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**Chapter 17 Answers to questions 15, 16, 17, 18 and 20**

AP Statistics B

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**Overview This is the introduction to using the binomial theorem**

I will do some of the calculations by hand Others, 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

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Chapter 17, Exercise 15 This is the first time we’ve been using the binomial theorem to calculate mean and standard deviation Set-ups and 15 and 16 are straightforward Remember Exercise 13 on lefties (lefthanded-ness, not political orientation): p=0.13, q=0.87, n=5

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**Chapter 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:

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Chapter 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)

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**Chapter 17, Exercise 15(c) (continued)**

Answer: μ = 1/p = 1/.13 = 7.69 people

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Chapter 17, Exercise 16 Virtually the identical set-up as Exercise 15. I am therefore not going to repeat the workout, but just give you the answers. μ= 4.8 σ=0.98 1.25 shots Confused? me!

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Chapter 17, Exercise 17 Same set-up as 15, except we increase n from 5 to 12 p 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 group So p becomes q and q become p, since success is finding RIGHT-HANDERS

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**Chapter 17, Exercise 17(a) Right-handed mean and standard deviation**

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Chapter 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

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Chapter 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.

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**Chapter 17, Exercise 17(c) Alternatives to brute-force calculation**

We do what we did before: find the percentage MORE than 10, and calculate that It’s considerably easier, since we only have two binomial coefficients to calculate: The second is just 1, so let’s calculate the first:

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**Chapter 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?)

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**Chapter 17, Exercise 17(d) The probability of exactly 6 of each?**

So n=12, k=16, and we calculate:

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**Chapter 17, Exercise 17(e): Time to throw in the towel on computation**

OK, I surrender to technology Yes, 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:

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**Chapter 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 distribution On the next page are pictures of your calculators showing you what button to push

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**Calculating binomials with TI 83+/84+ Using the 2nd-Distr key combination**

Location of keys on TI 83+ Location of keys on TI 84+

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**Getting to the binomial distributions**

Keys to press Screen of the TI Push 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

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binompdf and binomcdf Here are the two functions we’ll be working with binompdf 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”

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**How 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-handed Answer: On the next slide, I’ll take you through how to do it on the calculator.

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binompdf n=6, p=0.87, x=6 Choosing binompdf gives you a screen like this Fill it in like this: n p x

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**binompdf Press ENTER and…. Pretty easy, huh?**

MUCH easier than doing it by hand. This is how we’ll do the rest of them

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Recap on binompdf Use 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)

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**Introduction to binomcdf**

binomcdf=binomial cumulative density function Think of this as being like what we did with the normal function, except easier Remember 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 failure Bad news: well, see the next slide

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**binomcdf: good news and bad**

Good news: easy to calculate IF set up right Bad 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)

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**binomcdf: Exercise 17(e)**

Choose binomcdf Enter values for n, p, and x and press enter Not so hard, is it? Except that we forgot to subtract this from 1!

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**binomcdf: subtracting**

Here’s the way you SHOULD set it up Set up, as usual, is harder and more challenging than the calculation…..

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**Redoing 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 binomcdf Take 5-6 minutes to practice. Share at your table, then share out to the class Somebody write the collective wisdom on the board, then go on to the next couple of slides for the answers

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**Redoing 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?

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**Answers 1-binompdf(12, .87, 12) or binomcdf(12, .87, 11)**

probability that no more than 10 righties = binomcdf(12, .87, 10)

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**Using 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=10 p=.80 (would Jonathan Zuniga or Jose Fuentes be this successful? Don’t believe them….) q=.20 Do the problems together using your calculators. Answers with screen shots are on next slides

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**Answers, Exercise 18(a)-(c)**

μ= σ=1.26 Probability that never misses=10 of 10 hits=use binompdf No more than 8 bull’s eyes: use binomcdf

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**Answers, Exercise 18(d)-(e)**

Exactly 8 bull’s eyes=use binompdf Hits bull’s-eye more often than she misses: use binomcdf, but subtract from 1: 1-binomcdf (5 successes)

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**Exercise 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 failure n= 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

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**Exercise 20(a) Two options, for once**

No frogs with traits… use either binomcdf or binompdf Why can you use either to get the same answer?

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**Exercise 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 trait We 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 condition Subtract this sum from 1 to get your answer

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**Exercise 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

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**Finally, 20(c) Sum of two probabilities**

“3 or 4 frogs” with the trait Easiest approach here is to use binompdf twice: one for exactly three frogs and one for exactly 4 frogs Answer: binompdf(3 frogs with trait)+binompdf(4 frogs with trait)

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