Lesson Objectives

1. Bernoulli and Binomial Distributions Lesson Objectives Chapter 6.2, 7.4 • Learn when to use the Binomial distribution. • Learn how to calculate probabilitiesfor the Binomial using the formulaand the two tables in the book, andin Excel.

2. Does a person have ESP? Experiment: • Two people in different rooms. • “A” is shown one of the five cards, selected randomly. “A” transmits his thoughts. • “B” selects the card she thinks is beingsent to her, and records it. • If the two cards match, a success occurs. P(X = 1) = ___ = p P(X = 0) = ___ = 1- p X = 1  success = 0  failure Bernoulli (p = ____ ) X ~

3. Bernoulli Distribution A discrete data distribution used to describe a population of binary variable values. Binary  one of only two outcomescan occur, coded as “0” or “1”

4. k 0 1 P(X=k) Bernoulli distribution,  “one” trial. 1 ­ pp Bernoulli has one parameter:p = the probability of success. Bernoulli (p) X ~ m = The mean of of the Bernoulli is s = The standard deviation is

5. k 0 1 P(X=k) 1 ­ pp  s = p(1 - p) Bernoulli distribution,  “one” trial. How are the “mean” and “standard deviation” determined? Use the same equations as the previous section. m = p 0•(1-p) + 1•p = a little algebra . . . s2 = (0 - p)2 • (1 - p) + (1 - p)2 • p = p2 • (1 - p) + (1 - p)2 • p = p(1 - p) • [ p + (1 - p) ] = p(1 - p) Once we know these results, we don’t need to derive it again.

6. Sex (male or female) Major (business or not business) Defective? (defective or non-defective) Response to a T-F question (true or false) Where student lives (on-campus or off-campus) Credit application result (accept or deny) Own home? (own or rent) Course result (Pass or Fail) Examples of Bernoulli Variables

7. Bar Chart of Population for ESP 1.0 1–p p 0 X 0 1

8. Binomial Distribution A discrete data distribution used to model a population of counts for “n” independent repetitions of a Bernoulli experiment. • Conditions: • a fixed number of trials, “n”. • all n trials must be independent of each other. • the same probability of success on each trial. • X = count of the number of successes.

9. Binomial distribution,  “n trials” k 0 1 2 3  n P(X = k) to be determined Binomial has two parameters:n = the fixed number of trials, p = the probability of success for each trial. X ~ Bino( n, p) m = The mean of of the Binomial is s = The standard deviation is

10. Formula gives P(X = k), the probability for exactly one value. Tables Table A.1 gives P(X = k),the probability for exactly one value. or Table A.2 gives P(X < k),the cumulative probability forX = 0 through X = k. Computing Binomial probabilities For selected values of n and p. • Excel (BINOMDIST function),gives either individual or cumulative.

11. A count of the number of females in a sampleof 80 fans at a Rolling Stones concert. A count of the number of defectives in sampleof 50 tires coming off a production line. A count of the number of the number of corrects answers on 10 true-false questionsfor which everybody guessed. A count of the number of credit applications that are denied from a sample of 200. Examples of Binomial Variables

12. Are these situation Binomial? • A count of the number of fans at a Stones concert needed until we find 50 females. • A count of the number of defectives tires coming off a production line in one year. • Count of people choosing Dr Pepper over Pepsi in a blind taste-test with 20 people. • A count of the number of playing cards that are diamonds in a sample of 13 cards. • A count of the number of credit applications that are accepted from a sample of 200.

13. If the population of X-values has a binomial data distribution, then the proportion of the population having the value k is given by: n k ( ) pk (1 – p)n–k P(X = k ) = for k = 0, 1, 2, ..., n This is also the probability that a single value of X will be exactly equal to x.

14. Little “k” is a “specific value”of Big X. Big “X” is the “random variable.” Example:P( X = k)P( Count = 4) n k ( ) pk (1 – p)n–k P(X = k ) =

15. n k () pk (1 – p)n–k P(X = k ) = ( ) n k = n! = 0! =

16. These are the possible values of X;each value has its own probability. ( ) n x px (1 – p)n–x P(X = x ) = x = 0, 1, 2, ..., n

17. Does a person have ESP? Experiment: Repeat the experiment 5 times. X = a count of the number of successes.X ~ Bino(n=5, p =.20) Find the probability of exactlyone success. 1 4 5 1 .20 .80 P(X=1) = P(X=3) =  = 5 • .20 • .4096 = .4096

18. 0 5 1 4 2 3 5 0 5 1 5 2 + + = .20 .80 .20 .80 .20 .80    P(X < 2) = ESP Experiment Bino(n=5, p =.20) Find the probability of two or fewer successes. P(X = 0) + P(X = 1) + P(X = 2) = 1• 1•.32768 + 5•.20•.4096 + 10•.04•.5120 = .32768 + .4096 + .2048 = .94208

19. Table A.2 gives P( X < k ) p 0.2 p 0.2 n = 5 n = 5 k: k: .3277 .4096 .2048 .0512 .0064 .0003 0 1 2 3 4 5 0 1 2 3 4 5 P( X < 2 ) = ESP Experiment Bino(n=5, p =.20) Cumulative Individual Table A.1 gives P( X = k ) .3277 .7373 .9421 .9933 .9997 1.0000 P( X = 2 ) = .2048 .9421

20. TableA.2 gives P( X£ k ) Bino(n=8, p =.20) For the multiple choice test with 8 questions, 5 choices for each, find the probability of getting two or fewer correct. 0.2 k .1678 .5033 .7969 .9437 .9896 .9988 .9999 1.000- 1.0000 0 1 2 3 4 5 6 7 8 About 80% of the classshould have two orfewer correct; about 20% should havethree or more correct.

21. 0.2 k .1678 .5033 .7969 .9437 .9896 .9988 .9999 1.000- 1.0000 0 1 2 3 4 5 6 7 8 TableA.2 gives P( X£ k ) Bino(n=8, p =.20) For the multiple choice test, find the probability of more the one but no more than three correct. Same question as . . . “two or more but less than four;”or “exactly two or three.”

22. 0.2 k .1678 .5033 .7969 .9437 .9896 .9988 .9999 1.000- 1.0000 0 1 2 3 4 5 6 7 8 TableA.2 gives P( X£ k ) Bino(n=8, p =.20) For the multiple choice test find the probability of three or more. k P( X3) = 1.000 – P(X£2) = 1.000 – .7969 = .2031 Same question as . . . “more than two”

23. 0.2 k .1678 .5033 .7969 .9437 .9896 .9988 .9999 1.000- 1.0000 0 1 2 3 4 5 6 7 8 TableA.2 gives P( X£ k ) Bino(n=8, p =.20) P( 3 or morecorrect) = P(more than 3 correct) = Watch the wording!

24. 0.2 k .1678 .5033 .7969 .9437 .9896 .9988 .9999 1.000- 1.0000 0 1 2 3 4 5 6 7 8 TableA.2 gives P( X£ c ) Bino(n=8, p =.20) Find the probability ofat least one correct.

25. Does a person have ESP? Experiment: Repeat the experiment 20 times. X = a count of the number of successes.X ~ Bino(n=20, p =.20) The probability of observing a result this extreme is 0.0026. Beth and Mike matched 10 times? Is this unusual? This is a rare event! The evidence indicatesthat they do better thanguessing. P(X10) = = =

26. Does a person have ESP? X = a count of the number of successes.X ~ Bino(n=20, p =.20) Expected valuem = _______ Standard deviations =

27. Free throws • Are shots independent? • Evidence says yes. • Suppose probability of Mo making one free throw is .70. • Mo will shoot 9 shots.

28. So, let Y = count of misses. Y~ B ( ) Let X = count of shots made, X ~ B ( ) n = 9, p = .70 Find probability he misses fewer than three. n = 9, p = _____

29. Count p Y 0 1 2 3 4 5 6 7 8 9 MissesX 9 8 7 6 5 4 3 2 1 0 Hits ? .70 P( Y£ 2 ) P( Y < 3 ) = = 1 - P( X£ 6 ) P( X > 6 ) = = = Lookup in Tables.

30. Moral: Pay attention to what you are counting!