Presentation on theme: "Chapter 17 Answers to questions 15, 16, 17, 18 and 20"— Presentation transcript:
1Chapter 17 Answers to questions 15, 16, 17, 18 and 20 AP Statistics B
2Overview This is the introduction to using the binomial theorem I will do some of the calculations by handOthers, we will use the binomial function on your calculator (“about time, “I hear you say, though it was in the text which you could have read…..)I will show you how the calculator works by doing some examples that we first will work through by hand
3Chapter 17, Exercise 15This is the first time we’ve been using the binomial theorem to calculate mean and standard deviationSet-ups and 15 and 16 are straightforwardRemember Exercise 13 on lefties (lefthanded-ness, not political orientation):p=0.13, q=0.87, n=5
4Chapter 17, Exercise 15(a) and (b) “How many lefties do you expect?” in (a) asks for the mean. (b) is expressly the standard deviations.Formulas:Mean:SD:
5Chapter 17, Exercise 15(c)“If we keep picking people until we find a lefty, how long do you expect it will take?”TERRIBLE wording of the question: where have we been dealing with TIME? (“how long….”)What they REALLY want is the expected value, which is what for the binomial distribution? (next slide has answer, but quiz yourselves first and come to a consensus)
7Chapter 17, Exercise 16Virtually the identical set-up as Exercise 15. I am therefore not going to repeat the workout, but just give you the answers.μ= 4.8σ=0.981.25 shotsConfused? me!
8Chapter 17, Exercise 17Same set-up as 15, except we increase n from 5 to 12p still is 0.13, q is 0.87, n=12.Is this correct?Well, not exactly….because the questions asks for mu and sigma (mean and standard deviation) for the number of RIGHT-HANDERS in the groupSo p becomes q and q become p, since success is finding RIGHT-HANDERS
9Chapter 17, Exercise 17(a) Right-handed mean and standard deviation
10Chapter 17, Exercise 17(b)What’s the probability that they’re not all right-handed?Trick: find the percentage of all right-handers, and subtract from 1.First calculate:Then subtract from 1: =0.812
11Chapter 17, Exercise 17(c)“What’s the probability that there are no more than 10 righties?”Think before calculating.Do you really want to calculate all the following binomial coefficients? (The answer is “no”, in case you’re wondering.) Discuss among yourselves alternatives to such massive calculations, and when you reach a consensus, go on to the next slide.
12Chapter 17, Exercise 17(c) Alternatives to brute-force calculation We do what we did before: find the percentage MORE than 10, and calculate thatIt’s considerably easier, since we only have two binomial coefficients to calculate:The second is just 1, so let’s calculate the first:
13Chapter 17, Exercise 17(c) Putting everything together We have the binomial coefficients, so all we have to do is connect them to the terms they belong with.Don’t forget to subtract this from 1! (Why?)
14Chapter 17, Exercise 17(d) The probability of exactly 6 of each? So n=12, k=16, and we calculate:
15Chapter 17, Exercise 17(e): Time to throw in the towel on computation OK, I surrender to technologyYes, we COULD calculate the answer to this. But we’d have to calculate each of the following binomial coefficients (or generate Pascal’s triangle for n=12), and pair them with the appropriate exponentiated variables, then add them up:
16Chapter 17: calculating binomials with the TI 83+/84+ calculator As Monsanto used to say, “better living through chemistry”We’re doing better living through calculators (for once)You need to learn about the binomial distributionOn the next page are pictures of your calculators showing you what button to push
17Calculating binomials with TI 83+/84+ Using the 2nd-Distr key combination Location of keys on TI 83+Location of keys on TI 84+
18Getting to the binomial distributions Keys to pressScreen of the TIPush the 2nd-Distr keys together (Distr is on the VARS key, just below the cursor keys)You should get a screen that looks like this.Move to the binom distributions by pushing the top cursor key to get this screen
19binompdf and binomcdfHere are the two functions we’ll be working withbinompdf is “binomial probability density function” which we use to determine probabilities of a single parameter (e.g., 5 out of 12)binomcdf=“binomial cumulative density function”, used to find things like “find 4 or fewer hits out of 12”
20How to use binompdf Let’s redo Exercise 17(d) n=12, p=0.87, and x=6 (i.e., a population of 12, p of 0.87, and 6 out of 12 are right-handedAnswer:On the next slide, I’ll take you through how to do it on the calculator.
21binompdf n=6, p=0.87, x=6Choosing binompdf gives you a screen like thisFill it in like this:npx
22binompdf Press ENTER and…. Pretty easy, huh? MUCH easier than doing it by hand.This is how we’ll do the rest of them
23Recap on binompdfUse to solve questions like “exactly 6 of 13 have are left-handed”Access through VARS key, but be sure you press 2nd-VARS (actually 2nd-Distr)Syntax is binompdf(n, p, x)
24Introduction to binomcdf binomcdf=binomial cumulative density functionThink of this as being like what we did with the normal function, except easierRemember how we’d say on the normal model that the percentage of the population below Z was x%?It will be similar here: this function answers questions like “what’s the probability of finding 5 left-handers or less in a population of 9?”Good news: we don’t have to calculate all the binary coefficients and hook them up with the probability products of success and failureBad news: well, see the next slide
25binomcdf: good news and bad Good news: easy to calculate IF set up rightBad news: not always set up right.For example, Exercise 17(e): “What is the probability that the MAJORITY is right handed?”This is the equivalent of asking for 1-(Probability of 6 or fewer right-handers)Good news (redux): we can set up the equation like this: answer=1-binomcdf(12, 0.87, 6)
26binomcdf: Exercise 17(e) Choose binomcdfEnter values for n, p, and x and press enterNot so hard, is it? Except that we forgot to subtract this from 1!
27binomcdf: subtracting Here’s the way you SHOULD set it upSet up, as usual, is harder and more challenging than the calculation…..
28Redoing Exercises 17(b) and (c) using the TI Look back on 17(b) and (c)See if you can write out ways of calculating those answers by using either binompdf or binomcdfTake 5-6 minutes to practice.Share at your table, then share out to the classSomebody write the collective wisdom on the board, then go on to the next couple of slides for the answers
29Redoing Exercises 17(b) and (c) using the TI (answer to set-up) probability that not all are right-handers = 1-binompdf(12, .87, 12) OR binomcdf(12, .87, 11)probability that no more than 10 righties = binomcdf(12, .87, 10)we’ve already redone (d), right?
30Answers 1-binompdf(12, .87, 12) or binomcdf(12, .87, 11) probability that no more than 10 righties = binomcdf(12, .87, 10)
31Using the TI for Exercise 18 Enough of working by hand. Technology is a tool. You shouldn’t try to hammer a nail in with your hand or your neighbor’s head.Data for Exercise 18 on our archer Diana:n=10p=.80 (would Jonathan Zuniga or Jose Fuentes be this successful? Don’t believe them….)q=.20Do the problems together using your calculators. Answers with screen shots are on next slides
32Answers, Exercise 18(a)-(c) μ= σ=1.26Probability that never misses=10 of 10 hits=use binompdfNo more than 8 bull’s eyes: use binomcdf
33Answers, Exercise 18(d)-(e) Exactly 8 bull’s eyes=use binompdfHits bull’s-eye more often than she misses: use binomcdf, but subtract from 1: 1-binomcdf (5 successes)
34Exercise 20 Set-up on the TI Success=presence of the trait; failure=lack of trait (or vice versa, as we’ve seen)Success=1 out of 8. What’s that in percentages? 12.5% or for success, for failuren= 12(a), (b), and (d) are fairly straightforward, though you can do (a) a couple of different ways.(c) is something we haven’t done yet, using the TI, anyway, so you should stop at that slide and watch carefully.Except for(c), I’m just going to show you what to use and how to set it up, and give you screens shots of the answers
35Exercise 20(a) Two options, for once No frogs with traits… use either binomcdf or binompdfWhy can you use either to get the same answer?
36Exercise 20(b) Subtraction and reframing again… “at least 2 frogs” means we could have 2, 3, 4, 5,6,7, 8, 9, 10, 11 or 12 frogs with the unusual traitWe already have the figure from 20(a) for 0 frogs with condition (=.201). Add that to the probability of 1 frog with condition, and you have the probability of exactly 1 frog having the conditionSubtract this sum from 1 to get your answer
37Exercise 20(d) Straightforward “plug and chug” (d): “no more than 4 frogs with the (1 out of 8) trait” Obvious and straightforward application of binomcdf
38Finally, 20(c) Sum of two probabilities “3 or 4 frogs” with the traitEasiest approach here is to use binompdf twice: one for exactly three frogs and one for exactly 4 frogsAnswer: binompdf(3 frogs with trait)+binompdf(4 frogs with trait)