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Probability and Statistics Binomial Distributions Introduction to Binomial Distributions Let X = the number of right handed students out of 4 randomly selected. Fill in the following table. X Sample points 0 1 P(X) Number of ways that X can occur 3 = 0 Probability of one sample point corresponding to X (number of ways that X can occur) x (probability of each sample point) 3 = 1 2 3 Letting p = the probability that one observes a right-handed student, q = the probability that one observes a left handed student, and n = the total number of students observed, can you create a general formula for P(X)? P(X) = _____________________________________ Introduction to Binomial Distributions Let X = the number of right handed students out of 4 randomly selected. X Sample points 0 1 P(X) LLLL .134 = 0.0003 RLLL LRLL LLRL LLLR (.87)(.133 ) = 0.0019 (Times 4) 0.0076 (.872)(.132 ) LLRR RRLL = 0.0128 2 RLLR LRRL (Times 6) RLRL LRLR 0.0768 LRRR (.873)(.13) RLRR = 0.0856 3 RRLR (Times 4) RRRL 0.3424 4= .87 RRRR 4 0.5729 Number of Probability of one sample ways that X point corresponding to X can occur 4 0x 4 = 1 .87 .13 0 4 = 1 (number of ways that X can occur) x (probability of each sample point) 4 .870 x 4 .13 0 (.87)(.133 ) 4 (.87)(.133 ) 1 4 =6 2 (.872 )(.132 ) 4 (.872 )(.132 ) 2 4 =4 3 (.873 )(.13) 4 (.873 )(.13) 3 4 =1 4 .874 𝑥 .130 4 .874 𝑥 .130 4 4 The Binomial Probability Distribution Formula (number of ways that x can occur) x (probability of each sample point) P(x = 0) = 4 .870 x 4 .13 0 4 (.87)(.133 ) P(x = 1) = 1 P(x = 2) = 4 (.872 )(.132 ) 2 4 (.873 )(13) P(x = 3) = 3 P(x = 4) = 4 .874 𝑥 .130 4 Letting p = the probability that one observes a right-handed student, q = the probability that one observes a left handed student, and n = the total number of students observed, can you create a general formula for P(X)? In General: 𝑛 P(x) = (𝑝 𝑥 )(𝑞 𝑛−𝑥 ) 𝑥 The Binomial Probability Distribution Formula 𝑛 𝑥 𝑛−𝑥 P(x) = (𝑝 )(𝑞 ) 𝑥 This same probability can be found on the calculator: You can get this same value on the TI-84 by the following steps: 1) 2nd VARS 2) Down to binompdf (and press enter) 3) binompdf(n,p,x), so fill in n, p and x. or with the newer calculators: trials: n p: p x value: value of x that you want P(x) for. Example 1: Use the calculator to find at least 4 of the probabilities above. X Sample points 0 1 P(X) LLLL .134 = 0.0003 RLLL LRLL LLRL LLLR (.87)(.133 ) = 0.0019 (Times 4) 0.0076 (.872)(.132 ) LLRR RRLL = 0.0128 2 RLLR LRRL (Times 6) RLRL LRLR 0.0768 LRRR (.873)(13) RLRR = 0.0856 3 RRLR (Times 4) RRRL 0.3424 4= .87 RRRR 4 0.5729 Number of Probability of one sample ways that X point corresponding to X can occur 4 0x 4 = 1 .87 .13 0 4 = 1 (number of ways that X can occur) x (probability of each sample point) 4 .870 x 4 .13 0 (.87)(.133 ) 4 (.87)(.133 ) 1 4 =6 2 (.872 )(.132 ) 4 (.872 )(.132 ) 2 4 =4 3 (.873 )(.13) 4 (.873 )(.13) 3 4 =1 4 .874 𝑥 .130 4 .874 𝑥 .130 4 4 Example 2: Use the formula to find the probability that one observes 5 H’s and 5 tails when flipping a coin 10 times. In other words, let X = the number of heads observed out of 10 tosses. Find P(5). 𝑛 P(x) = (𝑝 𝑥 )(𝑞 𝑛−𝑥 ) 𝑥 P(x=5) = 10 (𝑝 𝑥 )(𝑞 𝑛−𝑥 ) 5 Means flip coin Each having 10 times and observe 5 heads HHTTHHTHTT or HHHTTTHTHT or how many other ways? 252 ways = 252 x 0.000977 = 0.246 probability of (.55 )(.55 ) = 0.000977 P(5) = binompdf(10, .5, 5) = 0.246 Binomial Probability Distributions Many experiments only have TWO possible outcomes. Which of the experiments have only two outcomes for each trial? a) Flipping a coin 20 times Two outcomes for each trial: Each flip can only result in Heads or Tails Binomial Probability Distributions Many experiments only have TWO possible outcomes. Which of the experiments have only two outcomes for each trial? b) Rolling a single die 5 times, recording the result SIX outcomes for each trial: Each roll can result in 1, 2, 3, 4, 5 or 6. c) Rolling a single die 5 times and counting how many times it landed on one. Two outcomes for each trial: Each roll can only result in 1 or not Binomial Probability Distributions Many experiments only have TWO possible outcomes. Which of the experiments have only two outcomes for each trial? d) Guess at 6 multiple choice questions with 4 possible answers, record answer A, B, C, or D FOUR outcomes for each trial: Each questions can result in A, B, C, D e) Guess at 6 multiple choice questions with 4 possible answers, record correct or incorrect Two outcomes for each trial: Each question is correct or incorrect The Binomial probability distribution can be applied if: n 1) The experiment consists of _____ identical trials. only two possible 2) There are ____________________outcomes for each trial. success s = ______________ or f =_______________ failure probability of success/failure remains the same 3) The _____________ for each trial. independent 4) The trials are ____________________ WHY? The __________________________________ is the Binomial random variable number of successes in n trials. Example – If we consider the random variable X = the number of right-handed students out of 4 randomly chosen, then let’s see if we can use the binomial formula to find the probability distribution of X. Asking 4 students the identical question 1)______________________________________________ 2 possible outcomes, left or right 2)______________________________________________ P(s) = .87 P(f) = .13 remains the same for each trial 3)______________________________________________ The occurrence of one student being left or right handed doesn’t affect the probability of the next student being left or right handed, therefore independent. 4)______________________________________________ Do you get it? – If we consider the random variable X = the number of Aces out of 3 cards (without replacement), can we use the Binomial probability distribution? The trials are not identical, one card will be drawn from a full deck and others will not 1)______________________________________________ 2 possible outcomes, either an Ace or not 2)______________________________________________ P(ACE) = ? P(Not and ACE) = ? Changes for each trial 3)______________________________________________ The occurrence of one card drawn does affect the probability of the next card, since without replacement, therefore NOT independent. 4)______________________________________________ Notation for Binomial Probability Distributions s and f _______________ (success and failure) are the two possible outcomes. p and q _______________ denotes the probabilities or s and f. P(s) = ________ = probability of success in one of the n trials p q P(f) = ________ = probability of failure in one of the n trials n = number of trials ____ ____ x = number of successes n! P( x) p x q n x (n x)! x! P(x) = binompdf(n, p, x) is the probability of getting exactly x successes in n trials. is the probability of getting exactly x successes in n trials. Mean and Standard Deviation for the Binomial Random Variable n p n pq Example 3 Overbooking Flights: Air America has a policy of booking as many as 42 people on an airplane that can seat only 40. Past studies have revealed that only 85% of the booked passengers actually arrive for the flight. a) Find the probability that if Air America books 42 people, not enough seats will be available. Interpret this result by indicating what percent of the time we expect one of these flights to be overbooked. The plane will be overbooked if 41 or 42 of the booked passengers show up. We need to find the probability of 41 or 42 passengers showing up. Find P(41) and P(42) and add them together. P(41) = binompdf(42, .85, 41) .008 + .001 = .009 P(42) = binompdf(42, .85, 42) Less than a 1% chance. Example 3 Overbooking Flights: Air America has a policy of booking as many as 42 people on an airplane that can seat only 40. Past studies have revealed that only 85% of the booked passengers actually arrive for the flight. b) Use the two standard deviation interval to determine what is the least number of passengers we could expect to (usually) arrive at one of these flights? n p n pq µ = np = 42 x 0.85 = 35.7 𝜎 = 𝑛𝑝𝑞 = 42 𝑥 0.85 𝑥 0.15 = 2.3 𝜇 ± 2𝜎 = 35.7 ± 2 2.3 = (31.1, 40.3) The least number we expect to arrive is 32 and most is 40. Example 4 A statistics quiz consists of 10 multiple-choice questions, each with four possible answers. Let the random variable x = the number of questions correct out of 10. a) Find the following: 0, 1, 2, …10 Domain of X = __________________ 0.25 p = _____ 10 n = _______ b) Create a probability distribution for X in the form of a table. X P(X) 0 1 2 3 4 5 6 7 8 9 10 .06 .18 .28 .25 .15 .06 .02 .00 .00 .00 .00 You can find the entire distributions by binompdf(10, .25). c) What is the probability that a student gets exactly 3 correct? ____________ X P(X) 0 1 2 3 4 5 6 7 8 9 10 .06 .18 .28 .25 .15 .06 .02 .00 .00 .00 .00 P(x = 3) = .25 or a 25% chance. d) For someone who makes random guesses for all of the answers, find the probability of failing if the minimum passing grade is 70%. X P(X) 0 1 2 3 4 5 6 7 8 9 10 .06 .19 .28 .25 .15 .06 .02 .00 .00 .00 .00 P(x < 7) = 1 or a 100% chance. e) If the entire class guesses at 10 questions, what is the Expected value for the average number of questions one would get correct? X P(X) 0 1 2 3 4 5 6 7 8 9 10 .06 .19 .28 .25 .15 .06 .02 .00 .00 .00 .00 µ = np = 10 x 0.25 = 2.5 f) How many questions correct would be usual (find the 2 standard deviation interval)? X P(X) 0 1 2 3 4 5 6 7 8 9 10 .06 .18 .28 .25 .15 .06 .02 .00 .00 .00 .00 µ = np = 10 x 0.25 = 2.5 𝜎 = 𝑛𝑝𝑞 = 10 𝑥 0.25 𝑥 0.75 = 1.37 𝜇 ± 2𝜎 = 2.5 ± 2 1.37 = (−0.24, 5.24) The usual outcomes are 0, 1, 2, 3, 4, 5 questions correct. µ = np = 10 x 0.25 = 2.5 𝜎 = 𝑛𝑝𝑞 = 10 𝑥 0.25 𝑥 0.75 = 1.37 𝜇 ± 2𝜎 = 2.5 ± 2 1.37 = (−0.24, 5.24) The usual outcomes are 0, 1, 2, 3, 4, 5 questions correct. X P(X) 0 1 2 3 4 5 6 7 8 9 10 .06 .19 .28 .25 .15 .06 .02 .00 .00 .00 .00 We will now do this experiment and verify the above. 1) ___ 2) __ 3) ___ 4) ____ 5) ___ 6) ___ 7) ____8) __ 9) ___ 10)___ Example 5 Rotator cuff surgery has a 90% chance of success. The surgery is performed on five patients. Out of the five patients, three of them were successful. Is this usual given the 90% success rate, or should we steer clear of this doctor? X 0 P(x) .00001 1 .00045 2 .0081 3 .0729 4 5 .32805 .59049 P(x = 3) = .0729, so not unusual, but very unlikely μ = 4.5 σ = 0.67 2 standard deviations = (3.16, 5.84) Using the two standard deviation interval, x = 3 is unusual. Example 1 Find the probability distribution of X = the number of heads observed if you toss a coin 4 times. From Last section we would: S = {HHHH THHH HTHH HHTH HHHT TTTT HTTT THTT TTHT TTTH HHTT TTHH THHT HTTH THTH HTHT} X = { 0, 1, 2, 3, 4} X P(X) 0 1/16 = .0625 1 4/16 = .25 2 6/16 = .375 3 4/16 = .25 4 1/16 = .0625 Using Section 4.4 n = _______ 4 THE NUMBER OF TRIALS p = _______ THE PROBABILITY OF SUCCESS IN ONE OF THE TRIALS 0.5 q = 1 – p = _______THE PROBABILITY OF FAILURE IN ONE OF THE TRIALS 0.5 If we want to find the probability that out of 4 tosses 2 are heads we calculate P(2) as follows n! p x q n x Use P( x) (n x)! x! n = 4, p = .5, q=.5 = Binompdf(4, .5, 2) P(2) = __________________________________________ Or by the calculator 1) 2nd VARS 2) Down to Binompdf (and press enter) 3) Binompdf(4, .5, 2) = .375 x 0 P(0) = 4! .50.540 .0625 binompdf(4 ,.5,0) (4 0)!0! 1 4! 1 4 1 P(1) = .5 .5 .25 binompdf(4 ,.5,1) (4 1)!1! 2 P(2) = 4! 2 4 2 .5 .5 .375 binompdf(4 ,.5,2) (4 2)!2! 4! 3 4 3 .5 .5 .25 binompdf(4 ,.5,3) 3 P(3) = (4 3)!3! 4! 4 4 4 .5 .5 .0625 binompdf(4 ,.5,4) 4 P(4) = (4 4)!4! Mean and Standard Deviation for the Binomial Random Variable n p n pq Example 2 Suppose we have each student in the class toss a coin 15 times. If we listed the sample points like in chapter 3 we there would be ______________ of them. WOW! TOO MANY. a) What is the mean and standard deviation? b) How many head’s would be unusual to get when this experiment is performed? c) What is the probability of that you get 11 or more heads when you toss a coin 15 times? d) We will not do this experiment and verify the above. Example 3 Using Directory Assistance Use the binomial probability formula to find the probability of getting exactly 3 correct responses among 5 different requests from AT&T directory assistance. Assume in that in general, AT&T is correct 90% of the time. We use the following formula n! P( x) p x q n x = binompdf(n, p, x) (n x)! x! 5 n = __________ The number of correct responses out of 5 x = _______________________________________________________ 0.9 p = ___________( 90% of the time directory assistance gives correct number) 0.1 q = ___________( 10% of the time directory assistance gives wrong number) Find the probability distribution for X X P(X) 0 1 .00001 .00045 2 3 .0081 .0729 4 5 .32805 .59049 X P(X) 0 1 .00001 .00045 2 3 .0081 .0729 4 5 .32805 .59049 a) Find and interpret P(3). P(3) = .0729 There is a 7.29% chance that AT&T will give exactly 3 correct responses out of 5, or in other words 7 out of 100 times when calling directory assistance 5 times, we would get exactly 3 correct responses. b) Is this low? Why? Yes this is low. AT&T is correct 90% of the time, 3 out of 5 means correct 60% of the time, so not that likely, but NOT unusual since above 5%. c)- What is the probability of getting at least 3 correct responses. So either 3, 4, or 5 out of 5 calls? P(at least 3 correct responses) = P(3) + P(4) + P(5) = .0729 + .32805 + .59049 = .99144 d)What is the mean and standard deviation? Why is the mean not 2.5, the middle of the values of the random variable X? How could the mean be 2.5? μ = np = 5 (.9) = 4.5 npq 5(.9)(.1) .671 e)What are the usual values of X? Show two ways to answer this question. Example 4 A statistics quiz consists of 10 multiple-choice questions, each with five possible answers. For someone who makes ransom guesses for all of the answers, find the probability of passing if the minimum passing grade is 60%. Is the probability high enough to make it worth the risk of trying to pass by making random guesses instead of studying? Let X = ________________________________ p = _____________________ n = _________________ Create a probability distribution for X in the form of a table and then answer the question. What is the average number of questions one would get correct if they guessed at all 10 questions? How many questions correct would be usual? We will now do this experiment and verify the above. Example 5 Suppose that I draw 5 cards from a deck of cards and count the number of hearts drawn. a) Find the probability that 3 of the 5 cards are hearts. b) Can we use the binomial distribution to find the above probability?