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AP Statistics February 27, 2012 AP Statistics B warm-up Monday, February 27, 2012 1. Write out the general addition rule of Chapter 15. How does it differ from the addition rule of Chapter 14? 2. 30% of all Americans are overweight, 35% have high bloodpressure, and 23% are both overweight and suffer from high blood pressure a) What percentage of Americans have high blood pressure but are not overweight? b) What percentage of Americans either have high bloodpressure or are overweight, but are not suffering from both? c) What percentage of Americans do not suffer from either high blood-pressure or overweightness (which probably isn’t a word)? Answers are on the next slide. Slide 2 of 49 Answers to warmup Monday, February 27, 2012 P( A B) P( A) P( B) P( A B) 1. . 2. Draw a diagram (“Make a picture, make a picture, make a picture!”) (C’mon, somebody go to the board and draw a Venn diagram before I give you the answers!) Push the space bar to display the answers. a) 12% (which = 35% who have high blood-pressure minus 23% who are both overweight and suffer from high blood pressure b) 19% (Add the 12% from 2 a) to 7% who are overweight but do not have high blood pressure (subtract 23% (both overweight and high blood-pressure) from 30% who are overweight)) c) 58% (42% have some combination of being overweight and having high blood-pressure) Slide 3 of 49 Outline for today Monday, February 27, 2012 • Ms. Thien reports that you’d like to review problems 14, 15, 16, and 24 from Chapter 15. Most of today will be spent going through these. • I working at getting smaller sound files to embed in the PowerPoints. Friday’s was 185 megabytes. I’m striving to get them small enough (5-10 megs) so that I can post them on the Garfield web site and you can download them with the sound. • Please text or email me comments….what you liked, what you hated, just so that’s it’s feedback. • You can stop the display, but make sure you go to the final slide (#49) by the end of the period, since it has tomorrow’s homework assignment. Slide 4 of 49 Problem 14: birth order, take 2 Birth Order First or only Second or later Total Arts and Sciences 34 23 57 Agriculture 52 41 93 Human Ecology 15 28 43 Other 12 18 30 Total 113 110 223 Slide 5 of 49 How do we determine conditional probabilities using a table? (problem 14(b) 1. Pick the row (or column) that follows “given” or “for A…..” or similar statement E.g., here: “Among Arts and Sciences, students…..” (Ch. 15, problem 14(b)) 2. Pick the column (or row) that corresponds to the “probability” Here, “the probability that a student was second child or more”? 3. The intersection of the column and row in the table is the numerator 4. The “total” at the end of the first column or row that you picked in #1 above is the denominator Slide 6 of 49 Yellow=first row we pick “Arts & Sciences” Red=column we pick next 23=numerator 57=denominator Birth Order First or only Second or later Total Arts and Sciences 34 23 57 Agriculture 52 41 93 Human Ecology 15 28 43 Other 12 18 30 Total 113 110 223 Answer: 23/57 = 40.35% or 0.4035 Slide 7 of 49 We could also do it by picking a column first (yellow) and then a row (red) • E.g., Given that a person is a second child, what is the probability that they will be enrolled in Agriculture? (this is similar to problem 14(c)) First or only Second or later Total Arts and Sciences 34 23 57 Agriculture 52 41 93 Human Ecology 15 28 43 Other 12 18 30 Total 113 110 223 So here, the probability is 41/110, which equals 0.3727, or 37.27%. Slide 8 of 49 Distinguishing probabilities • What we’ve just done is a conditional probability. • The problems also ask for absolute probabilities, i.e., the probabilities that you will draw any specific person at random. • For this, the analysis is a bit different. Slide 9 of 49 Absolute (universal) probabilities • Somebody read problem 14(a) for the class. • Here, we need to establish the probabilities for EVERYTHING inside the table. • The key is that we divide each entry of the table by the total number of students (that is, the entry in the bottom row, last column)…here, 223. Slide 10 of 49 So here’s what your table of probabilities should look like Birth Order First or only Second or later Total Arts and Sciences 34 (34/223) 15.25% 23 (23/223) 10.31% 57 Agriculture 52 (52/223) 23.32% 41 (41/223) 18.39% 93 Human Ecology 15 (15/223) 6.73% 28 (28/223) 12.56% 43 Other 12 (12/223) 5.38% 18 (18/223) 30 Total 113 110 223 Note: it’s easier and faster to calculate this on Excel than with a calculator! Slide 11 of 49 Problems 14 (c), (d), and (e) • (No audio on this slide) • Take 5 minutes or so and recalculate the probabilities for Problems (c), (d) and (e) • When you’re done, push the space bar and I’ll show you another way of calculating these. Slide 12 of 49 Problem 14(c) • Read problem, someone! • Calculate probabilities of second children • Pick A&S (yellow) Birth Order First or only Second or later Total Arts and Sciences 34 23 (23/110) 20.9% 57 Agriculture 52 41 (41/110) 37.27% 93 Human Ecology 15 28 (28/110) 25.45% 43 Other 12 18 (18/110) 16.36% 30 Total 113 110 223 Slide 13 of 49 Problem 14(d) • Pick 1st child column • Calculate probability for Agriculture First or only Second or later Total 34 23 57 52 (52/113) 46.02% 41 93 Human Ecology 15 28 43 Other 12 18 30 Total 113 110 223 Arts and Sciences Agriculture Slide 14 of 49 Problem 14: birth order, take 2 Birth Order First or only Second or later Total Arts and Sciences 34 23 57 Agriculture 52 41 93 Human Ecology 15 28 43 Other 12 18 30 Total 113 110 223 Slide 15 of 49 Problem 14(e) What is the probability that an Agriculture student is a first or only child? Answer: 52/93, or 55.91% Birth Order First or only Second or later Total Arts and Sciences 34 23 57 Agriculture 52 41 93 Human Ecology 15 28 43 Other 12 18 30 Total 113 110 223 Slide 16 of 49 Notation in terms of P(B|A) Let’s rephrase problem 14 in terms of conditional probability: 14(b): P(2nd child|Arts and Science) 14(c): P(Arts & Science|2nd child) 14(d): P(Agriculture|1st or only child) 14(e): P(1st or only child|Agriculture) Slide 17 of 49 Comparison of P(B|A) with language of problems (14(b)) P(B|A) • 14(b): P(2nd child|Arts and Science) Problem • Among the Arts and Sciences students, what’s the probabilibty a student was a second child (or more)? Comparison of P(B|A) with language of problems (14(c)) P(B|A) • 14(c): P(Arts & Science|2nd child) Problem • Among second children (or more), what’s the probability a student is enrolled in the Arts and Sciences? Comparison of P(B|A) with language of problems (14(d)) P(B|A) • 14(d): P(Agriculture|1st or only child) Problem • What’s the probability that a first or only child is enrolled in the Agriculture College? Comparison of P(B|A) with language of problems (14(e)) P(B|A) • 14(e): P(1st or only child|Agriculture) Problem • What is the probability that an Agriculture student is a first or only child? Problem 15 Sick Kids. Seventy percent of kids who visit a doctor have a fever, and 30% of kids with a fever have sore throats. What’s the probability that a kid who goes to the doctor has a fever and a sore throat? Problem 15, sick kids: make a picture, make a picture, make a picture! 70% of patients have fever; 30% of those with fever have sore throat. % of both? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Slide 23 of 49 Problem 15: sick kids 70% of patients have fever; 30% of those with fever have sore throat. % of both? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 Slide 24 of 49 Problem 15: sick kids 70% of patients have fever; 30% of those with fever have sore throat. % of both? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 30% of patients who do not have fever This includes those who have sore throats but no fever Slide 25 of 49 Problem 15: sick kids 70% of patients have fever; 30% of those with fever have sore throat. % of both? 10% 10% 10% 10% 10% 10% 30% of patients who do not have fever This includes those who have sore throats but no fever 10% 10% 10% 10% Slide 26 of 49 Problem 15: sick kids 70% of patients have fever; 30% of those with fever have sore throat. % of both? 30% of patients who do not have fever This includes those who have sore throats but no fever Red=fever AND sore throat Blue= fever alone Slide 27 of 49 Problem 15 solution: just count the boxes (each box = 1%) = 21% 70% of patients have fever; 30% of those with fever have sore throat. % of both? 30% of patients who do not have fever This includes those who have sore throats but no fever 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Red=fever AND sore throat Blue= fever alone Slide 28 of 49 How can we solve this faster? Answer: the General Multiplication Rule, i.e., P( A B) P( A) P( B | A) Slide 29 of 49 General Rule of multiplication P(AΩB) is what we are looking for, which=? P(AΩB) = P(A) × P(B|A) P(A)= probability of having a fever, which is? 70% P(B|A) = probability of having a sore throat GIVEN that you have a fever, which is? 30% So the answer is 30% of 70%..... ………or 0.3 × 0.7 = 0.21 = 21%. QED. Slide 30 of 49 Problem 16 Sick Cars. Twenty percent of cars that are inspected have faulty pollution control system. The costs of repairing a pollution control system exceeds $100 about 40% of the time. When a driver takes a car in for inspection, what’s the probability that she will end up paying more than $100 to repair the pollution control system? Let’s draw a diagram (like problem 15) Percentage of cars Cars needing repairs (20%) Cars that need no repairs (80%) Slide 32 of 49 Let’s draw a diagram (like problem 15) Need repairs Don’t need repairs >$100 <$100 Answer: 8 cells out of 100, or 8% (0.08) Slide 33 of 49 Redoing problem 16 in language/formulas • The driver has a 20% chance of having to repair the pollution control system (P(A)). • If she has to repair the pollution control system, she has a 40% chance of paying more than $100 (P(B|A)) • The general multiplication rule has us multiply the two together: 0.2 × 0.4 = 0.08 = 8%. • P(AΩB)=P(A) × P(B|A) Slide 34 of 49 Problem 24 On the road again. According to Ex. 4, the probability that a U.S. resident has traveled to Canada is 0.18, to Mexico is 0.09, and to both countries is 0.04. a) What is the probability that someone who has traveled to Mexico has visited Canada, too? b) Are travel to Mexico and Canada disjoint events? Explain. c) Are travel to Mexico and Canada independent events? Explain. Draw the Venn diagram 0.77 Mexico Canada 0.18-0.04= 0.14 0.04 0.09-0.04 = 0.05 Slide 36 of 49 Another approach Travel to Mexico Travel to Canada Yes No Total Yes 0.04 0.14 0.18 No 0.05 0.77 0.82 Total 0.09 0.91 1.00 Slide 37 of 49 Problem 24(a) • What’s the probability that someone who has travelled to Mexico has travelled to Canada? • Use either Venn diagram or table – Total percent of people travelling to Mexico? 9% – Total percent of those travelling to Mexico also travelling to Canada? 4% • 4%/9%=0.444=44.4% Slide 38 of 49 Problem 24(b) • Are travel to Mexico and Canada disjoint events? Explain. • Disjoint=if A occurs, B cannot occur. – Example: getting a grade. If you get an A, you can’t get a B. You can have a probability of getting an A, or a B, but you can’t get both • Travel is not disjoint because – You can go to both – 4% of the population HAS been to both countries! Slide 39 of 49 Problem 24(c) • Are travel to Mexico and Canada independent events? Explain. • Independent=one event doesn’t alter the probability of the other occurring • Travel is not independent because – 18% of all U.S. residents have been to Mexico. – 44.4% of those have also been to Canada – If the events were independent, the probabilities would be equal Slide 40 of 49 Today’s lesson • “Drawing without replacement” • Usually done with decks of cards, but lots of other uses as well • Examples is on pp.354-55 of text Slide 41 of 49 Visualizing the problem • Draw a picture! • 3 Gold (desirable), 4 Silver (less desirable), 5 Wood (we don’t like) Gold Silver Silver Brown Gold Silver Brown Brown Gold Silver Brown Brown Slide 42 of 49 You pick one and get gold • • • • Your odds were 3/12 = ¼. 1 less gold (only 2 left) 1 less option (now only 11 choices) What are the odds of your friend selecting gold? Answer: 2/11 Silver Silver Brown Gold Silver Brown Brown Gold Silver Brown Brown Slide 43 of 49 Multiplication rule applies • Your odds of getting a Gold were ¼. • The remaining odds of your friend getting Gold on the next draw were 2/11. • Multiply both together to get 2/44 or 1/22 = 0.045. Slide 44 of 49 Problem 17 (p. 364) • Rules: you draw three cards, one at a time. • We are looking at probabilities of specific outcome. • Trick: well, not a trick, exactly, but another way of looking at things: recast the question not as the odds of GETTING something, but of NOT getting it. Slide 45 of 49 Example: 17(a) (first heart you get is the third card dealt) • The odds of getting a heart are ¼ for the first card dealt (13 hearts out of 52 cards in all) • But the probability is that you DIDN’T get a heart. That’s ¾, or .75 (because there are 39 cards in the deck that aren’t hearts) • So the odds of NOT getting a heart on the first draw is 0.75. Slide 46 of 49 Second step (still not getting a heart on the second draw) • Same analysis as before, except the numbers have changed: – The odds of NOT drawing a heart are the number of non-hearts left in the deck, divided by 51 (total cards remaining) – There are still 13 hearts, but now only 38 cards that aren’t hearts. • So new odds of not getting a heart are 38/51. Slide 47 of 49 Third step (odds of getting a heart on the third draw after twice not getting any hearts) • Similar, except we now are calculating the odds of GETTING a heart. • Still 13 hearts left, but only 50 cards in the deck. • So odds are 13/50. • Total probabilities = product of all probabilities: (3/4)(38/51)(13/50)= (0.75)(.745)(.26)= 0.147 = 14.7% (or about 1 out of 7 times) Slide 48 of 49 Homework for Tuesday 1. Complete problem 17 and SHOW WORK. 2. Problems 18-20. 3. Read Chapter 16 (it’s short, and we’ll start it tomorrow) 4. Bring your Globe at Night data tomorrow: we need to enter it by Wednesday if you want extra credit. Slide 49 of 49