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Hypergeometric Binomial Poisson

Finding the Mean , the Variance , and Approximate Answers for Problems involving the Special Discrete Probability Distributions. Hypergeometric Binomial Poisson. Finding the Mean and the Variance for the Special Discrete Probability Distributions. Hypergeometric Binomial Poisson.

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Hypergeometric Binomial Poisson

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  1. Finding the Mean, the Variance,andApproximate Answersfor Problems involving theSpecial Discrete Probability Distributions • Hypergeometric • Binomial • Poisson

  2. Finding the Meanand the Variancefor the Special Discrete Probability Distributions • Hypergeometric • Binomial • Poisson

  3. Mean and Variance Hypergeometric and Binomial

  4. Sampling w/o replacement it is hypergeometric, and the probability distribution is given by: X P(X) 0 1 2 3 4 Consider the example problem with the six white balls and the four black balls from which we select n = 4. Success = White Success = White Sampling with replacement it is binomial, and the probability distribution is given by: X P(X) 0 1 2 3 4

  5. Sampling with replacement it is binomial, and the probability distribution is given by: X P(X) 0 .0256 1 .1536 2 .3456 3 .3456 4 .1296 Sampling w/o replacement it is hypergeometric, and the probability distribution is given by: X P(X) 0 .00476 1 .11429 2 .42857 3 .38095 4 .07143 Consider the example problem with the six white balls and the four black balls from which we select n = 4. Success = White Success = White

  6. Sampling with replacement it is binomial, and the expected value is given by: XP(X) 0  .0256 1  .1536 2  .3456 3  .3456 4  .1296  = 2.4 Sampling w/o replacement it is hypergeometric, and the expected value is given by: X P(X) 0  .00476 1  .11429 2  .42857 3  .38095 4  .07143  = 2.4 In the sample of 4 balls, what is the expected # of white balls?

  7. Sampling with replacement it is binomial, and the variance is given by: (X- )2P(X) (0-2.4)2 .0256 (1-2.4)2 .1536 (2-2.4)2 .3456 (3-2.4)2 .3456 (4-2.4)2 .1296 2 = .96 Sampling w/o replacement it is hypergeometric, and the variance is given by: (X- )2P(X) (0-2.4)2 .00476 (1-2.4)2 .11429 (2-2.4)2 .42857 (3-2.4)2 .38095 (4-2.4)2 .07143 2 = .64 In the sample of 4 balls, what is the variance of the number of white balls in the sample?

  8. It is a lot of work to get the mean and the variance from the definition, but there is a much easier and faster way. Sampling with replacement it is binomial, and the expected value is given by: XP(X) 0  .0256 1  .1536 2  .3456 3  .3456 4  .1296  = np = 4(.6) = 2.4 Sampling w/o replacement it is hypergeometric, and the expected value is given by: X P(X) 0  .00476 1  .11429 2  .42857 3  .38095 4  .07143 Let p = S/N and  = np = 4(6/10) = 2.4

  9. In the sample of 4 balls, what is the variance of the number of white balls in the sample? Sampling w/o replacement it is hypergeometric, and the variance is given by: (X- )2P(X) (0-2.4)2 .00476 (1-2.4)2 .11429 (2-2.4)2 .42857 (3-2.4)2 .38095 (4-2.4)2 .07143 Let p = S/N and 2 = np(1-p)(N-n)/(N-1) 2 = 4(.6)(1-.6)(10-4)/9 = .64 Sampling with replacement it is binomial, and the variance is given by: (X- )2P(X) (0-2.4)2 .0256 (1-2.4)2 .1536 (2-2.4)2 .3456 (3-2.4)2 .3456 (4-2.4)2 .1296 2 = np(1-p) = 4(.6)(1-.6) = .96

  10. Mean and Variance Poisson

  11. For the Poisson problems λ=  = E(X) = mean and λ= σ2= E(X-)2 = variance

  12. Approximationsto Discrete Probability Distributions

  13. Notation condition A B means A may be approximated by B if condition is true

  14. DISCRETE TO DISCRETEAPPROXIMATIONS n  20 and n  0.05N p  0.05 HYPERGEOMETRIC BINOMIAL POISSON

  15. DISCRETE TO CONTINUOUSAPPROXIMATIONS np  5 and n(1-p)  5 BINOMIAL NORMAL Let p = S/N np  5 and n(1-p)  5 HYPERGEOMETRIC NORMAL  > 20 POISSON NORMAL

  16. Approximation Examples

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