2AP Statistics B warm-up Monday, February 27, 2012 Write out the general addition rule of Chapter 15. How does it differ from the addition rule of Chapter 14?30% of all Americans are overweight, 35% have high blood-pressure, and 23% are both overweight and suffer from high blood pressurea) What percentage of Americans have high blood pressure but are not overweight?b) What percentage of Americans either have high blood-pressure 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.
3Answers to warmup Monday, February 27, 2012 .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 pressureb) 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)
4Outline 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 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.
5Problem 14: birth order, take 2 First or onlySecond or laterTotalArts and Sciences342357Agriculture524193Human Ecology152843Other121830113110223
6How do we determine conditional probabilities using a table How do we determine conditional probabilities using a table? (problem 14(b)Pick the row (or column) that follows “given” or “for A…..” or similar statementE.g., here: “Among Arts and Sciences, students…..” (Ch. 15, problem 14(b))Pick the column (or row) that corresponds to the “probability”Here, “the probability that a student was second child or more”?The intersection of the column and row in the table is the numeratorThe “total” at the end of the first column or row that you picked in #1 above is the denominator
7Yellow=first row we pick “Arts & Sciences” Red=column we pick next 23=numerator 57=denominator Birth OrderFirst or onlySecond or laterTotalArts and Sciences342357Agriculture524193Human Ecology152843Other121830113110223Answer: 23/57 = 40.35% or
8We 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 onlySecond or laterTotalArts and Sciences342357Agriculture524193Human Ecology152843Other121830113110223So here, the probability is 41/110, which equals , or 37.27%.
9Distinguishing 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.
10Absolute (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.
11So here’s what your table of probabilities should look like Birth OrderFirst or onlySecond or laterTotalArts and Sciences34 (34/223)15.25%23 (23/223)10.31%57Agriculture52 (52/223)23.32%41 (41/223)18.39%93Human Ecology15 (15/223)6.73%28 (28/223)12.56%43Other12 (12/223)5.38%18 (18/223)30113110223Note: it’s easier and faster to calculate this on Excel than with a calculator!
12Problems 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.
13Problem 14(c) Read problem, someone! Calculate probabilities of second childrenPick A&S (yellow)Birth OrderFirst or onlySecond or laterTotalArts and Sciences3423 (23/110)20.9%57Agriculture5241 (41/110)37.27%93Human Ecology1528 (28/110)25.45%43Other1218 (18/110)16.36%30113110223
14Problem 14(d) Pick 1st child column Calculate probability for AgricultureFirst or onlySecond or laterTotalArts and Sciences342357Agriculture52 (52/113)46.02%4193Human Ecology152843Other121830113110223
15Problem 14: birth order, take 2 First or onlySecond or laterTotalArts and Sciences342357Agriculture524193Human Ecology152843Other121830113110223
16Problem 14(e)What is the probability that an Agriculture student is a first oronly child? Answer: 52/93, or 55.91%Birth OrderFirst or onlySecond or laterTotalArts and Sciences342357Agriculture524193Human Ecology152843Other121830113110223
17Notation 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)
18Comparison of P(B|A) with language of problems (14(b)) 14(b): P(2nd child|Arts and Science)Among the Arts and Sciences students, what’s the probabilibty a student was a second child (or more)?
19Comparison of P(B|A) with language of problems (14(c)) 14(c): P(Arts & Science|2nd child)Among second children (or more), what’s the probability a student is enrolled in the Arts and Sciences?
20Comparison of P(B|A) with language of problems (14(d)) 14(d): P(Agriculture|1st or only child)What’s the probability that a first or only child is enrolled in the Agriculture College?
21Comparison of P(B|A) with language of problems (14(e)) 14(e): P(1st or only child|Agriculture)What is the probability that an Agriculture student is a first or only child?
22Problem 15Sick 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?
23Problem 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?123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100
24Problem 15: sick kids70% of patients have fever; 30% of those with fever have sore throat. % of both?123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100
25Problem 15: sick kids70% of patients have fever; 30% of those with fever have sore throat. % of both?123456730% of patients who do not have feverThis includes those who have sore throats but no fever8910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061626364656667686970
26Problem 15: sick kids70% of patients have fever; 30% of those with fever have sore throat. % of both?10%30% of patients who do not have feverThis includes those who have sore throats but no fever
27Problem 15: sick kids70% of patients have fever; 30% of those with fever have sore throat. % of both?30% of patients who do not have feverThis includes those who have sore throats but no feverRed=fever AND sore throat Blue= fever alone
28Problem 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 feverThis includes those who have sore throats but no fever123456789101112131415161718192021Red=fever AND sore throat Blue= fever alone
29How can we solve this faster? Answer: the General Multiplication Rule, i.e.,
30General 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.
31Problem 16Sick 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?
32Let’s draw a diagram (like problem 15) Percentage of carsCars needing repairs (20%)Cars that need no repairs (80%)
33Let’s draw a diagram (like problem 15) Need repairs Don’t need repairs>$100<$100Answer: 8 cells out of 100, or 8% (0.08)
34Redoing 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)
35Problem 24On 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.What is the probability that someone who has traveled to Mexico has visited Canada, too?Are travel to Mexico and Canada disjoint events? Explain.Are travel to Mexico and Canada independent events? Explain.
37Another approach Travel to Mexico Travel to Canada Yes No Total 0.04 0.140.180.050.770.820.090.911.00Travel to Canada
38Problem 24(a)What’s the probability that someone who has travelled to Mexico has travelled to Canada?Use either Venn diagram or tableTotal 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%
39Problem 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 bothTravel is not disjoint becauseYou can go to both4% of the population HAS been to both countries!
40Problem 24(c)Are travel to Mexico and Canada independent events? Explain.Independent=one event doesn’t alter the probability of the other occurringTravel is not independent because18% of all U.S. residents have been to Mexico.44.4% of those have also been to CanadaIf the events were independent, the probabilities would be equal
41Today’s lesson “Drawing without replacement” Usually done with decks of cards, but lots of other uses as wellExamples is on pp of text
42Visualizing the problem Draw a picture!3 Gold (desirable), 4 Silver (less desirable), 5 Wood (we don’t like)GoldSilverBrown
43You 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: 2/11SilverBrownGold
44Multiplication 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 =
45Problem 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.
46Example: 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.
47Second 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.
48Third 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)= = 14.7% (or about 1 out of 7 times)
49Homework for Tuesday Complete problem 17 and SHOW WORK. ProblemsRead Chapter 16 (it’s short, and we’ll start it tomorrow)Bring your Globe at Night data tomorrow: we need to enter it by Wednesday if you want extra credit.