a.) …he makes his 1st homerun on the 5th at bat. WARM – UP EXAMPLE: Let X be the number of ‘at bats’ needed for a college baseball player to make his first homerun. The player has probability 0.15 of making a random homerun. Find the probability that… a.) …he makes his 1st homerun on the 5th at bat. b.) …it takes 4 or fewer at bats to make a homerun. c.) …at least 3 at bats are required to make a homerun. d.) …the player will hit 3 homeruns in a row before he misses one? e.) What is the expected # of at bats for making 1st HR? P(x = 5) = (0.854 ∙ 0.15) = 0.0783 P(x ≤ 4) = geometcdf(0.15, 4) = 0.4780 P(x ≥ 3) = 1 – P(x ≤ 2) = 1 – geometcdf(0.15, 2) = 0.7225 (0.15)3·(0.85) = 0.0029 1/0.15 = 6.667

Ch. 17 - The Binomial Distribution The Bernoulli Trials : 1. Each observation must fall into one of just Two Categories (“Success” or “Failure”). 2. All observation must be Independent. 3. The probability of success, pi, must be the same for each observation. 4B. There must be a Fixed Number, n, of observations. EXAMPLE: What is the probability that 3 out of 8 students will pass the AP exam?

B(n,p) = Binomial Distribution with parameters, n and p The random variable, X, has possible outcomes which are whole numbers from 0 to n. B(n,p) = Binomial Distribution with parameters, n and p EXAMPLE: A couple plans to have only 4 children. Let X = the number of boys. Each child has a prob. 0.5 of being a boy. a.) Is this a Binomial Setting? What are the parameters? b.) List the Binomial Distribution. YES! B(4, 0.5) BBBB BBBG BBGB BBGG BGBB BGBG BGGB BGGG GBBB GBBG GBGB GBGG GGBB GGBG GGGB GGGG X 1 2 3 4 P(X=k) 0.0625 0.25 0.375 0.25 0.0625

Computing Binomial Probabilities Combinations EXAMPLE: A couple plans to have only 4 children. Let X = the number of boys. Each child has a prob. 0.5 of being a boy. a.) P(x = 2) = b.) P(x = 3) = BBBB BBBG BBGB BBGG BGBB BGBG BGGB BGGG GBBB GBBG GBGB GBGG GGBB GGBG GGGB GGGG

P(x ≤ k) = P(x = 0) + P(x = 1) + … + P(x = k) The Probability Distribution Function: (The Probability of EXACTLY k successes) P(x = k) = binompdf (n, p, k) EXAMPLE: Let X = the number of boys out of 4 children. P(x = 2) = binompdf (4, .5, 2) = 0.375 The Cumulative Distribution Function: (At Most) P(x ≤ k) = P(x = 0) + P(x = 1) + … + P(x = k) P(x ≤ k) = binomcdf (n, p, k) At Least k successes out of n: P(x ≥ k) = 1 – P(x ≤ k-1) 0 1 2 3 4 5 6 7 8 9… k-1 k k+1 … n

a.) All students pass the exam. b.) Exactly 10 pass the exam. EXAMPLE: Let X be the number of students that pass the AP Statistics Exam out of a class of 20. The probability that a random student will pass is 0.62. Find the probability that:: a.) All students pass the exam. b.) Exactly 10 pass the exam. c.) Six or fewer pass the exam. d.) At most 15 pass the exam. e.) At least 12 pass the exam. P(x = 20) = binompdf(20, 0.62, 20) = 0.00007 P(x = 10) = binompdf(20, 0.62, 10) = 0.0974 P(x ≤ 6) = binomcdf(20, 0.62, 6) = 0.0037 P(x ≤ 15) = binomcdf(20, 0.62, 15) = 0.9274 P(x ≥ 12) = 1 – P(x ≤ 11) = 1 – binomcdf(20, 0.62, 11) = 0.6659

Example #1 A college basketball player has an 82% probability of making Free throws. During a typical game he usually attempts 12 free throws. Find: 1. The prob. of making exactly 5 baskets. 2. The prob. of making at most 5 baskets. 3. The prob. of making at least 9 baskets. 4. The prob. of making all 12. P(x = 5) = binompdf(12, 0.82, 5) = 0.0018 P(x ≤ 5) = binomcdf(12, 0.82, 5) = 0.0021 P(x ≥ 9) = 1 – P(x ≤ 8) = 1 – binomcdf(12, 0.82, 8) = 0.8448 P(x = 12) = binompdf(12, 0.82, 12) = 0.0924

HW: Page 399: 19-20

Binomial Mean and Standard Deviation A Binomial Distribution, B(n, p) has: Mean = μ = np Standard Deviation = σ = √np(1 – p) EXAMPLE: Let X be the number of students that pass the AP Statistics Exam out of a class of 20. The probability that a random student will pass is 0.62. Find: 1. The mean number of students expected to pass. The Standard Deviation. μ = np = 20 · 0.62 μ = 12.4 σ = √np(1 – p) = √ 20(0.62)(1 – 0.62) σ = 2.171

Example #2 There is a 22% chance of making a “5” on the AP Statistics Exam. In a class of 30, what is the probability that… …half of the class will get 5’s? …a third of the class get 5’s? …at least 6 students get 5’s? P(x = 15) = binompdf(30, 0.22, 15) = 0.000511 P(x = 10) = binompdf(30, 0.22, 10) = 0.0554 P(x ≥ 6) = 1 – P(x ≤ 5) = 1 – binomcdf(30, 0.22, 5) = 0.6739