advertisement

Chapter 6 6.1 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. 6.2 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. 6.3 a x 0 1 2 3 4 5 P(x) 1218/101,501 = .012 32,379/101,501 = .319 37,961/101,501 = .374 19,387/101,501 = .191 7714/101,501 = .076 2842/101,501 = .028 b (i) P(X 2) = P(0) + P(1) + P(2) = .012 + .319 + .374 = .705 (ii) P(X > 2) = P(3) + P(4) + P(5) = .191 + .076 + .028 = .295 (iii) P(X 4) = P(4) + P(5) = .076 + .028 = .104 6.4 P(0) = P(1) = P(2) = . . . = P(10) = 1/11 = .091 6.5aP(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 6.6 a P(X < 2) = P(0) + P(1) = .05 + .43 = .48 b P(X > 1) = P(2) + P(3) = .31 + .21 = .52 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 6.7 a P(HH) = .25 b P(HT) = .25 c P(TH) = .25 d P(TT) = .25 6.8 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 6.9 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 6.10 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 6.11a = E(X) = 2 = V(X) = xP (x) = 0(.4) + 1(.3) + 2(.2) + 3(.1) = 1.0 ( x ) 2 P( x ) = (0–1.0) 2 (.4) + (1–1.0) 2 (.3) + (2–1.0) 2 (.2) + (3–1.0) 2 (.1) = 1.0 = b. 2 1.0 = 1.0 x 0 1 2 3 y 2 5 8 11 P(y) .4 .3 .2 .1 yP( y) = 2(.4) + 5(.3) + 8(.2) + 11(.1) = 5.0 =V(Y) = ( y ) P( y) = (2 – 5) (.4) + (5 – 5) (.3) + (8 – 5) c. E(Y) = 2 = 2 2 2 2 2 (.2) + (11 – 5) (.1) = 9.0 2 9.0 = 3.0 d. E(Y) = E(3X + 2) = 3E(X) + 2 = 3(1) + 2 = 5.0 2 = V(Y) = V(3X + 2) = V(3X) = 3 2 V(X) = 9(1) = 9.0. = 2 9.0 = 3.0 The parameters are identical. 6.12 a P(X 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 6.13 a P(X > 1) = P(2) + P(3) + P(4) = .17 + .06 + .01 = .24 b P(X = 0) = .45 c P(1 X 3) = P(1) + P(2) + P(3) = .31 + .17 + .06 = .54 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 6.14 = E(X) = xP (x) = 0(.10) + 1(.20) + 2(.25) + 3(.25) + 4(.20) = 2.25 ( x ) 2 = V(X) = 2 2 2 2 P( x ) = (0–2.25) (.10) + (1–2.25) (.20) + (2–2.25) (.25) + (3–2.25) 2 (.13) 2 + (4–2.25) (.20) = 1.59 = 2 1.59 = 1.26 6.15 E(X) = xP (x) = 1,000,000(1/10,000,000) + 200,000(1/1,000,000) + 50,000(1/500,000) + 10,000(1/50,000) + 1,000(1/10,000) = .1 + .2 + .1 + .2 + .1 = .7 Expected payoff = 70 cents. 6.16 = E(X) = 2 = V(X) = xP (x) = 1(.05) + 2(.12) + 3(.20) + 4(.30) + 5(.15) + 6(.10) + 7 (.08) = 4.00 ( x ) 2 2 2 2 2 2 P( x ) = (1–4.0) (.05) + (2–4.0) (.12) + (3–4.0) (.20) + (4–4.0) (.30) 2 2 + (5–4.0) (.15) +(6–4.0) (.10) + (7–4.0) (.08) = 2.40 6.17 Y = .25X; E(Y) = E(.25X) = .25E(X) = .25(4.0) = 1.0 2 V(Y) = V(.25X) = (.25) V(X) =.0625(2.40) = .15 6.18 P(X = x) = n! p x (1 p) n x x! (n x )! a P(X = 3) = 10! (.3) 3 (1 .3)103 = .2668 3! (10 3)! b P(X = 5) = 10! (.3) 5 (1 .3)105 = .1029 5! (10 5)! c P(X = 8) = 10! (.3) 8 (1 .3)108 = .0014 8!(10 8)! 3) – P(X 2) = .6496 – .3828 = .2668 b P(X = 5) = P(X 5) – P(X 4) = .9527 – .8497 = .1030 c P(X = 8) = P(X 8) – P(X 7) = .9999 – .9984 = .0015 6.19 a P(X = 3) = P(X 6.20 a .26683 b .10292 c .00145 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 6.21 P(X = x) = n! p x (1 p) n x x! (n x )! a P(X = 2) = 6! (.2) 2 (1 .2) 62 = .2458 2! (6 2)! b P(X = 3) = 6! (.2) 3 (1 .2) 63 = .0819 3! (6 3)! c P(X = 5) = 6! (.2) 5 (1 .2)5 = .0015 5! (6 5)! 2) – P(X 1) = .9011 – .6554 = .2457 b P(X = 3) = P(X 3) – P(X 2) = .9830 −.9011 = .0819 c P(X = 5) = P(X 5) – P(X 4) = .9999 – 9984 = .0015 6.22 a P(X = 2) = P(X 6.23 a .24576 b .08192 c .00154 6.24 Table 1 with n = 25 and p = .3: P(X 10) = .9022 6.25 Table 1 with n = 25 and p = .75: P(X 15) = 1 – P(X 14) = 1 – .0297 = .9703 6.26 Table 1 with n = 25 and p = .10 0) = .0718 b P(X < 5) = P(X 4) = .9020 c P(X > 2) = P(X 3) = 1 – P(X 2) = 1 – .5371 = .4629 a P(X = 0) = P(X 6.27 Excel with n = 100 and p = .20: P(X > 25) = P(X 26) = 1 – P(X 25) = 1 – .91252 = .08748 6.28 a P(X = 0) = e 2 2 0 e x = = .1353 0! x! b P(X = 3) = e 2 2 3 e x = = .1804 3! x! c P(X = 5) = e 2 2 5 e x = = .0361 5! x! © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use. 6.29 a P(X = 5 with = 14/3) = b. P(X = 1 with = 14/3) = e 1 / 3 (1 / 3)1 e x = = .2388 1! x! 6.30 a Table 2 with = 5: P(X b Table 2 with = 10: P(X 3) = 1 – P(X 2) = 1 – .73062 = .26938 X 15) = P(X 15) – P(X 9) = .9780 – .5874 = .3906 6.32 a Table 2 with = 5: P(X b Excel with = 25: P(X 10) = 1 – P(X 9) = 1 – .9682 = .0318 20) = 1 – P(X 19) = 1 – .9965 = .0035 6.31 a Excel with = 1.8: P(X b Table 2 with = 9: P(10 e 14 / 3 (14 / 3) 5 e x = = .1734 5! x! 10) = 1 – P(X 9) = 1 – .9682= .0318 25) = 1 – P(X 24) = 1 – .47340 = .52660 6.33 a Table 2 with = 1.5: P(X 2) = 1 – P(X 1) = 1 – .5578 = .4422 b Table 2 = 6: P(X < 4) = P(X 3) = .1512 6.34 a P(X = 0 with = 1.5) = e x e 1.5 (1.5) 0 = = .2231 0! x! b Table 2 with = 4.5: P(X 5) = .7029 c Table 2 with = 3.0: P(X 3) = 1 – P(X 2 = 1 – .4232 = .5768 © 2015 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part, except for use as permitted in a license distributed with a certain product or service or otherwise on a password-protected website for classroom use.