Stat 198 for 134 Instructor: Mike Leong

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1 Chapter 2: Repeated Trials ad Samplig Sectio 2.1 Biomial Distributio 2.2 Normal Approximatio: Method 2.3 Normal Approximatios: Derivatio (Skip) 2.4 Poisso Approximatio 2.5 Radom Samplig Chapter 2 Table of Cotets Chapter 2: Repeated Trials ad Samplig... 1 Chapter 2 Table of Cotets... 1 Chapter 2 Outlies Outlie: Biomial Distributio Outlie: Normal Approximatio: Method Outlie: Poisso Approximatio Outlie: Radom Samplig Chapter 2 Worksheets Worksheet: Biomial Distributio Worksheet: Normal Approximatio: Method Worksheet: Poisso Approximatio Worksheet: Radom Samplig Chapter 2 Aswers Aswers: Biomial Distributio Aswers: Normal Approximatio: Method Aswers: Poisso Approximatio Aswers: Radom Samplig Writte by Mike Leog, mleog@berkeley.edu Page 1 of 24 Chapter 2: Repeated Trials ad Samplig

2 Chapter 2 Outlies Sectio 2.1 Biomial Distributio Normal Approximatio: Method Normal Approximatios: Derivatio (Skip) Poisso Approximatio Radom Samplig... Writte by Mike Leog, mleog@berkeley.edu Page 2 of 24 Chapter 2: Repeated Trials ad Samplig

3 2.1 Outlie: Biomial Distributio Coutig factorial! = ( 1) 1 Ex: 3! = 6 order k terms () k = ( 1) ( k + 1) k terms =! ( k)! Ex: (10) 3 = = 720 choose k k =!, k = 0, 1,, k! ( k)! Ex: 10 10! = 3 3! 7! = 720 k = () k, k = 0, 1,, k! Ex: 10 3 = (10) 3 = 120 3! Values of k such that choose k evaluates to 0 = 0 if k 0, 1,, k Ex: 10 = 0, = 0 Symmetric idetity of choose k = k k Ex: 10 3 = 10 7 Special cases of choose k 0 = = 1 = 1 choose k 1,, k r Suppose k k r =.! k 1, k 2,, k =, r k 1! k 2! k r k 1, k 2,, k = r k k 1 k 1 k 2 k 1 k r 1 1 k 2 k 3 k r 10 Ex: 2, 3, 1, 4 = 10! 2! 3! 1! 4! = Ex: 2, 3, 1, 4 = = Writte by Mike Leog, mleog@berkeley.edu Page 3 of 24 Chapter 2: Repeated Trials ad Samplig

4 Pascal s Triagle (tilted) k = 0 k = 1 k = 2 k = 3 k = 4 k = 5 = = 1 = = = 1 = = = = 1 = = = = = 1 = = = = = = 1 = = = = = = = 1 Relatioships of Cosecutive Biomial Coefficiets Relatioships Let k = = k k + 1 k = 5 3 k = k k = k 1 k = k = 4 2 Iterpretatio Symbol Expressio Number! factorial # of ways to order distict elemets () k order k # of ways to order distict elemets i k positios k choose k elemets # of ways to order elemets where you have k elemets # of ways to choose k uordered elemets from distict k 1, k 2,, k, r choose k 1, k 2,, k r of oe type ad k elemets of aother type # of ways to choose k 1, k 2,, k r uordered elemets from distict elemets # of ways to order elemets where there are r types ad k i elemets of type i Writte by Mike Leog, mleog@berkeley.edu Page 4 of 24 Chapter 2: Repeated Trials ad Samplig

5 Permutatios 1! = 1 2! = 2 3! = 6 4! = 24 1 A 1 AB 1 ABC 1 ABCD 7 BACD 13 CABD 19 DABC 2 BA 2 ACB 2 ABDC 8 BADC 14 CADB 20 DACB 3 BAC 3 ACBD 9 BCAD 15 CBAD 21 DBAC 4 BCA 4 ACDB 10 BCDA 16 CBDA 22 DBCA 5 CAB 5 ADBC 11 BDAC 17 CDAB 23 DCAB 6 CBA 6 ADCB 12 BDCA 18 CDBA 24 DCBA Combiatios ad Number of Arragemets A B C D E A B C D E A B A C A D A E B C B D B E C D C E D E A B C A B D A B E A C D A C E A D E B C D B C E B D E C D E A B C D A B C E A B D E A C D E B C D E A B C D E ! = 0 0! 5! = 1 5 5! = 1 1! 4! = 5 5 5! = 2 2! 3! = ! = 3 3! 2! = ! = 4 4! 1! = 5 5 5! = 5 0! 5! = 1 Writte by Mike Leog, mleog@berkeley.edu Page 5 of 24 Chapter 2: Repeated Trials ad Samplig

6 Biomial Distributio Biomial Formula (a + b) = 2 = k k=0 k=0 k ak b k 0 = ( 1) k k k=0 Biomial Distributio Let X = # of successes i idepedet trials, where p = probability of success for each trial. X ~ BBB(, p) o {0, 1,, } P(X = x) = x px (1 p) x Coditios to use the Biomial Distributio i) is fixed. ii) p is costat. iii) Trials are idepedet. Cosecutive Odds Ratio P(X = x) R(x) = P(X = x 1) R(x) > 1 R(x) > 1 R(x) < 1 R(x) < 1 R(x) <1 R(x) > 1 R(x) > 1 R(x) = 1 R(x) < 1 R(x) < 1 To fid the mode, look for the largest x such that R(x) 1. Biomial Cosecutive Odds Ratio P(X = x) x + 1 R(x) = = p P(X = x 1) x q, x = 1, 2,, Mode Let m = + p. m, + p Z Mode(X) + = m, m 1, + p Z + Multiomial Distributio: Extesio of the Biomial Distributio X ~ MMMMMM(, p 1, p 2,, p k ) x P(X = x) = x 1, x 2,, x p 1 x k 1 p 2 x 2 p k k x = (x where 1, x 2,, x k ) x 1 + x x k = Writte by Mike Leog, mleog@berkeley.edu Page 6 of 24 Chapter 2: Repeated Trials ad Samplig

7 2.2 Outlie: Normal Approximatio: Method Stadard Normal Z ~ N(0, 1) φ(z) = 1 2π e 1 2 z2 The iflectios are at: z = ±1 Symmetry: φ( z) = φ(z) Iflectio poits 1 2π Cumulative Stadard Normal z Φ(z) = 1 2π e 1 2 t2 dd Middle Area of the Stadard Normal P( 1 < Z < 1) 68%, P( 2 < Z < 2) 95%, P( 3 < Z < 3) 99.7% Commet: More accurately, P( 2 < Z < 2) 95.4% ad P( 1.96 < Z < 1.96) 95% Areas of a Stadard Normal Curve P(Z < z) = Φ(z) P(Z > z) = 1 Φ(z) P(a < Z < b) = Φ(b) Φ(a) P( z < Z < z) = 2Φ(z) 1 Normal X ~ N(μ, σ 2 ) φ(z) = 1 2πσ e 1 2 x μ σ 2 = 1 2πσ 2 e (x μ) 2 2σ 2 Writte by Mike Leog, mleog@berkeley.edu Page 7 of 24 Chapter 2: Repeated Trials ad Samplig

8 Square Root Law for Biomial Suppose X ~ BBB(, p). SS(X) = (1 p) Let p = 1 X. p(1 p) SS(p ) = Normal Approximatio to the Biomial X ~ Bi(, p) o {0, 1,, }, µ = E ( X ) = p, σ = SD( X ) = p(1 p) 1 1 b + 2 µ a Φ Φ 2 µ P ( a X b) σ σ Coditios whe the Normal Approximatio to the Biomial is appropriate Large σ: 3 As icreases, the σ icreases. As p gets closer to 0.5 from either directio (less skewed), the σ icreases. Law of Large Numbers (Weak Law of Numbers) ε > 0, lim P( p p < ε) = 1 Higher Orders of the Stadard Normal φ (z) = zz(z) (2.2-15a) φ (z) = (z 2 1)φ(z) (2.2-15b) φ (z) = ( z 3 + 3z)φ(z) (2.2-16a) Writte by Mike Leog, mleog@berkeley.edu Page 8 of 24 Chapter 2: Repeated Trials ad Samplig

9 2.4 Outlie: Poisso Approximatio Taylor Series of e x e x = x + 1 1! 2! x2 + Poisso Distributio X ~ PPPP(μ) o {0, 1, 2, } μx μ P(X = x) = e x! P(X = 0) = e μ P(X 1) = 1 e μ Commet: The rage of a Biomial distributio is x = 0, 1, 2,,. I a Poisso, the rage is from 0 to. For a small μ, the Poisso probabilities are stacked at the lower values of x. This is a skewed right distributio. Poisso Approximatio to the Biomial (Guidelies) Ideal for large ad small p where μ = 3 However, if is too large ad p is ot small eough such that μ = > 9, the use a ormal approximatio. Cosecutive Odds Ratio P(X = x) R(x) = P(X = x 1) = μ, x = 1, 2, x Mode Let m = μ. m, m Z Mode(X) = + m, m 1, m Z + Coutig the Complemet If is large ad p is large, cout the complemet evet. Writte by Mike Leog, mleog@berkeley.edu Page 9 of 24 Chapter 2: Repeated Trials ad Samplig

10 2.5 Outlie: Radom Samplig Hypergeometric Distributio X ~ HHHHHH(, N, G) o {max(0, + G N),, mi(, G)} G G N P(X = x) = x x N G x 0 G x x mi(, G) x 0 x x 0 N G + x 0 x 0 x max(0, + G N) x + G N Relatig the Hypergeometric Probability to a Sequece of Probabilities G g B b P(X = x) = N = g (G) g(b) b (N) where G + B = N, g + b = Calculatig the Hypergeometric probability from either directio i a 2 by 2 table. Good Bad Sample x x Ur G x B + x or N N G + x G B N I a sample of, we wat x good elemets ad x bad elemets. G g B b P(X = x) = N Of the G good elemets, assig x to the sample ad G x to the ur. N x P(X = x) = G x N G Extesio of the Hypergeometric Distributio N 1 N 2 N k x P(X = x) = 1 x 2 x k N x = (x 1, x 2,, x k ) where x 1 + x x k = N 1 + N N k = N Writte by Mike Leog, mleog@berkeley.edu Page 10 of 24 Chapter 2: Repeated Trials ad Samplig

11 Chapter 2 Worksheets Biomial Distributio Normal Approximatio: Method Poisso Approximatio Radom Samplig Writte by Mike Leog, mleog@berkeley.edu Page 11 of 24 Chapter 2: Repeated Trials ad Samplig

12 2.1 Worksheet: Biomial Distributio 1. How may ways are there to arrage the followig elemets? a) ABCDE b) AABBB c) ABCDEFGHIJ d) AABBBCCCCD 2. Out of 10 studets, how may ways are there to cout the followig? a) Select 3 studets as officers. b) Select 3 studets as officers where there is a presidet, a vice-presidet, ad a treasurer. c) Assig umerical scores from 1, 2,, 10 to all 10 studets ad there are o ties. d) Assig grades of 2 A s, 3 B s, 3 C s, 1 D ad 1 F. 3. Out of 10 people, how may ways are there to choose the followig? Assume all groups formed are distiguishable. a) Form a group of 3. b) Form 2 groups of 5. c) Form 2 groups of 3, a group of 2, ad 2 groups of Out of 32 people, how may ways are to form idistiguishable groups? a) Form a group of 15 ad a group of 17. b) Form 2 groups of 16 each. c) Form 3 groups of 2, 4 groups of 5, ad a group of Toss a p-coi 5 times ad record the outcome of each trial. Fid the probability of the followig evets. a) HHHTT b) HTHTH c) 3 heads d) At least 1 head e) Not all heads f) A eve umber of heads 6. Roll a fair die 10 times. Let X be the umber of oes. Fid: a) average of X b) mode of X c) modes of X if the die is rolled 11 times 7. Suppose X ~ BBB(, p). a) Prove the mea of a biomial distributio. b) Prove the cosecutive odds ratio (or ratio of cosecutive probabilities) of a biomial distributio. c) Prove the mode of a biomial distributio. Writte by Mike Leog, mleog@berkeley.edu Page 12 of 24 Chapter 2: Repeated Trials ad Samplig

13 8. Simplify each of the followig summatio expressios. Also match each summatio to a problem that ivolves rollig a die 10 times. a) x px (1 p) x x=0 b) x px (1 p) x x=1 c) x p 1 x p 2 x x=0 1 d) x p 1 x p 2 x x=1 A. Fid the probability that you get at least 1 six. B. Fid the probability that you get oes ad twos ad o other values i your 10 trials. C. Fid the probability that you get either a oe or a two o each trial. D. Fid the probability that you get 0 to 10 sixes. 9. A licese plate is made of 3 letters ad 5 umbers. A letter or umber ca be used more tha oce uless the umbers ad letters are specified. Fid the umber of licese plates that ca be created uder each coditio. a) give 8 umbers ad letters: ABC12345 b) give 8 umbers ad letters: AAB11122 c) 3 give letters, ABC, followed by the 5 umbers, d) 3 give letters, AAB, followed by the 5 umbers, e) 3 letters followed by 5 umbers f) 3 uique letters followed by 5 uique umbers g) 3 uique letters i alphabetical order followed by 5 uique umbers i ascedig order h) 3 letters ad 5 umbers (Ex S777E86E) i) 3 uique letters ad 5 uique umbers (Ex S123L98C) j) 3 uique letters i alphabetical order ad 5 uique umbers i ascedig order (Ex A135C68E) 10. Roll a die 20 times. Let X = # of sixes i the 20 trials. Let X 1 = # of sixes i the first 5 trials, X 2 = # of sixes i the last 15 trials, ad X 3 = # of sixes i the last 19 trials. Fid: a) P(4 sixes) b) P(4 sixes 1 ss trial is a six) c) P(4 sixes at least 1 six) d) P(2 sixes i the first 5 trials 1 ss five showed up o the 6 th trial) e) P(4 sixes at least 1 five) f) P(1 six i the first 5 trials ad 3 sixes i the last 15 trials) g) P(1 six i the first 5 trials 4 sixes) h) P(more tha 3 sixes i the 1 ss 5 trials ad more tha 5 sixes total) 11. Roll a die util you see 3 sixes. Let X = # of trials whe you stop. a) P(X = 10) b) P(X > 10) 12. Roll a fair die 12 times. Let X be the umber of oes, Y be the umber of twos or threes, ad Z be the umber of fours, fives, or sixes. Fid: a) P(2 oes, 3 twos or threes, 7 fours, fives, or sixes) b) P(2 oes give there are o twos ad o threes) Writte by Mike Leog, mleog@berkeley.edu Page 13 of 24 Chapter 2: Repeated Trials ad Samplig

14 2.2 Worksheet: Normal Approximatio: Method 1. Suppose Z ~ N(0, 1). Give Φ(1.28) = 0.90, fid the followig: a) Φ( 1.28) b) P( 1 < Z < 2) 2. Roll a die 600 times. Let X = # of sixes. Approximate the followig probabilities with a ormal distributio. a) P(more tha 120 sixes) b) P(more tha 20% sixes) c) P(betwee 90 ad 115 sixes iclusive) 3. A studet guesses o all 100 questios of a multiple choice exam. There are 5 choices per questio, oly 1 of which is correct. Let X = # of questios that he guesses correctly. Use the followig methods to fid the probability he gets 20 correct. a) Biomial b) Cumulative desity fuctio (CDF) of ormal curve c) Normal curve desity fuctio d) Stirlig s Approximatio to the Biomial (optioal)! 2ππ e 4. A studet guesses o all 100 questios of a multiple choice exam. There are 5 choices per questio, oly 1 of which is correct. Suppose the studet gets 4 poits for each correct aswer ad loses 1 poit for each wrog aswer. Let Y = # of total poits. Use the followig methods to fid the probability he gets 0 poits. a) Cumulative desity fuctio (CDF) of ormal curve b) Normal curve desity fuctio 5. Roll a die 600 times. Let X = # of sixes. The probability of gettig 100 ± d sixes is approximately 90%. Fid d. 6. Roll a die util you see 100 sixes. Let X = # of trials whe you stop. Fid the probability exactly usig a computer program, ad approximate with the ormal distributio. a) P(X = 613) b) P(X > 612) Writte by Mike Leog, mleog@berkeley.edu Page 14 of 24 Chapter 2: Repeated Trials ad Samplig

15 2.4 Worksheet: Poisso Approximatio 1. Derive the followig properties of the Poisso distributio. a) Prove the sum of all the probabilities over its rage is 1. μx μ e = 1 x=0 x! b) Derive the Poisso distributio usig the limit defiitio of e. e = lim c) Derive the Poisso distributio usig cosecutive odds ratio. R(x) = P(x) P(x 1) = μ, x = 1, 2, x 2. I a statistics class of 1000 studets, each studet will flip a coi 10 times. Fid the followig probabilities exactly usig the biomial distributio ad approximately usig the Poisso distributio. a) P(o oe will get all heads) b) P(at least 1 studet will get all heads) c) P(3 studets will get all heads) d) P(less tha 3 studets will get all heads) 3. Prove the mode of a Poisso distributio has the followig properties. There is either a sigle mode or a double mode. A double mode occurs whe μ is a iteger. Let m = μ. m, m Z Mode(X) = + m, m 1, m Z + 4. Suppose that every day that this circuit system is tured o is cosidered a idepedet trial. Also assume the circuits are idepedet. Let p i be the probability that the i th circuit is workig. I a year, fid the probability that the system is workig for more tha 361 days. p 1 = 0.9 p 2 = 0.8 p 3 = 0.7 Writte by Mike Leog, mleog@berkeley.edu Page 15 of 24 Chapter 2: Repeated Trials ad Samplig

16 2.5 Worksheet: Radom Samplig 1. How may ways are there to choose 3 elemets from {ABCDEFGHIJ} uder each coditio? (The order does ot matter.) a) A elemet ca be chose oce. b) A elemet ca be chose more tha oce. There are 3 of a kid. c) A elemet ca be chose more tha oce. There are 1 of a kid ad 2 of a kid. d) A elemet ca be chose more tha oce. There are 3 distict elemets (same as part a)). e) A elemet ca be chose more tha oce. 2. Draw cards without replacemet from a stadard deck. Fid the probability distributio of X ad state its rage. a) Draw 5 cards from a deck. Let X = # of hearts. b) Draw 5 cards from a deck. Let X = # of aces. c) Draw 20 cards from a deck. Let X = # of umeric cards. A umeric card is defied to be ay card with a rak from a ace through te. d) Draw 40 cards from a deck. Let X = # of hearts. 3. I a ur of 40 gree ad 60 blue marbles, draw 8 marbles without replacemet. a) Place the appropriate umbers ito a 2 by 2 table. Gree Blue Sample x x Ur G x N G + x N G B N b) Fid the probability of gettig 3 gree marbles. 4. I a ur of 40 gree, 60 blue, ad 50 red marbles, draw 12 marbles without replacemet. a) Place the appropriate umbers ito a 2 by 3 table. Gree Blue Red Sample x y z Ur G x B y R z N G B R N b) Fid the probability of gettig 3 gree, 5 blue, ad 4 red marbles. 5. Deal i cards from a stadard deck to 4 players. Let X i = # of hearts that the i th player gets. Fid: Player 1 Player 2 Player 3 Player 4 Hearts x 1 = 2 x 2 = 3 x 3 = 1 x 4 = 7 G = 13 No-Hearts 1 x 1 = 8 2 x 2 = 17 3 x 3 =5 4 x 4 = 9 B = 39 1 = 10 2 = 20 3 = 6 4 = 16 N = 52 a) P(X 1 = 2) b) P(X 1 = 2, X 2 = 3, X 3 = 1, X 4 = 7) Writte by Mike Leog, mleog@berkeley.edu Page 16 of 24 Chapter 2: Repeated Trials ad Samplig

17 6. Out of a total of N = 52 raffle tickets, each of 4 players buys 13 tickets. There are 3 prizes. Fid: a) P(1 triple wier) b) P(1 sigle wier ad 1 double wier) c) P(3 sigle wiers) 7. There are 4 wiig tickets out of 200. Distribute the 200 tickets evely to 10 idividuals. Let X = # of idividuals that have wo. a) Fid the distributio of X. b) If there are exactly 2 wiers, fid the probability they have the same umber of wiig tickets. 8. Draw = 10 cards from a deck. Fid the probability of: a) o large cards {Large cards: 10, J, Q, K, A}; b) at least oe ace, but o other large cards; c) o aces, but at least 1 other large card; d) at most oe kid of large card. 9. I a poker had of = 5 cards, fid the followig probabilities. (Deal 5 cards without replacemet.) a) P(3 of a kid, 2 of a kid) {raks: 3a, 2b} b) P(2 pairs) {raks: 2a, 2b, c} 10. I a poker had of = 7 cards, fid the followig probabilities. a) P(4 of a kid, 3 of a kid) {raks: 4a, 3b} b) P(4 of a kid, a pair) {raks: 4a, 2b, c} c) P(3 of a kid, 2 pairs) {raks: 3a, 2b, 2c} d) P(2 pairs) {raks: 2a, 2b, c, d, e} 11. Deal 13 cards each from a stadard deck to 4 players. Fid the probability of each evet: a) each player has a ace b) 1 player has a pair of aces, ad 2 other players have 1 each c) exactly 2 players have a pair of aces d) 1 player has 3 aces ad aother player has 1 ace Writte by Mike Leog, mleog@berkeley.edu Page 17 of 24 Chapter 2: Repeated Trials ad Samplig

18 Chapter 2 Aswers Biomial Distributio Normal Approximatio: Method Poisso Approximatio Radom Samplig Writte by Mike Leog, mleog@berkeley.edu Page 18 of 24 Chapter 2: Repeated Trials ad Samplig

19 2.1 Aswers: Biomial Distributio 1. a # = 5! b # = 5! 2! 3! = 5 2 = c # = 10! d # = 2, 3, 4, 1 2. a # = 10 3 b # = (10) 3 10 c # = 10! d # = 2, 3, 3, 1, 1 3. a # = 10 3 b # = 10 5 c 10 # = 3, 3, 2, 1, 1 4. a # = b # = 1 2! c # = 1 3! 4! 32 2, 2, 2, 5, 5, 5, 5, 6 5. a P(HHHTT) = p 3 q 2 b P(HTHTH) = p 3 q 2 c P(X = 3) = 5 3 p3 q 2 d P(X 1) = 1 q 5 e P(X 5) = 1 p 5 f 6. a μ = 10 6 c mode(x) = 2, 1 P(X {2, 4, 6}) = 5 0 q p2 q p4 q b mode(x) = 1 7. a μ = b R(x) = c m, + p Z Mode(X) + = m, m 1, + p Z + x + 1 x p q 8. a x px (1 p) x = 1, D b x px (1 p) x = 1 q, A x=0 c x p 1 x x p 2 x=0 d x p 1 x x p 2 = (p 1 + p 2 ) x=1, C = (p 1 + p 2 ) p 1 p 2, B 9. a # = 8! b 8 8! # = = 2, 1, 3, 2 2! 3! 2! c # = 3! 5! d # = 3! 2! 5! 3! e # = f # = (26) 3 (10) 5 g # = h # = 8! 3! 5! x=1 1 i # = 8! 3! 5! (26) 3 (10) 5 = 8! j # = 8! 3! 5! Writte by Mike Leog, mleog@berkeley.edu Page 19 of 24 Chapter 2: Repeated Trials ad Samplig

20 10. a P(X = 4) = b P(X = 4 I 1 = 1) = c 20 P(X = 4 X 1) = d P(X 1 = 2 W = 6) = e P(X = 4 T 1 1) 20 = g P(X 1 = 1 X = 4) = f P(X 1 = 1, X 2 = 3) = h P(X 1 4, X 6) = a P(X = 10) = b P(X > 10) = a P(X = 2, Y = 3, Z = 7) = 12 2, 3, b P(X = 2 Y = 0) = Writte by Mike Leog, mleog@berkeley.edu Page 20 of 24 Chapter 2: Repeated Trials ad Samplig

21 2.2 Aswers: Normal Approximatio: Method 1. a Φ( z) = 0.10 b P( 1 < Z < 2) = Φ(2) Φ( 1) a P(X > 120) 1 Φ(2.246) b P(p > 0.2) 1 Φ(2.246) c P(90 X 115) Φ(1.70) Φ( 1.15) a P(X = 20) = b P(X = 20) 2Φ c P(X = 20) 2π d P(X = 20) 2π a P(Y = 0) 2Φ b P(Y = 0) 1 8 2π d = 15 P(X = 613) = a P(X = 613) Φ Φ P(X > 612) = 612 b y 1 6 y y y=0 P(X > 612) Φ 2.5 = Writte by Mike Leog, mleog@berkeley.edu Page 21 of 24 Chapter 2: Repeated Trials ad Samplig

22 2.4 Aswers: Poisso Approximatio μx μ x=0 x! 1. a e 2. a 3 = 1 b P(X = x) = μx x! e μ c R(x) = P(x) P(x 1) = μ, x = 1, 2, x c d P(X = 0) = b P(Y = 0) = e 1000/ P(X = 3) = P(Y = 3) = e 1000/1024 (1000/1024) ! 2 P(X < 3) = i 1024 i i P(Y < 3) = i=0 2 i=0 (1000/1024)i 1000/1024 e i! Let m = μ. m, m Z Mode(X) = + m, m 1, m Z + 4 P(X > 361) e ! ! ! P(X 1) = P(Y 1) = 1 e 1000/ Writte by Mike Leog, mleog@berkeley.edu Page 22 of 24 Chapter 2: Repeated Trials ad Samplig

23 2.5 Aswers: Radom Samplig 1. a # = 120 b # = 10 c # = 90 d # = 120 e # = a P(X = x) = x 5 x , x = 0, 1,, 5 b P(X = x) = x 5 x 5 52, x = 0, 1,, 4 5 c P(X = x) = x 20 x 52, x = 8, 9,, 20 d P(X = x) = x 40 x 52, x = 1, 2,, a Sample Ur Gree x = 3 G x = 37 G = 40 Blue x = 5 N G + x = 55 B = 60 = 8 N = 92 N = b P(X = 3) = a Gree Blue Red Sample x y z = 12 Ur G x = 40 x B y = 60 y R z = 50 z N = 138 G = 40 B = 60 R = 50 N = b P(X = 3, Y = 5, Z = 4) = 3 5. a P(X 1 = 2) 13 = = b = P(X 1 = 2, X 2 = 3, X 3 = 1, X 4 = 7) 13 = a P(3A) = b P(A, 2B) = (4) c P(A, B, C) = Writte by Mike Leog, mleog@berkeley.edu Page 23 of 24 Chapter 2: Repeated Trials ad Samplig

24 7. a x P(X = x) b P(same umber of wiig tickets X = 2) = a P(X + Z = 0) = b P(X 1, Z = 0) = c P(X = 0, Z 1) = d P(X + Z = 0) + 5P(X 1, Z = 0) = P(3 A, 2 B) P(2 A, 2 B, 1 C) 9. a = (13) b = a P(4 A, 3 B) b P(4 A, 2 B, 1 C) = (13) = (13) c P(3 A, 2 B, 2 C) d P(2 A, 2 B, 1 C, 1 D, 1 E) = = a 13 b P(X 1 = 1, X 2 = 1, X 3 = 1, X 4 = 1) = P(2 A, 1 B, 1 C) = c P(2 A, 2 B) = d P(3 A, 1 B) = (4) Writte by Mike Leog, mleog@berkeley.edu Page 24 of 24 Chapter 2: Repeated Trials ad Samplig

Math for Liberal Studies

Math for Liberal Studies We want to measure the influence each voter has As we have seen, the number of votes you have doesn t always reflect how much influence you have In order to measure the power of