ISM Chapter 07

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

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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 4) = P(4) + P(5) = .06 + .02 = .08 157
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7.8 a P(2 X 5) = P(2) + P(3) + P(4) + P(5) = .310 + .340 + .220 + .080 = .950 P(X > 5) = P(6) + P(7) = .019 + .001 = .020 P(X < 4) = P(0) + P(1) + P(2) + P(3) = .005 + .025 + .310 + .340 = .680 b. b. E(X) = ) x ( xP = 0(.005) + 1(.025) + 2(.310) + 4(.340) +5(.080) + 6(.019) + 7(.001) = 3.066 c. 2 σ = V(X) = μ - ) x ( P ) x ( 2 = (0–3.066) 2 (.005) + (1–3.066) 2 (.025) + (2–3.066) 2 (.310) + (3–3.066) 2 (.340) + (4–3.066) 2 (.220) + (5–3.066) 2 (.080) + (6–3.066) 2 (.019) + (7–3.066) 2 (.001) = 1.178 σ = 178 . 1 2 = σ = 1.085 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 1) = P(2) + P(6) + P(8) = .3 + .4 + .1 = .8 c P(X 2) = P(2) + P(6) + P(8) = .3 + .4 + .1 = .8 d P(2 X 5) = P(2) = .3 7.11a P(3 X 6) = P(3) + P(4) + P(5) + P(6) = .04 + .28+ .42 + .21 = .95 b. P(X > 6) = P(X 7) = P(7) + P(8) = .02 + .02 = .04 c. P(X < 3) = P(X 2) = P(0) + P(1) + P(2) = 0 + 0 + .01 = .01 7.12 P(Losing 6 in a row) = 6 5 . = .0156 7.13 a P(X < 2) = P(0) + P(1) = .05 + .43 = .48 b P(X > 1) = P(2) + P(3) = .31 + .21 = .52 7.14 158
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a P(HH) = .25 b P(HT) = .25 c P(TH) = .25 d P(TT) = .25 7.15 a P(0 heads) = P(TT) = .25 b P(1 head) = P(HT) + P(TH) = .25 + .25 = .50 c P(2 heads) = P(HH) = .25 d P(at least 1 head) = P(1 head) + P(2 heads) = .50 + .25 = .75 7.16 7.17 a P(2 heads) = P(HHT) + P(HTH) + P(THH) = .125 + .125 + .125 = .375 b P(1 heads) = P(HTT) + P(THT) = P(TTH) = .125 + .125 + .125 = .375 c P(at least 1 head) = P(1 head) + P(2 heads) + P(3 heads) = .375 + .375 + .125 = .875 d P(at least 2 heads) = P(2 heads) + P(3 heads) = .375 + .125 = .500 7.18a.
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This homework help was uploaded on 04/18/2008 for the course STAT 212 taught by Professor Holt during the Spring '08 term at UVA.

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

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