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Chapter 6: Binomial Probability Distributions

Chapter 6: Binomial Probability Distributions. In Chapter 6:. 6.1 Binomial Random Variables 6.2 Calculating Binomial Probabilities 6.3 Cumulative Probabilities 6.4 Probability Calculators 6.5 Expected Value and Variance of Binomial Random Variables

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Chapter 6: Binomial Probability Distributions

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  1. Chapter 6: Binomial Probability Distributions

  2. In Chapter 6: 6.1 Binomial Random Variables 6.2 Calculating Binomial Probabilities 6.3 Cumulative Probabilities 6.4 Probability Calculators 6.5 Expected Value and Variance of Binomial Random Variables 6.6 Using the Binomial Distribution to Help Make Judgments

  3. §6.1 Binomial Random Variables • Binomial = a family of discrete random variables • Binomial random variable ≡ the random number of successes in n independent Bernoulli trials (a Bernoulli trials has two possible outcomes: “success” or “failure” • Binomials random variables have two parametersn  number of trialsp  probability of success of each trial

  4. Binomial Example • Consider the random number of successful treatments when treating four patients • Suppose the probability of success in each instance is 75% • The random number of successes can vary from 0 to 4 • The random number of successes is a binomial with parameters n = 4 and p = 0.75 • Notation: Let X ~b(n,p) represent a binomial random variable with parameters n and p. The illustration variable is X ~ b(4, .75)

  5. §6.2 Calculating Binomial Probabilities Formula for binomial probabilities: where nCx≡ the binomial coefficient (next slide) p≡ probability of success for each trial q≡ probability of failure = 1 – p

  6. Binomial Coefficient Formula for the binomial coefficient: where ! represents the factorial function, calculated: x!= x  (x – 1)  (x – 2)  …  1 For example, 4! = 4  3  2  1 = 24 By definition 1! = 1 and 0! = 1 For example:

  7. Binomial Coefficient The binomial coefficient is called the “choose function” because it tells you the number of ways you could choose x items out of n nCx the number of ways to choose x items out of n For example, 4C2 = 6 means there are six ways to choose two items out of four

  8. Binomial Calculation – Example Recallthe “Four patients example”. Four patients; probability of success of each treatment = .75. The number of success is the binomial random variable X ~ b(4,.75). Note q = 1 −.75 = .25. What is the probability of observing 0 successes under these circumstances?

  9. X~b(4,0.75), continued Pr(X = 1) = 4C1· 0.751 · 0.254–1 = 4 · 0.75 · 0.0156 = 0.0469 Pr(X = 2) = 4C2· 0.752 · 0.254–2 = 6 · 0.5625 · 0.0625 = 0.2106

  10. X~b(4, 0.75) continued Pr(X = 3) = 4C3· 0.753 · 0.254–3 = 4 · 0.4219 · 0.25 = 0.4219 Pr(X = 4) = 4C4· 0.754 · 0.254–4 = 1 · 0.3164 · 1 = 0.3164

  11. pmf for X~b(4, 0.75)Tabular and graphical forms

  12. Pr(X = 2) =.2109 × 1.0 Area Under The Curve Recall the area under the curve (AUC) concept. AUC = probability!

  13. §6.3: Cumulative Probability Pr(X  2) on X ~b(4,.75) • Recall the cumulative probability concept • Cumulative probability ≡ the probability of that value or less • Pr(X  x) • Corresponds to left tail of pmf

  14. Pr(X  1) = Pr(X =0) + Pr(X = 1) = .0039 + .0469 = 0.0508 Pr(X  2) = Pr(X =0) + Pr(X = 1) + Pr(X = 2) Pr(X  4) = Pr(X =0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3) Pr(X  4) = Pr(X =0) + Pr(X = 1) + … + Pr(X = 4) Cumulative Probability Function • Cumulative probability function lists cumulative probabilities for all possible outcome • Example: The cumulative probability function for X~b(4, 0.75) Pr(X  0) = 0.0039 Pr(X  1) = 0.0508 Pr(X  2) = 0.2617 Pr(X  3) = 0.6836 Pr(X  4) = 1.0000

  15. §6.5: Expected Value and Variance for Binomials • The expected value (mean) μof a binomial pmf is its “balancing point” • The variance σ2is its spread • Shortcut formulas:

  16. Expected Value and Variance, Binomials, Illustration For the “Four patients” pmf of X~b(4,.75) μ = n∙p = (4)(.75) = 3 σ2 = n∙p∙q = (4)(.75)(.25) = 0.75

  17. §6.6 Using the Binomial • Suppose we observe 2 successes in the “Four patients” example • Note μ = 3, suggesting we should see 3 success on average • Does the observation of 2 successes cast doubt on p = 0.75? • No, because Pr(X  2) = 0.2617 is not too unusual

  18. StaTable Probability Calculator • Calculates probabilitiesfor many types of random variables • This figure shows probabilities for X~b(4,0.75) • Available in Java, Windows, and Palm versions (download from website) x = 2 p = .75 n = 4 Pr(X = 2) = .2109 Pr(X≤ 2) = .2617

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