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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
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. μ = E(X) = ) x ( xP = –2(.59) +5(.15) + 7(.25) +8(.01) = 1.40 2 σ = V(X) = μ - ) x ( P ) x ( 2 = (–2–1.4) 2 (.59) + (5–1.4) 2 (.15) + (7–1.4) 2 (.25) + (8–1.4) 2 (.01) = 17.04 159

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b. x –2 5 7 8 y –10 25 35 40 P(y) .59 .15 .25 .01 c. E(Y) = ) y ( yP = –10(.59) + 25(.15) + 35(.25) + 40(.01) = 7.00 V(Y) = μ - ) y ( P ) y ( 2 = (–10–7.00) 2 (.59) + (25–7.00) 2 (.15) + (35–7.00) 2 (.25) + (40–7.00) 2 (.01) = 426.00 d. E(Y) = E(5X) = 5E(X) = 5(1.4) = 7.00 V(Y) = V(5X) = 5 2 V(X) = 25(17.04) = 426.00. 7.19a μ = E(X) = ) x ( xP = 0(.4) + 1(.3) + 2(.2) + 3(.1) = 1.0 2 σ = V(X) = μ - ) x ( P ) x ( 2 = (0–1.0) 2 (.4) + (1–1.0) 2 (.3) + (2–1.0) 2 (.2) + (3–1.0) 2 (.1) = 1.0 σ = 0 . 1 2 = σ = 1.0 b. x 0 1 2 3 y 2 5 8 11 P(y) .4 .3 .2 .1 c. E(Y) = ) y ( yP = 2(.4) + 5(.3) + 8(.2) + 11(.1) = 5.0 2 σ =V(Y) = μ - ) y ( P ) y ( 2 = (2 – 5) 2 (.4) + (5 – 5) 2 (.3) + (8 – 5) 2 (.2) + (11 – 5) 2 (.1) = 9.0 σ = 0 . 9 2 = σ = 3.0 d. E(Y) = E(3X + 2) = 3E(X) + 2 = 3(1) + 2 = 5.0 2 σ = V(Y) = V(3X + 2) = V(3X) = 3 2 V(X) = 9(1) = 9.0. σ = 0 . 9 2 = σ = 3.0 The parameters are identical. 7.20a. P(X 2) = P(2) + P(3) = .4 + .2 = .6 b. μ = E(X) = ) x ( xP = 0(.1) + 1(.3) + 2(.4) + 3(.2) = 1.7 2 σ = V(X) = μ - ) x ( P ) x ( 2 = (0–1.7) 2 (.1) + (1–1.7) 2 (.3) + (2–1.7) 2 (.4) + (3–1.7) 2 (.2) = .81 7.21 E(Profit) = E(5X) = 5E(X) = 5(1.7) = 8.5 V(Profit) = V(5X) = 5 2 V(X) = 25(.81) = 20.25 7.22 a P(X > 4) = P(5) + P(6) + P(7) = .20 + .10 + .10 = .40 160
b P(X 2) = 1– P(X 1) = 1 – P(1) = 1 – .05 = .95 7.23 μ = E(X) = ) x ( xP = 1(.05) + 2(.15) + 3(.15) + 4(.25) + 5(.20) + 6(.10) + 7(.10) = 4.1 2 σ = V(X) = μ - ) x ( P ) x ( 2 = (1–4.1) 2 (.05) + (2–4.1) 2 (.15) + (3–4.1) 2 (.15) + (4–4.1)

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

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