Chapter 7 solution manual

# Chapter 7 solution manual - Chapter 7 7.1 a b c d 0 1,2 Yes...

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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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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
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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## This note was uploaded on 05/17/2011 for the course ECO 220 taught by Professor Tanaka during the Spring '11 term at University of Toronto.

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Chapter 7 solution manual - Chapter 7 7.1 a b c d 0 1,2 Yes...

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