Report

Chapter 17: The binomial model of probability Part 2 AP Statistics 1 The binomial model: jumping in with both feet • There’s a close connection between tree diagrams, combinations, expanding binomial expressions, and the binomial model • I’ve debated about giving you a lot of background on it to show you the connections FIRST • Instead, I think it will be more effective to show you how to solve the problems, and then show you the connections, which then may be more meaningful. 2 The binomial model: formulae • So here they are, the three formulae you need to recognize and use. • Let’s do the easiest two first, mean and standard deviation – Check out the Math Box on p. 392 if you want to see the derivations • Mean: • Standard deviation: 3 The binomial model: probability of exactly k successes in n trials • This is the key equation to recognize (memorizing it is hard, at least at first) • Remember that p=probability of success, q=probability of failure, and the binomial coefficient is the number of combinations that satisfy k successes in n trials • Here it is……wait for it……. 4 The binomial model: how to dissect the binomial formula • It’s the number of combinations you can get from taking k samples from a population of n • That, in turn, is equal to • These are the calculations we’ll be working on today 5 The binomial model/Exercise 13(d): Starting with Exercise 13(d) • We were not able to do Exercises 13(d)-(f) or 14(d)-(f) because we needed to apply the binomial model. • So let’s start with Exercise 13(d), which asks for what percentage of the populations has exactly 3 lefties. • Before going on to the next slide, work for 2-3 minutes and see how many different combinations you can make using three lefties and two righties. 6 The binomial model/Exercise 13(d): combinations of 3 lefties, 2 righties • Here’s the distribution on the right. • It took me about 5 minutes to derive and check it; YMMV. • But even if you do it faster than I did, what about 3 out 10? Or 3 out of 15? Or even 3 out of 8? • It will take way too long. • We need a better (faster) method. 1. L 2. L 3. L 4. L 5. L 6. L 7. R 8. R 9. R 10. R L L L R R R L L L R L R R L L R L R L L R L R L R L L L R L R R L R L L R L L L 7 The binomial model/Exercise 13(d): Using the binomial coefficient • The binomial coefficient will save us from the tedium of finding such distributions • For n=5 and k = 3 (the conditions of our problem), we can calculate the number of combinations directly: 8 The binomial model/Exercise 13(d): Applying the binomial coefficient to the probabilities • We have the binomial coefficient, which is 10 for this problem. • Now, how to apply the q and p probabilities? • The best thing to do is to create a table. We’re going to spend some time doing this, and it should help to make everything clear. 9 The binomial model/Exercise 13(d): Making a probability table • Because the calculations are complicated, we’re going to make a couple of tables to help make things clear. • Here’s the form of the tables to copy for this part of the exercise. Make two of them: 5-k k (5 k) 0 1 2 3 4 5 Totals: q5-n pn (5 k)q5-npn Lefties 0 1 2 3 4 5 10 The binomial model/Exercise 13(d): Filling in the 1st probability table with formulae • Note that the table k=0 through k=5. We’ll discuss why in a bit as we fill it in. • Now, let’s fill in the formulas before you do the calculations. You’ll do the calculations in the 2nd table. Fill in the blanks as indicated: 5-k 5-0=5 5-1=4 k (5 k) q5-n 0 5!/5!0! 0.87^5 1 5!/1!4! 0.87^4 2 5!/2!3! 3 4 5 Totals: pn 0.13^0 0.13^1 (5 k)q5-npn Lefties 0 1 2 3 4 5 11 The binomial model/Exercise 13(d): What the 1st table should look like when you’re done 5-k 5-0=5 5-1=4 5-2=3 5-3=2 5-4=1 5-5=0 k 0 1 2 3 4 5 Totals: (5 k) 5!/5!0! 5!/1!4! 5!/2!3! 5!/3!2! 5!/4!1! 5!/5!0! 5-n q 0.87^5 0.87^4 0.87^3 0.87^2 0.87^1 0.87^0 n 5-n n p (5 k)q p Lefties 0.13^0 0 0.13^1 1 0.13^2 2 0.13^3 3 0.13^4 4 0.13^5 5 12 The binomial model/Exercise 13(d): Starting the 2nd table • Your second table will contain the calculations using the formulas in the 1st table. As you start, it should look like this: 5-k k (5 k) 0 1 2 3 4 5 Totals: q5-k pk (5 k)q5-kpk Lefties 0 1 2 3 4 5 13 The binomial model/Exercise 13(d): Completing the 2nd table • Using the formulas in Table 1, do the calculations. Be sure to fill in the calculations in the second column from the right that I’ve grayed out. • By the end, you should have a table of all the values possible where there are 0 lefties, 1 lefty, …., 4 lefties, and 5 lefties. • It will probably take about 10 minutes to complete this. Be sure to complete the totals for the columns headed by (5 n) and by (5 k)q5-npn . 14 The binomial model/Exercise 13(d): What the second table should like and its meaning 5-k 5 4 3 2 1 0 k (5 k) 0 1 1 5 2 10 3 10 4 5 5 1 Totals: 32 q5-k 0.4984209 0.5728976 0.6585030 0.7569000 0.8700000 1.0000000 pk (5 k)q5-kpk Lefties 1.0000000 0.4984209 0 0.1300000 0.3723834 1 0.0169000 0.1112870 2 0.0021970 0.0166291 3 0.0002856 0.0012424 4 3.71E-05 3.713E-05 5 1 15 The binomial model/Exercise 13(d): Important points 1. (5 k)q5-kpk is the probability that you will get k events in 5 trials. 2. The column headed by (5 k)q5-kpk adds up to 1, which tells us that this is the TOTAL probability model. 3. Note that we start with k=0….we DO have the probability that we will get zero lefties. 4. If you have k trials, you will have k+1 entries in your table. 16 The binomial model/Exercise 13 (e): Using the table to determine summed probabilities • 13(e): at least 3 lefties in the group. • That means there could be 3….or 4…..or 5. • Find the total by adding the areas marked: 5-k 5 4 3 2 1 0 k (5 k) 0 1 1 5 2 10 3 10 4 5 5 1 Totals: 32 q5-k 0.4984209 0.5728976 0.6585030 0.7569000 0.8700000 1.0000000 pk (5 k)q5-kpk Lefties 1.0000000 0.4984209 0 0.1300000 0.3723834 1 0.0169000 0.1112870 2 0.0021970 0.0166291 3 0.0002856 0.0012424 4 3.71E-05 3.713E-05 5 1 17 The binomial model/Exercise 13 (f): Using the table to determine summed probabilities (2) • 13(d): no more than 3 lefties in the group. • That means there could be 3, 2, 1, or 0 • Find the total by adding the areas marked: 5-k 5 4 3 2 1 0 k (5 k) 0 1 1 5 2 10 3 10 4 5 5 1 Totals: 32 q5-k 0.4984209 0.5728976 0.6585030 0.7569000 0.8700000 1.0000000 pk (5 k)q5-kpk Lefties 1.0000000 0.4984209 0 0.1300000 0.3723834 1 0.0169000 0.1112870 2 0.0021970 0.0166291 3 0.0002856 0.0012424 4 3.71E-05 3.713E-05 5 1 18 The binomial model: Answers to Exercises 13(e) and 13(f) • Exercise 13(e):0.982 – Does this answer make sense? • Exercise 13(f): 0.0179 – Does THIS answer make sense? 5-k 5 4 3 2 1 0 k (5 k) 0 1 1 5 2 10 3 10 4 5 5 1 Totals: 32 q5-k 0.4984209 0.5728976 0.6585030 0.7569000 0.8700000 1.0000000 pk (5 k)q5-kpk Lefties 1.0000000 0.4984209 0 0.1300000 0.3723834 1 0.0169000 0.1112870 2 0.0021970 0.0166291 3 0.0002856 0.0012424 4 3.71E-05 3.713E-05 5 1 19 The binomial model/Exercise 14: Preparing the chart • Same as 13, except we now have 6 arrows, not 5 people • Do the tables again, starting with a blank table like this: Go to the next slide when done. 6-k 1 2 3 4 5 6 7 TOTAL: k 0 1 2 3 4 5 6 (6 k) q6-k pk (6 n)q6-kpk bull's-eyes 0 1 2 3 4 5 6 20 The binomial model/Exercise 14: The complete chart • Here’s what you should have come up with. • The answers are similar to what you did for 13. Use your chart to calculate them. Answers on next slide. 6-k 1 2 3 4 5 6 7 k 0 1 2 3 4 5 6 (6 k) 1 6 15 20 15 6 1 64 q6-k pk (6 n)q6-kpk bull's-eyes 0.000064 1 0.000064 0 0.00032 0.8 0.001536 1 0.0016 0.64 0.015360 2 0.008 0.512 0.081920 3 0.04 0.4096 0.245760 4 0.2 0.32768 0.393216 5 1 0.262144 0.262144 6 1 21 The binomial model/Exercise 14: Answers d) 0.246 (yellow shading) e) 0.0179 (adding numbers circled in red f) 0.345 (adding numbers circled in green) 6-k 6 5 4 3 2 1 0 k 0 1 2 3 4 5 6 (6 k) 1 6 15 20 15 6 1 64 q6-k pk (6 n)q6-kpk bull's-eyes 0.000064 1 0.000064 0 0.00032 0.8 0.001536 1 0.0016 0.64 0.015360 2 0.008 0.512 0.081920 3 0.04 0.4096 0.245760 4 0.2 0.32768 0.393216 5 1 0.262144 0.262144 6 1 22 The binomial model: Review of how to calculate combinations • Remember the formula: • k is the number of events you’re looking for in the population n • Be sure you put the k with the p, which, as you’ve seen, may change from problem to problem. • Don’t forget to calculate all three parts of the equation and do the multiplication 23 The binomial model: Practice: Exercise 19 • Similar to what we’ve done already • Tennis player has successful first serve 70% of the time. • Assuming independence and 6 serves (n=6), calculate the following probabilities: a. b. c. d. All 6 serves go in Exactly 4 serves go in At least 4 serves go in No more than 4 serves go in • Spend 5-10 minutes calculating (a) through (c), then advance to the next slide for the analysis. 24 The binomial model: Practice: Exercise 19(a) • All six serves go in. • n=6, k=number of good serves=6, p=0.7, so q=0.3 • Calculate combinations first: • Next, pkqn-k=(0.7)6(0.3)6-6=(0.7)6(0.3)0=(0.7)6=0.118 25 The binomial model: Exercise 19(b) • Exactly 4 serves go in. • Similar to (a), but with different numbers: n still 6, but k=4 • Binomial coefficient: • Next, pkqn-k=(0.7)4(0.3)6-4 = (0.7)4(0.3)2 =(0.24)(0.09)=0.0216 • Multiply 15 times 0.0216 to get 0.324 26 The binomial model: Exercise 19(c) • At least 4 serves in means she got 4, 5 or 6 serves in. If we know probabilities, we can add them. • Already know 6 from (a) and 4 from (b) • Calculate probabilities of 5: • Calculate pkqn-k (continued on next slide) 27 The binomial model: Putting all of 19(c) together • Remember that the total probability of at least 4 serves going in is P(4)+P(5)+P(6) • From previous slides, we have – P(4)= 0.324 – P(5)=0.301=6×0.0504 (from slide 27) – P(6)=0.118 – TOT: 0.743 • Whenever you have multiple outcomes, calculate and add them together!! 28 The binomial model: Homework • Chapter 17, Exercises 15, 16, 17 18 and 20. 29