October 15

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

§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 parameters n  number of trials p  probability of success of each trial

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)

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

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: Formula for the binomial coefficient:

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 n C x  the number of ways to choose x items out of n For example, 4 C 2 = 6 means there are six ways to choose two items out of four

Binomial Calculation – Example Recall the “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?

X~b(4,0.75), continued Pr(X = 1) = 4 C 1 · · –1 = 4 · 0.75 · = Pr(X = 2) = 4 C 2 · · –2 = 6 · · =

X~b(4, 0.75) continued Pr(X = 3) = 4 C 3 · · –3 = 4 · · 0.25 = Pr(X = 4) = 4 C 4 · · –4 = 1 · · 1 =

pmf for X~b(4, 0.75) Tabular and graphical forms xPr(X = x)

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

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

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) = Pr(X  1) = Pr(X  2) = Pr(X  3) = Pr(X  4) = Pr(X  1) = Pr(X = 0) + Pr(X = 1) = = 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)

§6.5: Expected Value and Variance for Binomials The expected value (mean) μ of a binomial pmf is its “balancing point” The variance σ 2 is its spread Shortcut formulas:

Expected Value and Variance, Binomials, Illustration For the “Four patients” pmf of X~b(4,.75) μ = n∙p = (4)(.75) = 3 σ 2 = n∙p∙q = (4)(.75)(.25) = 0.75

§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) = is not too unusual

StaTable Probability Calculator Calculates probabilities for 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) Pr(X = 2) =.2109 Pr(X ≤ 2) =.2617 x = 2 p =.75 n = 4