Chapter 7 solution manual

Chapter 7 solution manual - Chapter 7 7.1 a. b. c. d 0, 1...

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Chapter 7 7.1 a. 0, 1,2, . .. b. Yes, we can identifY the first value (0), the second (I), 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 any value between 0 and several hundred miles No, because we cannot identifY the second value or any other value larger than O. No, uncountable means infinite. d. The variable is continuous. 7.3 The values in cents are 0 ,I ,2, '" b. Yes, because we can identifY the first ,second, etc. Yes, it is finite because students cannot earn an infinite amount of money. Technically, the variable is discrete. 7.4 .. , 100 b. Yes. Yes, there are 101 values. The variable is discrete because it is countable. 7.5 No the sum of probabilities is not equal to I. b. Yes, because the probabilities lie between 0 and I and sum to I. No, because the probabilities do not sum to l. 7.6 P(x) = 1/6 for x = 1,2,3,4,5, and 6 7.7 x P(x) 0 24,7501165,000 .15 37,9501165,000 = .23 2 59,4001165,000 .36 3 29,700/165,000 = .18 4 9,9001165,000 = .06 5 3,300/165,000 = .02 (i) P(X s 2) = P(O) + P(I) + 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 .08 173
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7.8 a. P(2 5 X 5 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(O) + P(l) + P(2) + P(3) = .005 + .025 + .310 + .340 .680 b. b. E(X) = L xP(x) = 0(.005) + 1(.025) + 2(.310) + 4(.340) +5(.080) + 6(.019) + 7(.001) = 3.0,66' C. (}'2=V(X)= L(X-Il)2 p (X) (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 (}' = J;;2 =.JlT78 = 1.085 7.9 P(O) = P(l) = P(2) ... = P(lO) = 1111 .091 7.10 a. P(X > 0) = P(2) + P(6) + P(8) =.3 +.4 +.1 =.8 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 5 X 5 5) = P(2) .3 7.11 P(3 5 X 5 6) P(3) + P(4) + P(5) + P(6) = .04 + .28+.42 + .21 = .95 P(X > 6) P(X:;:: 7) P(7) + P(8) = .02 + .02 = .04 P(X < 3) = P(X 5 2) P(O) + P(l) + P(2) 0 + 0 + .01 .01 7.12 P(Losing6inarow)= .5 6 =.0156 7.13 P(X<2)=P(0)+P(l) .05+.43 .48 P(X> 1) = P(2) +P(3) .31+.21 .52 7.14 Joint events Probabilities H 0.5 H 0.5 T 0.5 HH fiT (0.5)(0.5) = 0.25 (0,5)(0.5) = 0.25 T 0.5 H 0.5 T 0.5 TH IT (0.5)(0.5) = 0.25 (0.5)(0.S) = 0.25 174
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a. P(HH) .25 b. P(HT) = .25 c. P(TH) = .25 d. P(IT) 7.15 P(O heads) P(TT) .25 b. PO head) = P(HT) + P(TH) = .25 + .25 = .50 P(2 heads) = P(HH) = .25 peat least 1 head) P(l head) + P(2 heads) = .50 + .25 .75 7.16 Joint events Probabilities H 0.5 HHH 0.125 H 0.5 T 0.5 mrr 0.125 H 0.5 H 0.5 lITH 0.125 T T HIT 0.125 H 0.5 THH 0.125 H 0.5 T 0.5 THT 0.125 T 0.5 H 0.5 TTH 0.125 T T TIT 0.125 P(2 heads) P(HHT) + P(HTH) + P(THH) = .125 + .125 + .125 = .375 P(l heads) = P(HIT) + P(THT) P(ITH) .125 + .125 + .125 = .375 peat least 1 head) = PO head) + P(2 heads) + P(3 heads) .375 + .375 + .125 = .875 peat least 2 heads) P(2 heads) + P(3 heads) .375 + .125 = .500 Jl E(X) L xP(x) -2(.59) +5(.15) + 7(.25) +8(.01) = lAO cr 2 = VeX) = L (x - Jl)2 P(x) = (-2-1.4) 2 (.59) + (5-1.4) 2 (.15) + (7-104) 2 (.25) + (8-1.4) 2 (.01) 17.04 175
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b. x -2 5 7 8 y -10 25 35 40 P(y) .59 .15 .25 .01 c. E(Y) = LyP(y)=-1O(.59) + 25(.15) + 35(.25) + 40(.01) = 7.00 V(Y) = L (Y-Il)2 P(y) = (-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(5)0 = 5E(X) = 5(1.4) = 7.00 V(Y) = V(5X) = 5 2 VeX) = 25(17.04) = 426.00. 7.19 a. Il=E(X)= LXP(X) = 0(.4) + 1(.3)+2(.2)+3(.1)= 1.0 2 0 = VeX) = L (X -Ill P(x) = (0-1.0) 2 (.4) + (1-1.0) 2 (.3) + (2-1.0) 2 (.2) + (3-1.0) 2 (.1) = 1.0 o = .,J;;i -= Jlo = 1.0 b. X 0 2 3 y 2 5 8 11 .P(y).4 .3 .2 .1 7. E(Y) = LYP(y) = 2(.4) + 5(.3) + 8(.2) + 11(.1) = 5.0 0 2 =V(Y) = L(y-Il)2 P(y) = (2 - 5)2 (.4) +(5 - 5)2 (.3) + (8 5)2 (.2) + (11- 5)2 (.1) = 9.0 0= = J9.O = 3.0 E(Y) = E(3X + 2) 3E(X) + 2 = 3(1) + 2 = 5.0 2 0 = V(Y) = V(3X + 2) =V(3X) = 3 2 VeX) = 9(1) = 9.0. = = 3.0 The parameters are identical. 7.20 P(X ;::: 2) = P(2) + P(3) = 04+ .2 = .6 Il= E(X) = LXP(x) 0(.1) + 1(.3) + 2(04) + 3(.2) = 1.7 2 0 = = L -1l)2 P(x) = (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) = 3E(X) = 3(1.7) = 5.1 V(Profit) = V(3X) = 3 2 = 9(.81) = 7.29 7.22 P(X> 4) = P(5) + P(6) + P(7) = .20 +.10 +.10 =.40 P(X~2)=I·-P(X$; 1)=I-P(I)=I-.05=.95 176
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23 Il=E(X) = IxP(x) = 1(.05) + 2(.15) + 3(.15) +4(.25) + 5(.20) +6(.10) + 7(.10) =4.1 0'2 = VeX) = I (x -11)2 P(x) = (1-4.1) 2 (.05) + (2-4.1) 2 (.15) + (3-4.1) 2 (.15) + (4-4.1) 2 (.25) + (5-4.1)2 (.20) + (6-4.1)2 (.10) + (7-4.1) 2 (.10) = 2.69 1.24 Y = .25X; E(Y) = .25E(X) = .25(4.1) 1.025 V(Y) = V(.25X) = (.25) 2 (2.69) = .168 7.25 a.
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Chapter 7 solution manual - Chapter 7 7.1 a. b. c. d 0, 1...

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