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Review of Exercises from Chapter 17 Statistics, Spring 2012 1

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Outline for today: 1.Outline for next two weeks 2.Review homework assignment 3.Reading and notes on Chapter 18 2

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Outline for next two weeks We need to move really quickly You need to do the reading ahead of time I will provide you with exercises to do in class to cement what we’re studying After you take notes, I will present a QUICK review You MUST take notes and be ahead on the reading or you won’t make it We will be finished with Chapter 20 by next Friday 3

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Chapter 17, Exercise 24 Apples (set-up) Data: 6% of apples have to go for cider because of bruises or blemishes 300 trees What information are we missing or need to calculate? (actually the latter) 4

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Chapter 17, Exercise 24 Apples (finding missing variables) What are the variables in a binomial distribution? n, p, and q Which do we have? n and p. What’s q? Always, always, always 1-p. So q = 1-p =1.00-0.06 which equals 0.94 What can we calculate with these? 5

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Chapter 17, Exercise 24 Apples (choosing the model) So which model do we use? We could try the binomial model….OK since events are independent, so Binom (300,0.06) would work Since np>=10 and nq>=10, normal model OK, too. ?????WHERE DID THE NORMAL MODEL COME FROM????? Well, let’s see….. 6

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Chapter 17, Exercise 24 Apples (calculating μ and σ) Nothing new here, just plug-and-chug with our n, p, and q: 7

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Chapter 17, Exercise 24 Apples (normal model) Since the numbers are big, we can also do the Normal model: N(18, 4.11) Your answer for (a) is therefore that either the binomial model of Binom (300, 0.06) or the normal model of N(18, 4.11) would be appropriate. 8

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Chapter 17, Exercise 24(b) Apples (Probability that no more than a dozen cider apples) Do it first on your calculator by calculating Binom (300, 0.06). I’ll post the answer in a couple of slides (go back over your notes from before break; it was the last lecture we had) Let’s do the Normal model to remind ourselves how to do it (you can do the binomial by plugging it into your calculator) Our formula is our old friend, the z-equation: 9

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Chapter 17, Exercise 24(b) Apples (Probability that no more than a dozen cider apples) What is y for the equation? 12. Can we take it directly from the z-table, or do we need to subtract our result from 1? (talk about it; answer on next slide) 10

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Chapter 17, Exercise 24(b) Apples (Probability that no more than a dozen cider apples) You can read it off directly, since the problem says “no more than a dozen cider apples.” So let’s calculate: The z-value of -1.46 is about 0.072. Answer for the binomial is a little different: 0.085. Remember, the normal model IS an approximation! 11

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Chapter 17, Exercise 24(c) Apples (how likely are more than 50 cider apples ) Before you start plugging and chugging, think for a moment. We’ve seen this type of problem before. Actually, we don’t have to do any more calculations. When you see a problem like this, it does NOT require an exact answer. You can answer it from N(18, 4.11). Talk at your table and come up with an answer. Share out, and let’s see how yours compares to mine. (¡Cállate, Jason!) 12

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Chapter 17, Exercise 24(c) Apples (estimating) Whenever you see a question like this, just measure off the value they give you—here, 50—and find out how many standard deviations it is from the mean. If it’s more than 3σ, no way is it probable. Here you should get something like 7.8 standard deviation, so no way, José (there, you got mentioned!). 13

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Chapter 17, Exercise 26 Airline no-shows p=0.05 (success is a no-show!) q therefore is 0.95 n = number of tickets sold =275. Now, how do we set this up? We need the probability that FEWER than 9 people will cancel. (If 10 people cancel, we have 265 seats and 265 passengers, but if only 9 cancel, we have 265 seats and 266 (angry) pasajeros. 14

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Chapter 17, Exercise 26 Airline no-shows (set-up) Let’s calculate mean and standard deviation, since we’re going to use a normal model (also calculate the binomial model using your calculators; let Ms. Thien know if you’re having problems and I’ll review this with you tomorrow) 15

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Chapter 17, Exercise 26 Setting up the normal model So we have N(13.75, 3.61) Calculating our z-value: The area under the normal curve that corresponds with -1.32 is 0.0934. Binom should give you 0.116, which is the exact answer (your answers may be a bit different if you use the calculator) 16

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Chapter 17, Exercise 27 Annoying phone calls Like, there’s another kind of phone call? This problem is actually very much like 24(c). If you didn’t get it, take about 5 minutes and rework it, using 24(c) as a model. Or go on to the next page, and we’ll work it together. (Better if you try on your own….you’ll remember it longer!) 17

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Chapter 17, Exercise 27 Annoying phone calls (set up) n=200; p=0.12, so q = 0.88 Calculate mean and standard deviations for normal model: What does this mean? That we expect sales of 24, with a standard deviation of 4.6 18

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Chapter 17, Exercise 27 Annoying phone calls (solution) Once again, we want to evaluate 10 sales as being how for from the mean (i.e., expected value) of 24. If the standard deviation is 4.6, 10 is more than 3 standard deviations away. NOT a good sign. We can therefore conclude that he was probably misled. From a telemarketer? I’m shocked, SHOCKED to find that there’s gambling in here….now where are my winnings? 19

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Chapter 17, Exercise 34 A skilled archer, a new bow This is a simple question to answer, if you press the right buttons on your calculator Use the binompdf function. Instructions are on the next page, which you can skip if you know how to use the function (it’s discussed on p. 393 of the text) 20

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Chapter 17, Exercise 34 A skilled archer, a new bow..and TIs 83 and 84+ Access by pressing the 2 nd -Vars button to get distributions Scroll up until you get to binompdf WARNING! Binompdf is at the “A” label in the TI 83+ and at “0” in the TI 84+ Press enter 21

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Chapter 17, Exercise 34 A skilled archer, a new bow..and TIs 83 and 84+ You should get something that looks like the first screen on the right The syntax is binompdf(n, p, X), where X is what we want (here, 6 in a row) Answer is about 0.26 What does that mean? 22

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Chapter 17, Exercise 34 Analysis of the numerical answer from binompdf 0.26 means that a little more than 1 time out of 4 (26% of the time), you can expect a skilled archer like Diana to get 6 bull’s-eyes in a row. The result is therefore not really surprising, and doesn’t suggest that there’s anything special about the new bow. Note: why did we use the binomial distribution here instead of the normal distribution? 23

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Chapter 17, Exercise 36 More archery Same problem as 34, except the archer shoots 50 shots and gets 45 of them. Are you NOW convinced? Well, let’s see if you are. Let’s do our standard calculations first. Since we have 50 shots, let’s use a normal model. 24

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Chapter 17, Exercise 36 More archery Standard calculations for normal model: n=50, p=.8, q therefore is 0.2 What’s your take on this? Good, great, and unlikely? 25

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Chapter 17, Exercise 36 More archery For sure, the shooting is good, and above the 80% mark. But 45 is less than 2 standard deviations above the mean. Good…indeed, MUCH better than just 6 in a row, as in Exercise 34, but not SO spectacular that we’re going to give it credit. Close, but no cigar….. 26

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Where we go from here It’s really important for you guys to be reading ahead. I want you to take notes on Chapter 18 now. I expect full notes by tomorrow. That’s your homework, but Ms. Thien is going to check and grade your progress before you leave today, so you’ll have two grades entered. See the next slide for the conclusion of the sermon. 27

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End of sermon (almost) Tomorrow I’ll give you as good a lecture as I can about Chapter 18, and we’ll spend a couple of days doing problems. Read ahead, because we’ll be repeating this on Thursday for Chapter 19. Welcome to college and adulthood….you heard it here, first. Send me feedback on how we’re doing. --Hartley. 28

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