Chapter 07 - 112 Chapter 7 7.1 a 0, 1, 2, b Yes, we can...

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112 Chapter 7 7.1 a 0, 1, 2, & b Yes, we can identify the first value (0), the second (1), and so on. c It is finite, because the number of cars is finite. d The variable is discrete because it is countable. 7.2 a any value between 0 and several hundred miles b No, because we cannot identify the second value or any other value larger than 0. c No, uncountable means infinite. d The variable is continuous. 7.3 a The values in cents are 0 ,1 ,2, & b Yes, because we can identify the first ,second, etc. c Yes, it is finite because students cannot earn an infinite amount of money. d Technically, the variable is discrete. 7.4 a 0, 1, 2, &, 100 b Yes. c Yes, there are 101 values. d The variable is discrete because it is countable. 7.5 a No the sum of probabilities is not equal to 1. b Yes, because the probabilities lie between 0 and 1 and sum to 1. c No, because the probabilities do not sum to 1. 7.6 p(x) = 1/6 for x = 1, 2 ,3 ,4 ,5, and 6 7.7 a x p(x) 0 24,750/165,000 = .15 1 37,950/165,000 = .23 2 59,400/165,000 = .36 3 29,700/165,000 = .18 4 9,900/165,000 = .06 5 3,300/165,000 = .02 b (i) P(X £ 2) = p(0) + p(1) + p(2) = .15 + .23 + .36 = .74 (ii) P(X > 2) = p(3) + p(4) + p(5) = .18 + .06 + .02 = .26 (iii) P(X ³ 4) = p(4) + p(5) = .06 + .02 = .08 113 7.8 a P(2 £ X £ 5) = p(2) + p(3) + p(4) + p(5) = .310 + .340 + .220 + .080 = .950 b P(X > 5) = p(6) + p(7) = .019 + .001 = .020 c P(X < 4) = p(0) + p(1) + p(2) + p(3) = .005 + .025 + .310 + .340 = .680 7.9 p(0) = p(1) = p(2) = . . . = p(10) = 1/11 = .091 7.10 a P(X > 0) = p(2) + p(6) + p(8) = .3 + .4 + .1 = .8 b P(X ³ 1) = p(2) + p(6) + p(8) = .3 + .4 + .1 = .8 c P(X ³ 2) = p(2) + p(6) + p(8) = .3 + .4 + .1 = .8 d P(2 £ X £ 5) = p(2) = .3 7.11 P(X ³ 2) = p(2) + p(3) = .4 + .2 = .6 7.12 a P(X < 2) = p(0) + p(1) = .05 + . .43 = .48 b P(X > 1) = p(2) + p(3) = .31 + .21 = .52 7.13 a P(HH) = .25 b P(HT) = .25 c P(TH) = .25 d P(TT) = .25 7.14 a P(0 heads) = P(TT) = 1/4 b P(1 head) = P(HT) + P(TH) = 1/4 + 1/4 = 1/2 c P(2 heads) = P(HH) = 1/4 d P(at least 1 head) = P(1 head) + P(2 heads) = 1/2 + 1/4 = 3/4 114 7.15 7.16 a P(2 heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8 b P(1 heads) = P(HTT) + P(THT) = P(TTH) = 1/8 + 1/8 + 1/8 = 3/8
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c P(at least 1 head) = P(1 head) + P(2 heads) + P(3 heads) = 3/8 + 3/8 + P(HHH) = 3/8 + 3/8 + 1/8 = 7/8 d P(at least 2 heads) = P(2 heads) + P(3 heads) = 3/8 + 1/8 = 4/8 = 1/2 7.17 a P(X > 4) = p(5) + p(6) + p(7) = .20 + .10 + .10 = .40 b P(X ³ 2) = 1- P(X £ 1) = 1 = p(1) = 1 - .05 = .95 7.18 a P(4 books) = p(4) = .06 b P(8 books) = p(8) = 0 c P(no books) = p(0) = .35 d P(at least 1 book) = 1 ' p(0) = 1 - .35 = .65 7.19 a P(X ³ 20) = p(20) + p(25) + p(30) + p(40) + p(50) + p(75) + P(100) = .08 + .05 + .04 + .04 + .03 + .03 + .01 = .28 b P(X = 60) = 0 c P(X > 50) = P(75) + P(100) = .03 + .01 = .04 d P(X > 100) = 0 7.20 a P(X = 3) = p(3) = .21 b b P(X ³ 5) = p(5) + p(6) + p(7) + p(8) = .12 + .08 + .06 + .05 = .31 c P(5 £X £ 7) = p(5) + p(6) + p(7) = .12 + .08 + .06 = .26
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This note was uploaded on 06/06/2011 for the course ADMS 2320 taught by Professor Rochon during the Spring '08 term at York University.

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Chapter 07 - 112 Chapter 7 7.1 a 0, 1, 2, b Yes, we can...

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