Honors Statistics. Daily Agenda
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1 Honors Statistics Aug 23-8:26 PM Daily Agenda 3. Review 6.3 Notes Quiz Aug 23-8:31 PM 1
2 Jan 27-2:30 PM Dec 10-9:59 AM 2
3 May 15-6:15 PM in a randomly selected group of three? = = ( May 15-6:17 PM 3
4 5. Draw the histogram of the probability distribution function on the graph below. Number of foreign-born May 15-6:17 PM 6. Draw the cumulative frequency histogram of the probability distribution function on the graph below. Number of foreign-born May 15-6:18 PM 4
5 Dec 10-11:39 AM Jan 27-2:34 PM 5
6 All trials have same probability of success Dec 10-11:28 AM Seed Depot advertises that its new flower seeds have an 85% chance of germinating (growing). Suppose that the company's claim is true. Judy gets a packet with 20 randomly selected new flower seeds from Seed Depot and plants them in her garden. Let X = the number of seeds that germinate Dec 10-12:10 PM 6
7 6.73a. Random Digit Dialing Dec 10-12:20 PM Dec 10-12:24 PM 7
8 Jan 22-6:14 PM R R G G G G R R R G R G G R G R R G G R Jan 22-6:14 PM 8
9 = Dec 10-12:29 PM Dec 10-2:05 PM 9
10 Jan 28-9:41 AM B(10,0.27) Dec 10-12:28 PM 10
11 B(10,0.27) binomialpdf(10,.27,3) = binomialpdf(10,.27,2) = binomialpdf(10,.27,1) = binomialpdf(10,.27,0) = Dec 10-12:28 PM 2nd VARS (distr) binomialpdf(trials,p,x value) this computes P(X = k) Dec 10-2:05 PM 11
12 Dec 10-3:14 PM binomialcdf(10,.27,3) = or 2) = 1 - ( ) = 1 - (0.4666) = binomialcdf(10,.27,2) = = Dec 10-3:14 PM 12
13 2nd VARS (distr) binomialcdf(trials,p,x value) this computes P(X k) Dec 10-2:05 PM Elk: Biologists estimate that a baby elk has a 44% chance of surviving to adulthood. Assume this estimate is correct. Suppose researchers choose 7 baby elk at random to monitor. Let X = the number who survive to adulthood. Use the binomial probability formula to find P(X = 4). Interpret this result in context. There is a probability that exactly 4 of the 7 elk survive to adulthood. Jan 27-7:10 PM 13
14 Elk: Refer to Exercise 75. How surprising would it be for more than 4 elk in the sample to survive to adulthood? Calculate an appropriate probability to support your answer. Jan 27-7:14 PM ASKIPS #74 Jan 30-12:31 PM 14
15 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 Jan 27-7:10 PM 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 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 we are to assume independent Jan 27-7:10 PM 15
16 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. binomialpdf(10,.05,1) = There is a probability that exactly 1 of the 10 rhubarb plants will die before producing any rhubarb. Jan 27-7:10 PM 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) = P(Y 3) = 1-P(Y 2) = = Yes, I would be surprised because there is only a 1.15% chance that 3 or more plants will fail to produce rhubarb. Sad Face! There is a probability that exactly 4 of the 7 elk survive to adulthood Jan 27-7:10 PM 16
17 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) = P(W = 2) = (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) = = binomialcdf(6,.10,1) = % is very "probable" so I would not be surprised if the train arrives late on 2 or more days. But I would not be happy! Jan 27-7:10 PM May 13-10:39 AM 17
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