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M15- Binomial Distribution 1 Department of ISM, University of Alabama, Lesson Objectives Learn when to use the Binomial distribution. Learn how to calculate probabilities for the Binomial using the formula and the two tables in the book, and in Excel. Bernoulli and Binomial Distributions Chapter 6.2, 7.4

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M15- Binomial Distribution 2 Department of ISM, University of Alabama, 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 being sent to her, and records it. If the two cards match, a success occurs. Does a person have ESP? Experiment: X= 1 success = 0 failure Bernoulli ( = ____ ) P(X = 1) = ___ = P(X = 0) = ___ = 1- X ~

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M15- Binomial Distribution 3 Department of ISM, University of Alabama, A discrete data distribution used to describe a population of binary variable values. Binary one of only two outcomes can occur, coded as “0” or “1” Bernoulli Distribution

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M15- Binomial Distribution 4 Department of ISM, University of Alabama, Bernoulli distribution, “one” trial. k 0 1 k 0 1 P(X=k) = = The mean of of the Bernoulli is The standard deviation is Bernoulli has one parameter: = the probability of success. Bernoulli ( ) X ~ 1

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M15- Binomial Distribution 5 Department of ISM, University of Alabama, Bernoulli distribution, “one” trial. k 0 1 k 0 1 P(X=k) 1 2 = (0 - ) 2 (1 - ) + (1 - ) 2 = 2 (1 - ) + (1 - ) 2 = 2 (1 - ) + (1 - ) 2 = (1 - ) [ + (1 - ) ] = (1 - ) [ + (1 - ) ] = 1 - = 1 - = (1 - ) 0(1- ) + 1 = = How are the “mean” and “standard deviation” determined? Once we know these results, we don’t need to derive it again. Use the same equations as the previous section. a little algebra...

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M15- Binomial Distribution 6 Department of ISM, University of Alabama, Examples of Bernoulli Variables 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)

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M15- Binomial Distribution 7 Department of ISM, University of Alabama, Bar Chart of Population for ESP X 01 1– 0 1.0

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M15- Binomial Distribution 8 Department of ISM, University of Alabama, A discrete data distribution used to model a population of counts for “n” independent repetitions of a Bernoulli experiment. Binomial Distribution Conditions: 1.a fixed number of trials, “n”. 2.all n trials must be independent of each other. 3.the same probability of success on each trial. 4.X = count of the number of successes.

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M15- Binomial Distribution 9 Department of ISM, University of Alabama, Binomial distribution, “n trials” k n k n P(X = k) = = The mean of of the Binomial is The standard deviation is Binomial has two parameters: n = the fixed number of trials, = the probability of success for each trial. to be determined X ~ Bino( n, )

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M15- Binomial Distribution 10 Department of ISM, University of Alabama, Computing Binomial probabilities 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. Excel(BINOMDIST function), gives either individual or cumulative. or Table A.2 gives P(X < k), the cumulative probability for X = 0 through X = k. For selected values of n and .

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M15- Binomial Distribution 11 Department of ISM, University of Alabama, Examples of Binomial Variables A count of the number of females in a sample of 80 fans at a Rolling Stones concert. A count of the number of defectives in sample of 50 tires coming off a production line. A count of the number of the number of corrects answers on 10 true-false questions for which everybody guessed. A count of the number of credit applications that are denied from a sample of 200.

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M15- Binomial Distribution 12 Department of ISM, University of Alabama, Are these situation Binomial? 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. A count of the number of fans at a Stones concert needed until we find 50 females.

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M15- Binomial Distribution 13 Department of ISM, University of Alabama, If the population of X-values has a binomial data distribution, then the proportion of the population having the value k is given by: This is also the probability that a single value of X will be exactly equal to x. P ( X = k ) = ( ) nknk p k (1 – p) n–k for k = 0, 1, 2,..., n

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M15- Binomial Distribution 14 Department of ISM, University of Alabama, P ( X = k ) = ( ) nknk p k (1 – p) n–k Big “X” is the “random variable.” Little “k” is a “specific value” of Big X. Example: P( X = k) P( Count = 4) Example: P( X = k) P( Count = 4)

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M15- Binomial Distribution 15 Department of ISM, University of Alabama, P ( X = k ) = ( )( ) nknk p k (1 – p) n–k ( ) nk = n! = 0! =

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M15- Binomial Distribution 16 Department of ISM, University of Alabama, P ( X = x ) = ( ) nxnx p x (1 – p) n–x These are the possible values of X; each value has its own probability. x = 0, 1, 2,..., n

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M15- Binomial Distribution 17 Department of ISM, University of Alabama, X = a count of the number of successes. X ~ Bino(n=5, =.20) Does a person have ESP? Experiment: Repeat the experiment 5 times. one Find the probability of exactly one success. P(X=1) = = =.4096 P(X=3) =

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M15- Binomial Distribution 18 Department of ISM, University of Alabama, Bino(n=5, =.20) ESP Experiment two or fewer Find the probability of two or fewer successes P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = = = =.94208

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Table A.1 gives P( X = k ) P( X = 2 ) = Bino(n=5, =.20) ESP Experiment P( X < 2 ) = Table A.2 gives P( X < k ) Cumulative Individual k: n = 5 k: n =

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M15- Binomial Distribution 20 Department of ISM, University of Alabama, k For the multiple choice test with 8 questions, 5 choices for each, find the probability of getting two or fewer correct. About 80% of the class should have two or fewer correct; about 20% should have three or more correct. About 80% of the class should have two or fewer correct; about 20% should have three or more correct. Table A.2 gives P( X £ k ) Bino(n=8, =.20)

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M15- Binomial Distribution 21 Department of ISM, University of Alabama, Table A.2 gives P( X £ k ) Bino(n=8, =.20) k 0.2 For the multiple choice test, f ind 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.”

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M15- Binomial Distribution 22 Department of ISM, University of Alabama, k P( X 3) = – P(X £ 2) = –.7969 =.2031 Table A.2 gives P( X £ k ) Bino(n=8, =.20) k 0.2 For the multiple choice test find the probability of three or more. Same question as... “more than two”

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M15- Binomial Distribution 23 Department of ISM, University of Alabama, P( 3 or more correct ) = P( more than 3 correct ) = k 0.2 Watch the wording! Table A.2 gives P( X £ k ) Bino(n=8, =.20)

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M15- Binomial Distribution 24 Department of ISM, University of Alabama, Table A.2 gives P( X £ c ) Bino(n=8, =.20) k 0.2 Find the probability of at least one correct.

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M15- Binomial Distribution 25 Department of ISM, University of Alabama, X = a count of the number of successes. X ~ Bino(n=20, =.20) Does a person have ESP? Experiment: Repeat the experiment 20 times. Beth and Mike matched 10 times? Is this unusual? P(X 10) = = = This is a rare event! The evidence indicates that they do better than guessing. The probability of observing a result this extreme is

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M15- Binomial Distribution 26 Department of ISM, University of Alabama, X = a count of the number of successes. X ~ Does a person have ESP? Bino(n=20, =.20) Expected value = _______ Standard deviation =

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M15- Binomial Distribution 27 Department of ISM, University of Alabama, Free throws Are shots independent? Evidence says yes. Suppose probability of Mo making one free throw is.70. Mo will shoot 9 shots.

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M15- Binomial Distribution 28 Department of ISM, University of Alabama, Let X = count of shots made, Find probability he misses fewer than three. So, let Y = count of misses. Y ~ B ( ) n = 9, =.70 n = 9, = _____ X ~ B ( )

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M15- Binomial Distribution 29 Department of ISM, University of Alabama, Y Misses X Hits P( Y < 3 ) = ?.70 Count P( Y £ 2 ) = P( X > 6 ) = 1 - P( X £ 6 ) == Lookup in Tables.

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M15- Binomial Distribution 30 Department of ISM, University of Alabama, Moral: Pay attention to what you are counting!

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