Honors Statistics Aug 23-8:26 PM Daily Agenda: 1. Review OTL C6#13 homework 2. Binomial mean and standard deviation 3. Discuss homework Jan 18-5:03 PM 1
Jan 30-12:31 PM Taking the train According to New Jersey Transit, the 8:00A.M. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let W = the number of days on which the train arrives late. Late not late 6? This is debatable (random says yes) 0.10 YES THIS IS BINOMIAL 2
Binomial setting? A binomial distribution will be approximately correct as a model for one of these two sports settings and not for the other. Explain why by briefly discussing both settings. (a) A National Football League kicker has made 80% of his field goal attempts in the past. This season he attempts 20 field goals. The attempts differ widely in distance, angle, wind, and so on. 1. Make field goal Miss field goal 20 3. 4. 0.80 NOT A BINOMIAL SETTING No attempts differ widely (b) A National Basketball Association player has made 80% of his free-throw attempts in the past. This season he takes 150 free throws. Basketball free throws are always attempted from 15 feet away with no interference from other players. 1. Make free throw Miss free throw 150 3. 4. 0.80 we are to assume independent Rhubarb Suppose you purchase a bundle of 10 bare-root rhubarb plants. The sales clerk tells you that 5% of these plants will die before producing any rhubarb. Assume that the bundle is a random sample of plants and that the sales clerk s statement is accurate. Let Y = the number of plants that die before producing any rhubarb. Use the binomial probability formula to find P(Y = 1). Interpret this result in context. 0.3151 binomialpdf(10,.05,1) = 0.3151 There is a 0.3151 probability that exactly 1 of the 10 rhubarb plants will die before producing any rhubarb. 3
Rhubarb Refer to Exercise 76. Would you be surprised if 3 or more of the plants in the bundle die before producing any rhubarb? Calculate an appropriate probability to support your answer. P(Y 3) = 1-P(Y 2) binomialcdf(10,.05,2) = 0.9885 P(Y 3) = 1-P(Y 2) = 1-0.9885 = 0.0115 Yes, I would be surprised because there is only a 1.15% chance that 3 or more plants will fail to produce rhubarb. There is a 0.2304 probability that exactly 4 of the 7 elk survive to adulthood Taking the train Refer to Exercise 72. (a) Find the probability that the train arrives late on exactly 2 days. Show your work. B(6, 0.10) binomialpdf(6,.10,2) = 0.09842 P(W = 2) = = 0.09842 (b) Would you be surprised if the train arrived late on 2 or more days? P(W 2) = 1-P(Y 1) P(W 2) = 1 - P(W 2) = 1-0.8857 = 0.1143 binomialcdf(6,.10,1) = 0.885735 11.43% is very "probable" so I would not be surprised if the train arrives late on 2 or more days. 4
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Now continue the front side of the beginning worksheet. May 15-6:15 PM Jan 28-8:12 PM 6
in a randomly selected group of three? = = ( May 15-6:17 PM May 13-10:11 AM 7
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May 12-2:52 PM Lie detectors A federal report finds that lie detector tests given to truthful persons have probability about 0.2 of suggesting that the person is deceptive. A company asks 12 job applicants about thefts from previous employers, using a lie detector to assess their truthfulness. Suppose that all 12 answer truthfully. Let X = the number of people who the lie detector says are being deceptive. Find P(X = 5). Interpret this result context. How surprising would it be to find more than 5 people who the lie detector says are being deceptive? Calculate an appropriate probability to support your answer. 9
B(, ) 3. Find and interpret µ x 4. Find and interpret σ x Jan 28-8:37 PM Now let Y = the number of people who the lie detector says are telling the truth. p=? B(, ) 5. Find P(Y 10). How is this related to P(X 2)? Explain. Jan 28-8:37 PM 10
Now let Y = the number of people who the lie detector says are telling the truth. p=? B(, ) 6. Calculate and How do they compare with and. Explain why this makes sense. Jan 28-8:37 PM Jan 30-12:31 PM 11