09 From Lecture 6: Random binomial variable. n = 4, p = .5. P (Y = 4) = (0.5) 4 = .0625 or 1 in 16 OK to do this with small or simple numbers but we also have a formula: For a binomial random variable Y, Pr { j success in n trials} = Pr {Y = j } = n C j p j (1 – p ) n – j Ex., 5 children, 2 girls, P (Y = 2) 1. n C j is the binomial coefficient, 5 C 2 We can use it to calculate probabilities. Number of ways j successes and n - j failures can be arranged. Number of ways j items can be arranged in n trials. ! = factorial notation. “n-factorial.” 0! = 1 by definition. 5 children, 2 girls, P (Y = 2), 5 C 2 Pr { j success in n trials} = Pr {Y = j } = n C j p j (1 – p ) n – j = 5 C 2 (.5) 2 (.5) 3 = 10 x .25 x .125 = .3125 2. Enumerate: 5 children, 2 girls, P (Y = 2 ). GGBBB GBGBB GBBGB GBBBG BGGBB BGBGB BGBBG BBGGB BBGBG BBBGG Number of possible outcomes for 5 children: 2 5 = 32 Enumerate these for practice.

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Lecturer 7 page 2 girls in 5 children} = = = .3125 3. Use Table 2, page 674. Pr{Y = 2, n = 5} 1+5+10+10+5+1 = 32 = possible outcomes with 5 children. 10 = possible outcomes for Y = 2. Pr(Y = 2, n = 5) = 10 / 32 = .3125.
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## This note was uploaded on 04/07/2009 for the course STAT 371 taught by Professor Koscik during the Spring '08 term at University of Wisconsin.

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Lecture 7S - Statistics 371-001 Spring 2009 Lecture#7 10...

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