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AP Statistics – Thursday December 3 Get in groups and work through the problems on this page together. The answers to the Review Problems will be online tonight. Quiz on Chapters 14 and 15 tomorrow. I. Check your homework against the answers in the back of the book if you haven’t already. Some of my solutions are below: II. Here’s another example of using tree diagrams. Go through this problem and make sure you understand it. (my solutions are on the next page) At Joe Schmoe University, 46% of the students are male. Among the male students, 34% are engineering majors, 40% are business majors, and the rest are liberal arts majors. Among the female students, 51% are engineering majors, 37% are business majors, and the rest are liberal arts majors. Draw a tree diagram to represent this situation. What is the probability that a randomly selected student is an engineering major? If we know that a student is a business major, what is the probability that he is a male? III. In chapter 14, events were always independent. In real life (and Chapter 15), they aren’t always independent. For example, if you are selecting items (such as cards) without replacing the card each time before you pick another, they are NOT independent. [think about the event ‘picking a red card’… every time I pick a card from the deck, the probability that the next card is red changes] Try these… remember that every time you pick a card, the denominator changes (my solutions are on the next page) You pick three cards at random from a standard deck. What is the probability that... a. the first heart you get is the third card dealt? b. your cards are all red? c. you get no spades? d. you have at least one ace? Solutions to parts II and III above Review Problems for Ch14/15 Quiz: 1. Are the events mutually exclusive? Roll two dice: sum is 6; doubles Randomly select a student at MHS: student is a freshman; student is in AP Calculus BC Randomly select a student at MHS: student takes English; student is in the Band Roll one die: a prime number; an even number Roll one die: an odd number; a number divisible by 3 Roll two dice: 6 on one die; sum is 5 2. Are the events independent? Roll one die and then flip a coin Roll one die 3 times Pick a card from a deck, do NOT put it back, and then pick another card Pick a card from a deck, put it back, and then pick another 3. If you roll two dice, find … a. P(a sum of 9) f. P(a sum of 8 and doubles) b. P(doubles) g. P(a sum of 8 | doubles) c. P(a sum of 9 or doubles) h. P(a sum of 9 | doubles) d. P(a sum of 8 or doubles) i. P(doubles | a sum of 8) e. P(a sum of 9 and doubles) 4. Roll a single die 3 times. Find… a. P(all rolls are even numbers) b. P(none are greater than or equal to 3) c. P(at least one number is a multiple of 3) d. P(the first 5 you get is on the third roll) 5. According to the Masterfoods company, the distribution of colors for M&M Mini’s is: 25% blue, 25% orange, 12% green, 13% yellow, 12% red, and the rest are brown. a. If you pick an M&M Mini at random, what is the probability that… i. it is brown? ii. a primary color (red, yellow, blue)? iii. it is not orange? b) If you pick three M&M Mini’s in a row, what is the probability that… i. they are all yellow? ii. none are red? iii. at least one is blue? iv. the third one is the first one that is brown? 6. Suppose you pick three cards in a row at random from a standard deck of 52 cards, without replacement. What is the probability that … i. you get all jacks? ii. you get no clubs? iii. you get at least one red card? 7. There are a total of 2.4 million drivers in the United States who own two-seater sports cars. Of them, 1.8 million are male, 2.1 million are single, and 1.6 million are male and single. Construct an appropriate diagram or table and then answer the questions below. What is the probability that a randomly selected driver of a two-seater sports car is… a. male, but not single? b. female? c. male or single? 8. A survey of families revealed that 58% of all families eat turkey at holiday meals, 44% eat ham, and 16% have both turkey and ham to eat at holiday meals. Make an appropriate diagram or table to represent the information above and then answer the questions. a. What is the probability that a family selected at random had neither turkey nor ham at their holiday meal? b. What is the probability that a family selected at random had ham but not turkey at their holiday meal? c. What is the probability that a randomly selected family that had turkey at their holiday meal also had ham? d. Are having turkey and having ham disjoint events? Explain in context. 9. For purposes of making on-campus housing assignments, a college classifies its students as Priority A (seniors), Priority B (juniors), and priority C (freshmen and sophomores). Of the students who choose to live on campus, 10% are seniors, 20% are juniors, and the rest are underclassmen. The most desirable dorm is the newly constructed Gold dorm, and 60% of the seniors living on campus elect to live there. Of the juniors living on campus, 15% of them live in the Gold dorm, along with only 5% of the on campus underclassmen. What is the probability that a randomly selected resident of the Gold dorm is a senior? Clearly show work to justify your answer. 10. 11. ANSWERS: 1. no 2. yes 7. a. 0.2/2.4=.083 yes yes b. 0.6/2.4=.25 no no c. 2.3/2.4=.958 no yes no 8. a. .14 yes b. .28 3. a. 4/36 f. 1/36 c. .16/.58=.2758 b. 6/36 g. 1/6 d. no.... 16% of c. 10/36 h. 0 families had both turkey and ham d. 10/36 i. 1/5 e. 0 9. 4. a. (1/2)(1/2)(1/2) = 1/8 b. (2/6)(2/6)(2/6) = 8/216 = 1/27 c. 1 - P(no mult of 3's) = 1 - (4/6)(4/6)(4/6) = 19/27 d. (5/6)(5/6)(1/6) = 25/216 5. a. i. .13 ii. .12 + .13 + .25 = .5 iii. 1 - .25 = .75 b. i. (.13)(.13)(.13) = .002197 ii. (.88)(.88)(.88) = .681472 10. a. 43/223 = .143 iii. 1 - P(none blue) b. 113/223 = .507 = 1 - (.75)(.75)(.75) = .578125 c. 15/223 = .067 iv. (.87)(.87)(.13) = .098397 d. 141/223 = .632 6. i. (4/52)(3/51)(2/50) = .00018 11. a. (16/20)(15/19) = .6316 ii. (39/52)(38/51)(37/50)= .41353 [not med , not med] iii. 1 - P(no red cards) b. (16/20)(15/19)(4/18) = .1404 1 - (26/52)(25/51)(24/50) = .88235 [not med, not med, med]

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AP Statistics, Advanced Placement, Placement Statistics, statistics course, the College Board, graphing calculator, Sampling Distributions, AP Exam, Data Analysis, Standard Deviation

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posted: | 4/30/2010 |

language: | English |

pages: | 4 |

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