Applied statistics and probability for engineers solutions

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Montgomery and Runger's bestselling engineering statistics text provides a practical approach oriented to engineering as well as chemical and physical sciences. By providing unique problem sets that reflect realistic situations, students learn how the material will be relevant in their careers. With a focus on how statistical tools are integrated into the engineering problem-solving process, all major aspects of engineering statistics are covered. Developed with sponsorship from the National Science Foundation, this text incorporates many insights from the authors' teaching experience along with feedback from numerous adopters of previous editions.

John Wiley & Sons, 9 aug. 2010 - 398 pagina's

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Montgomery and Runger's bestselling engineering statistics text provides a practical approach oriented to engineering as well as chemical and physical sciences. By providing unique problem sets that reflect realistic situations, students learn how the material will be relevant in their careers. With a focus on how statistical tools are integrated into the engineering problem-solving process, all major aspects of engineering statistics are covered. Developed with sponsorship from the National Science Foundation, this text incorporates many insights from the authors' teaching experience along with feedback from numerous adopters of previous editions.

CHAPTER 2 Section 2-1 2-1. Let "a", "b" denote a part above, below the specification { } S aaa aab aba abb baa bab bba bbb = , , , , , , , 2-2. Let "e" denote a bit in error Let "o" denote a bit not in error ("o" denotes okay) ⎪ ⎪ ⎭ ⎪ ⎪ ⎬ ⎫ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ ⎧ = oooo oeoo eooo eeoo oooe oeoe eooe eeoe ooeo oeeo eoeo eeeo ooee oeee eoee eeee S , , , , , , , , , , , , , , , 2-3. Let "a" denote an acceptable power supply Let "f" ,"m","c" denote a supply with a functional, minor, or cosmetic error, respectively. { } S afmc = ,, , 2-4. = set of nonnegative integers { S = 012 , , ,... } 2-5. If only the number of tracks with errors is of interest, then { } S = 012 24 , , ,..., 2-6. A vector with three components can describe the three digits of the ammeter. Each digit can be 0,1,2,...,9. Then S is a sample space of 1000 possible three digit integers, { } S = 000 001 999 , ,..., 2-7. S is the sample space of 100 possible two digit integers. 2-8. Let an ordered pair of numbers, such as 43 denote the response on the first and second question. Then, S consists of the 25 ordered pairs { } 1112 55 , ,..., 2-9. in ppb. { ,..., 2 , 1 , 0 = S } } } 2-10. in milliseconds { ,..., 2 , 1 , 0 = S 2-11. { } 0 . 14 , 2 . 1 , 1 . 1 , 0 . 1 K = S 2-12. s = small, m = medium, l = large; S = {s, m, l, ss, sm, sl, ….} 2-13 in milliseconds. { ,..., 2 , 1 , 0 = S 2-14. automatic transmission transmission standard without with air without air air air with white red blue black white red blue black white red blue black white red blue black 2-1

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Solution manual for the book: Applied statistics and probability for engineers montgomery && runger

Text of 38148911 solution-manual-for-applied-statistics-and-probability-for-engineers-c02to11

• 1. CHAPTER 2 Section 2-1 2-1. Let "a", "b" denote a part above, below the specification S = {aaa,aab,aba,abb,baa,bab,bba,bbb} 2-2. Let "e" denote a bit in error Let "o" denote a bit not in error ("o" denotes okay) eeee eoee oeee ooee , , , , = eeoo eooo oeoo oooo eeeo eoeo oeeo ooeo , , , , eeoe eooe oeoe oooe S , , , , , , , 2-3. Let "a" denote an acceptable power supply Let "f" ,"m","c" denote a supply with a functional, minor, or cosmetic error, respectively. S = {a,f,m,c} 2-4. S = {0,1,2,...} = set of nonnegative integers 2-5. If only the number of tracks with errors is of interest, then S = {0,1,2,...,24} 2-6. A vector with three components can describe the three digits of the ammeter. Each digit can be 0,1,2,...,9. Then S is a sample space of 1000 possible three digit integers, S = {000,001,...,999} 2-7. S is the sample space of 100 possible two digit integers. 2-8. Let an ordered pair of numbers, such as 43 denote the response on the first and second question. Then, S consists of the 25 ordered pairs {11,12,...,55} 2-9. in ppb. { ,..., 2 , 1 , 0 = S }} 2-10. S = {0,1,2,..., in milliseconds 2-11. S = {1.0,1.1,1.2,K14.0} 2-12. s = small, m = medium, l = large; S = {s, m, l, ss, sm, sl, .} 2-13 S = {0,1,2,..., } in milliseconds. 2-14. automatic transmission standard transmission without with air air air without air with red blue black white red blue black white red blue black white red blue black white 2-1
• 2. 2-15. PRESS CAVITY 1 2 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 2-16. memory disk storage 4 8 12 200 300 400 200 300 400 200 300 400 2-17. c = connect, b = busy, S = {c, bc, bbc, bbbc, bbbbc, } 2-18. S = {s,fs,ffs,fffS,fffFS,fffFFS,fffFFFA} 2-19 a.) b.) 2-2
• 3. c.) d.) e.) 2.20 a.) 2-3
• 4. b.) c.) d.) e.) 2-4
• 5. 2-21. a) S = nonnegative integers from 0 to the largest integer that can be displayed by the scale. Let X represent weight. A is the event that X > 11 B is the event that X 15 C is the event that 8 X 15. Therefore, B C would be the empty set. They have no outcomes in common or i) B C is the event 8 X 0} G 2.30 a) {ab, ac, ad, bc, bd, cd, ba, ca, da, cb, db, dc} b) {ab, ac, ad, ae, af, ag, bc, bd, be, bf, bg, cd, ce, cf, cg, ef, eg, fg, ba, ca, da, ea, fa, ga, cb, db, eb, fb, gb, dc, ec, fc, gc, fe, ge, gf} c) Let d = defective, g = good; S = {gg, gd, dg, dd} d) Let d = defective, g = good; S = {gd, dg, gg} 2.31 Let g denote a good board, m a board with minor defects, and j a board with major defects. a.) S = {gg, gm, gj, mg, mm, mj, jg, jm, jj} b) S={gg,gm,gj,mg,mm,mj,jg,jm} G E E Surface 2 E Edge 2 G E E E G G G E E G G E E E G E G G G E G E G 2-7
• 8. 2-32.a.) The sample space contains all points in the positive X-Y plane. b) 10 A c) 20 B d) 10 20 A B e) 10 20 A B 2-8
• 9. 2-33 a) b) c) d) 2-9
• 10. Section 2-2 2-34. All outcomes are equally likely a) P(A) = 2/5 b) P(B) = 3/5 c) P(A') = 3/5 d) P(AB) = 1 e) P(AB) = P()= 0 2-35. a) P(A) = 0.4 b) P(B) = 0.8 c) P(A') = 0.6 d) P(AB) = 1 e) P(AB) = 0.2 2-36. a) S = {1, 2, 3, 4, 5, 6} b) 1/6 c) 2/6 d) 5/6 2-37. a) S = {1,2,3,4,5,6,7,8} b) 2/8 c) 6/8 2-38. x x 20 = 0.3, = 6 2-39. a) 0.5 + 0.2 = 0.7 b) 0.3 + 0.5 = 0.8 2-40. a) 1/10 b) 5/10 2-41. a) 0.25 b) 0.75 2-42. Total possible: 1016, Only 108 valid, P(valid) = 108/1016 = 1/108 2-43. 3 digits between 0 and 9, so the probability of any three numbers is 1/(10*10*10); 3 letters A to Z, so the probability of any three numbers is 1/(26*26*26); The probability your license plate is chosen is then (1/103)*(1/263) = 5.7 x 10-8 2-44. a) 5*5*4 = 100 b) (5*5)/100 = 25/100=1/4 2-45. a) P(A) = 86/100 = 0.86 b) P(B) = 79/100 = 0.79 c) P(A') = 14/100 = 0.14 d) P(AB) = 70/100 = 0.70 e) P(AB) = (70+9+16)/100 = 0.95 f) P(AB) = (70+9+5)/100 = 0.84 2-46. Let A = excellent surface finish; B = excellent length a) P(A) = 82/100 = 0.82 b) P(B) = 90/100 = 0.90 c) P(A') = 1 0.82 = 0.18 d) P(AB) = 80/100 = 0.80 e) P(AB) = 0.92 f) P(AB) = 0.98 2-10
• 11. 2-47. a) P(A) = 30/100 = 0.30 b) P(B) = 77/100 = 0.77 c) P(A') = 1 0.30 = 0.70 d) P(AB) = 22/100 = 0.22 e) P(AB) = 85/100 = 0.85 f) P(AB) =92/100 = 0.92 2-48. a) Because E and E' are mutually exclusive events and EE = S 1 = P(S) = P( E E ) = P(E) + P(E'). Therefore, P(E') = 1 - P(E) b) Because S and are mutually exclusive events with S = S P(S) = P(S) + P(). Therefore, P() = 0 c) Now, B = A (A B) and the events A and A B are mutually exclusive. Therefore, P(B) = P(A) + P( AB ). Because P( A B ) 0 , P(B) P(A). Section 2-3 2-49. a) P(A') = 1- P(A) = 0.7 b) P ( A B ) = P(A) + P(B) - P( A B ) = 0.3+0.2 - 0.1 = 0.4 c) P( A B ) + P( A B ) = P(B). Therefore, P( AB ) = 0.2 - 0.1 = 0.1 d) P(A) = P( A B ) + P( A B ). Therefore, P( A B ) = 0.3 - 0.1 = 0.2 e) P(( A B )') = 1 - P( A B ) = 1 - 0.4 = 0.6 f) P( A B ) = P(A') + P(B) - P( AB ) = 0.7 + 0.2 - 0.1 = 0.8 A BC 2-50. a) P( ) = P(A) + P(B) + P(C), because the events are mutually exclusive. Therefore, P( ) = 0.2+0.3+0.4 = 0.9 A BC b) P ( A B C ) = 0, because A B C = c) P( A B ) = 0 , because A B = d) P( (A B)C ) = 0, because (A B)C = (AC)(B C) = e) P( A BC ) =1-[ P(A) + P(B) + P(C)] = 1-(0.2+0.3+0.4) = 0.1 2-51. If A,B,C are mutually exclusive, then P( A BC ) = P(A) + P(B) + P(C) = 0.3 + 0.4 + 0.5 = 1.2, which greater than 1. Therefore, P(A), P(B),and P(C) cannot equal the given values. 2-52. a) 70/100 = 0.70 b) (79+86-70)/100 = 0.95 c) No, P( A B ) 0 2-53. a) 350/370 b) 345 5 12 370 362 370 + + = c) 345 5 8 370 358 370 + + = d) 345/370 2-54. a) 170/190 = 17/19 b) 7/190 2-55. a) P(unsatisfactory) = (5+10-2)/130 = 13/130 b) P(both criteria satisfactory) = 117/130 = 0.90, No 2-56. a) (207+350+357-201-204-345+200)/370 = 0.9838 b) 366/370 = 0.989 c) (200+145)/370 = 363/370 = 0.981 d) (201+149)/370 = 350/370 = 0.946 Section 2-4 2-11
• 12. 2-57. a) P(A) = 86/100 b) P(B) = 79/100 c) P( AB) = 70 79 P A B ( ) = 70/100 = 79/100 P B ( ) d) P(BA ) = 70 86 P A B ( ) = 70/100 = 86/100 P A ( ) 2-58.a) 0.82 b) 0.90 c) 8/9 = 0.889 d) 80/82 = 0.9756 e) 80/82 = 0.9756 f) 2/10 = 0.20 2-59. a) 345/357 b) 5/13 2-60. a) 12/100 b) 12/28 c) 34/122 2-61. Need data from Table 2-2 on page 34 a) P(A) = 0.05 + 0.10 = 0.15 b) P(A|B) = P ( A B ) 0.04 0.07 = 0.153 0.72 ( ) + = P B c) P(B) = 0.72 d) P(B|A) = P ( A B ) 0.04 0.07 = 0.733 0.15 ( ) + = P B e) P(A B) = 0.04 +0.07 = 0.11 f) P(A B) = 0.15 + 0.72 0.11 = 0.76 2-62. a) 20/100 b) 19/99 c) (20/100)(19/99) = 0.038 d) If the chips are replaced, the probability would be (20/100) = 0.2 2-63. a) P(A) = 15/40 b) P(BA ) = 14/39 c) P( A B ) = P(A) P(B/A) = (15/40) (14/39) = 0.135 d) P( A B ) = P(A) + P(B) - P( A B ) = 15 40 14 39 15 40 14 39 + 0 599 = . 2-64. A = first is local, B = second is local, C = third is local a) P(A B C) = (15/40)(14/39)(13/38) = 0.046 b) P(A B C) = (15/40)(14/39)(25/39) = 0.085 2-65. a) 4/499 = 0.0080 b) (5/500)(4/499) = 0.000080 c) (495/500)(494/499) = 0.98 2-66. a) 3/498 = 0.0060 b) 4/498 = 0.0080 c) 5 500 4 499 3 498 4 82 10 7 = . x 2-67. a) P(gas leak) = (55 + 32)/107 = 0.813 b) P(electric failure|gas leak) = (55/107)/(87/102) = 0.632 c) P(gas leak| electric failure) = (55/107)/(72/107) = 0.764 2-12
• 13. 2-68. No, if B A , then P(A/B) = P A B P B P B P B ( ) ( ) ( ) ( ) = =1 A B 2-69. A B C Section 2-5 2-70. a) P(A B) = P(AB)P(B) = (0.4)(0.5) = 0.20 b) P(A B) = P(AB)P(B) = (0.6)(0.5) = 0.30 2-71. P A P A B P A B ( ) ( ) ( ) = + = P AB P B + P AB P B = + = + = ( ) ( ) ( ) ( ( 0 . 2 )( 0 . 8 ) ( 0 . 3 )( 0 . 2 ) 016 . 0 . 06 0 . 22 ) 2-72. Let F denote the event that a connector fails. Let W denote the event that a connector is wet. P(F) P(FW)P(W) P(FW )P(W ) = + = ( 0 . 05 )( 010 . ) + ( 0 . 01 )( 0 . 90 ) = 0 . 014 2-73. Let F denote the event that a roll contains a flaw. Let C denote the event that a roll is cotton. P(F) P(F C)P(C) P(F C )P(C ) = + = ( 0 . 02 )( 0 . 70 ) + ( 0 . 03 )( 0 . 30 ) = 0 . 023 2-74. a) P(A) = 0.03 b) P(A') = 0.97 c) P(B|A) = 0.40 d) P(B|A') = 0.05 e) P( A B ) = P(B A )P(A) = (0.40)(0.03) = 0.012 f) P( A B ') = P(B' A )P(A) = (0.60)(0.03) = 0.018 g) P(B) = P(BA )P(A) + P(BA ')P(A') = (0.40)(0.03) + (0.05)(0.97) = 0.0605 2-13
• 14. 2-75. Let R denote the event that a product exhibits surface roughness. Let N,A, and W denote the events that the blades are new, average, and worn, respectively. Then, P(R)= P(R|N)P(N) + P(R|A)P(A) + P(R|W)P(W) = (0.01)(0.25) + (0.03) (0.60) + (0.05)(0.15) = 0.028 2-76. Let B denote the event that a glass breaks. Let L denote the event that large packaging is used. P(B)= P(B|L)P(L) + P(B|L')P(L') = 0.01(0.60) + 0.02(0.40) = 0.014 2-77. Let U denote the event that the user has improperly followed installation instructions. Let C denote the event that the incoming call is a complaint. Let P denote the event that the incoming call is a request to purchase more products. Let R denote the event that the incoming call is a request for information. a) P(U|C)P(C) = (0.75)(0.03) = 0.0225 b) P(P|R)P(R) = (0.50)(0.25) = 0.125 2-78. a) (0.88)(0.27) = 0.2376 b) (0.12)(0.13+0.52) = 0.0.078 2-79. Let A denote a event that the first part selected has excessive shrinkage. Let B denote the event that the second part selected has excessive shrinkage. a) P(B)= P(BA )P(A) + P(BA ')P(A') = (4/24)(5/25) + (5/24)(20/25) = 0.20 b) Let C denote the event that the third chip selected has excessive shrinkage. P C P C A B P A B P C A B P A B ( ) = ( ) ( ) + ( ' ) ( ' ) P C A B P A B P C A B P A B + + 3 = 0.20 20 25 19 24 5 + 23 20 25 5 24 4 + 23 5 25 20 24 4 + 23 5 25 4 24 23 ( ' ) ( ' ) ( ' ' ) ( ' ' ) = 2-80. Let A and B denote the events that the first and second chips selected are defective, respectively. a) P(B) = P(B|A)P(A) + P(B|A')P(A') = (19/99)(20/100) + (20/99)(80/100) = 0.2 b) Let C denote the event that the third chip selected is defective. P ABC = P C AB P AB = P C AB P B A P A ( ) ( ) ( ) ( ) ( ) ( ) 19 18 = 0.00705 20 100 99 98 = Section 2-6 2-81. Because P( AB ) P(A), the events are not independent. 2-82. P(A') = 1 - P(A) = 0.7 and P( A' B ) = 1 - P( AB ) = 0.7 Therefore, A' and B are independent events. 2-83. P( A B ) = 70/100, P(A) = 86/100, P(B) = 77/100. Then, P( A B ) P(A)P(B), so A and B are not independent. 2-14
• 15. 2-84. P( A B ) = 80/100, P(A) = 82/100, P(B) = 90/100. Then, P( A B ) P(A)P(B), so A and B are not independent. 2-85. a) P( A B )= 22/100, P(A) = 30/100, P(B) = 75/100, Then P( A B ) P(A)P(B), therefore, A and B are not independent. b) P(B|A) = P(A B)/P(A) = (22/100)/(30/100) = 0.733 2-86. If A and B are mutually exclusive, then P( A B ) = 0 and P(A)P(B) = 0.04. Therefore, A and B are not independent. 2-87. It is useful to work one of these exercises with care to illustrate the laws of probability. Let Hi denote the event that the ith sample contains high levels of contamination. a) P(H' H' H' H' H' ) P(H' )P(H' )P(H' )P(H' )P(H' ) 1 2 3 4 5 = 1 2 3 4 5 by independence. Also, P(Hi' ) = 0.9 . Therefore, the answer is 0.95 = 0.59 b) A1 = (H1 H'2 H3' H4' H5' ) A2 = (H1' H2 H3' H4' H5' ) A3 = (H1' H2' H3 H4' H5' ) A4 = (H1' H2' H3' H4 H5' ) A5 = (H1' H2' H3' H4' H5 ) The requested probability is the probability of the union A1 A2 A3 A4 A5 and these events are mutually exclusive. Also, by independence P(Ai ) = 0.94(0.1) = 0.0656 . Therefore, the answer is 5(0.0656) = 0.328. c) Let B denote the event that no sample contains high levels of contamination. The requested probability is P(B') = 1 - P(B). From part (a), P(B') = 1 - 0.59 = 0.41. 2-88. Let Ai denote the event that the ith bit is a one. 1 10 2 a) By independence P(A1 A2 ... A10 ) P(A1)P(A2 )...P(A10 ) ( ) . = = = 0 000976 b) By independence, P A A A P A P A P Ac ( ' ' ... ' ) ( ' ) ( ' )... ( ) ( ) . 1 2 10 1 2 10 1 10 2 = = = 0 000976 c) The probability of the following sequence is P(A' A' A' A' A' A A A A A ) 1 2 3 4 5 6 7 8 9 10 1 10 2 =( ) , by independence. The number of sequences consisting of five "1"'s, and five "0"'s is ( ) 5 10 10 = ! = 252 5 5 ! ! . The answer is 252 1 2 0 246 10 = . 2-89. Let A denote the event that a sample is produced in cavity one of the mold. 1 5 8 a) By independence, P(A1 A2 A3 A4 A5) ( ) = = 0.00003 b) Let Bi be the event that all five samples are produced in cavity i. Because the B's are mutually exclusive, P(B1B2...B8 ) = P(B1) +P(B2 )+...+P(B8 ) From part a., P(Bi) = ( 1) 8 5 . Therefore, the answer is 8 1 ( )5 = 0.00024 8 c) By independence, P(A A A A A' ) ( ) ( ) 1 2 3 4 5 1 4 8 7 8 = . The number of sequences in which four out of five samples are from cavity one is 5. Therefore, the answer is 5 1 8 7 8 ( )4 ( ) = 0.00107 . 2-15
• 16. 2-90. Let A denote the upper devices function. Let B denote the lower devices function. P(A) = (0.9)(0.8)(0.7) = 0.504 P(B) = (0.95)(0.95)(0.95) = 0.8574 P(AB) = (0.504)(0.8574) = 0.4321 Therefore, the probability that the circuit operates = P(AB) = P(A) +P(B) P(AB) = 0.9293 2-91. [1-(0.1)(0.05)][1-(0.1)(0.05)][1-(0.2)(0.1)] = 0.9702 2-92. Let Ai denote the event that the ith readback is successful. By independence, P(A' A' A' ) P(A' )P(A' )P(A' ) ( . ) . = = 0 02 3 = 0 000008. 1 2 3 1 2 3 2-93. a) P(BA ) = 4/499 and P(B) = P(B A)P(A) + P(B A')P(A') = (4/ 499)(5/ 500) + (5/ 499)(495/500) = 5/ 500 Therefore, A and B are not independent. b) A and B are independent. Section 2-7 2-94. Because, P( AB ) P(B) = P( A B ) = P(B A ) P(A), 0.28 0.7(0.2) P A B P B ( ) = = = 0.5 ( ) ( ) P A ( ) P B A 2-95. Let F denote a fraudulent user and let T denote a user that originates calls from two or more metropolitan areas in a day. Then, 0.003 0.30(0.0001) P T F P F ( ) = 0.30(0.0001) 0.01(.9999) ( ) ( ) = P T F P F P T F P F ( ) ( ) ( ') ( ') + = + P F T 2-96. backup main-storage 0.25 0.75 life > 5 yrs life > 5 yrs life < 5 yrs life < 5 yrs 0.95(0.25)=0.2375 0.05(0.25)=0.0125 0.995(0.75)=0.74625 0.005(0.75)=0.00375 a) P(B) = 0.25 b) P( AB ) = 0.95 c) P( AB ') = 0.995 d) P( A B ) = P( AB )P(B) = 0.95(0.25) = 0.2375 e) P( A B ') = P( AB ')P(B') = 0.995(0.75) = 0.74625 f) P(A) = P( A B ) + P( A B ') = 0.95(0.25) + 0.995(0.75) = 0.98375 g) 0.95(0.25) + 0.995(0.75) = 0.98375. h) 0.769 0.05(0.25) P A B P B ( ') = 0.05(0.25) 0.005(0.75) ( ' ) ( ) = P A B P B P A B P B ( ' ) ( ) ( ' ') ( ') + = + P B A 2-16
• 17. 2-97. Let G denote a product that received a good review. Let H, M, and P denote products that were high, moderate, and poor performers, respectively. a) P(G) P(G H)P(H) P(G M)P(M) P(G P)P(P) = + + = + + = 0 95 0 40 0 60 0 35 0 10 0 25 0 615 . ( . ) . ( . ) . ( . ) . b) Using the result from part a., P GH P H P HG = = 0 95 0 40 = . P G ( ) ( ) ( ) ( ) . ( . ) 0 . 615 0 618 c) P HG P G H P H 0 05 0 40 = 1 0 615 = = . P G ( ') ( ' ) ( ) ( ') . ( . ) . 0 052 2-98. a) P(D)=P(D|G)P(G)+P(D|G)P(G)=(.005)(.991)+(.99)(.009)=0.013865 b) P(G|D)=P(GD)/P(D)=P(D|G)P(G)/P(D)=(.995)(.991)/(1-.013865)=0.9999 2-99. a) P(S) = 0.997(0.60) + 0.9995(0.27) + 0.897(0.13) = 0.9847 b) P(Ch|S) =(0.13)( 0.897)/0.9847 = 0.1184 Section 2-8 2-100. Continuous: a, c, d, f, h, i; Discrete: b, e, and g Supplemental Exercises 2-101. Let Di denote the event that the primary failure mode is type i and let A denote the event that a board passes the test. The sample space is S = {A,A'D1,A'D2,A'D3,A'D4,A'D5 } . 2-102. a) 20/200 b) 135/200 c) 65/200 d) A B 25 20 90 2-17
• 18. 2-103. a) P(A) = 19/100 = 0.19 b) P(A B) = 15/100 = 0.15 c) P(A B) = (19 + 95 15)/100 = 0.99 d) P(A B) = 80/100 = 0.80 e) P(A|B) = P(A B)/P(B) = 0.158 2-104. Let Ai denote the event that the ith order is shipped on time. a) By independence, ( ) ( ) ( ) ( ) (0.95)3 0.857 1 2 3 1 2 3 P A A A = P A P A P A = = b) Let B = A ' A A B = A A A B = A A A 1 1 2 3 ' 2 1 2 3 ' 3 1 2 3 ( ) = + + 1 2 3 1 2 3 ' ' ' ' 3 0 95 2 0 05 0135 = 3 = = ' ' = ' ' ' ( ) Then, because the B's are mutually exclusive, P(B B B ) P(B ) P(B ) P B ( . ) ( . ) . = = c) Let B A A A B A A A B A A A B A A A 1 1 2 2 1 2 3 3 1 2 3 4 1 2 3 Because the B's are mutually exclusive, P(B B B B ) = P(B ) + P(B ) + P(B ) + P B 1 2 3 4 1 2 3 4 3 0 05 2 0 95 0 05 3 0 00725 ( . ) ( . ) ( . ) . = + = 2-105. a) No, P(E1 E2 E3) 0 b) No, E1 E2 is not c) P(E1 E2 E3) = P(E1) + P(E2) + P(E3) P(E1 E2) - P(E1 E3) - P(E2 E3) + P(E1 E2 E3) = 40/240 d) P(E1 E2 E3) = 200/240 e) P(E1 E3) = P(E1) + P(E3) P(E1 E3) = 234/240 f) P(E1 E2 E3) = 1 P(E1 E2 E3) = 1 - 0 = 1 2-106. (0.20)(0.30) +(0.7)(0.9) = 0.69 2-18
• 19. 2-107. Let Ai denote the event that the ith bolt selected is not torqued to the proper limit. a) Then, P A A A A P A A A A P A A A ( ) = ( ) ( ) 1 2 3 4 4 1 2 3 1 2 3 P A A A A P A A A P A A P A ( ) ( ) ( ) ( ) = 0.282 15 20 14 19 13 18 12 17 4 1 2 3 3 1 2 2 1 1 = = b) Let B denote the event that at least one of the selected bolts are not properly torqued. Thus, B' is the event that all bolts are properly torqued. Then, P(B) = 1 - P(B') = 1 15 20 14 19 13 18 12 17 0 718 = . 2-108. Let A,B denote the event that the first, second portion of the circuit operates. Then, P(A) = (0.99)(0.99)+0.9-(0.99)(0.99)(0.9) = 0.998 P(B) = 0.9+0.9-(0.9)(0.9) = 0.99 and P( A B ) = P(A) P(B) = (0.998) (0.99) = 0.988 2-109. A1 = by telephone, A2 = website; P(A1) = 0.92, P(A2) = 0.95; By independence P(A1 A2) = P(A1) + P(A2) - P(A1 A2) = 0.92 + 0.95 - 0.92(0.95) = 0.996 2-110. P(Possess) = 0.95(0.99) +(0.05)(0.90) = 0.9855 2-111. Let D denote the event that a container is incorrectly filled and let H denote the event that a container is filled under high-speed operation. Then, a) P(D) = P(DH)P(H) + P(D H')P(H') = 0.01(0.30) + 0.001(0.70) = 0.0037 b) P D H P H ( ) = = 0.01(0.30) = 0.8108 0.0037 ( ) ( ) P D ( ) P H D 2-112. a) P(E T D) = (0.995)(0.99)(0.999) = 0.984 b) P(E D) = P(E) + P(D) P(E D) = 0.005995 2-113. D = defective copy 2 73 72 + 73 2 72 + 73 72 2 = a) P(D = 1) = 0.0778 73 74 75 73 74 75 73 74 75 2 1 73 2 73 1 73 2 1 = + + b) P(D = 2) = 0.00108 73 74 75 73 74 75 73 74 75 c) Let A represent the event that the two items NOT inspected are not defective. Then, P(A)=(73/75)(72/74)=0.947. 2-114. The tool fails if any component fails. Let F denote the event that the tool fails. Then, P(F') = 0 9 by independence and P(F) = 1 - 0 9 = 0.0956 . 910 . 910 2-115. a) (0.3)(0.99)(0.985) + (0.7)(0.98)(0.997) = 0.9764 0.02485(0.30) P E route P route ( 1) ( 1) b) ( 1 ) = 0.3159 1 0.9764 ( ) = = P E P route E 2-19
• 20. 2-116. a) By independence, 0.155 = 7.59 105 b) Let Ai denote the events that the machine is idle at the time of your ith request. Using independence, the requested probability is P(A A A A A ' or A A A A ' A or A A A ' A A or A A ' A A A or A ' A A A A ) 1 2 3 4 5 1 2 3 45 123 4 5 1 2 3 4 5 1 2 3 4 5 4 4 4 4 4 015 085 015 085 015 085 015 085 015 085 5 015 085 0 0022 . ( . ) . ( . ) . ( . ) . ( . ) . ( . ) ( . )( . ) . = + + + + = 4 = c) As in part b,the probability of 3 of the events is P A A A A A or A A A A A or A A A A A or A A A A A or A A A A A or A A A A A or A A A A A or A A A A A or A A A A A or A A A A A 10(0.15 )(0.85 ) 0.0244 = So to get the probability of at least 3, add answer parts a.) and b.) to the above to obtain requested probability. Therefore the answer is ) ( 3 2 3 4 5 '2 '1 4 5 '3 2 '1 5 '4 2 3 '1 '5 2 3 4 '1 4 5 '3 '2 1 5 '4 3 '2 1 '5 3 4 '2 5 1 '4 '3 1 2 '5 4 '3 1 2 '5 '4 1 2 3 = 0.0000759 + 0.0022 + 0.0244 = 0.0267 2-117. Let Ai denote the event that the ith washer selected is thicker than target. 30 29 28 = a) 0.207 8 49 50 b) 30/48 = 0.625 c) The requested probability can be written in terms of whether or not the first and second washer selected are thicker than the target. That is, P ( A ) P ( A A A orA A ' A orA ' A A orA ' A ' A ) = = + + + 3 1 2 3 1 2 3 1 2 3 1 2 3 ' ' ( )( ) ( )( ) ( ' ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ( )( ) ( P A 3 A 1 A 2 P A 1 A 2 P A 3 A 1 A 2 P A 1 A 2 P A 3 A 1 A ' ' ' ' ' 2 P A 1 A 2 P A 3 A 1 A 2 P A 1 A 2 P A A A P A A P A P A A A ' P A ' A P A P A A A P A A P A = + 3 1 2 2 1 1 3 12 21 ' ' 1 3 1 2 2 1 + ) 1 3 12 2 1 28 48 30 50 29 49 29 48 20 50 30 49 29 48 20 50 30 49 30 48 20 50 19 49 0 60 ' ) ( ' ' ) ( ' ' ) ( ) . + = + + + = P A A A P A A P A1 ' 2-118. a) If n washers are selected, then the probability they are all less than the target is 20 50 19 49 20 1 50 1 ... n + + n . n probability all selected washers are less than target 1 20/50 = 0.4 2 (20/50)(19/49) = 0.155 3 (20/50)(19/49)(18/48) = 0.058 Therefore, the answer is n = 3 b) Then event E that one or more washers is thicker than target is the complement of the event that all are less than target. Therefore, P(E) equals one minus the probability in part a. Therefore, n = 3. 2-20
• 21. 2-119. ) ( ) 112 68 246 0.262 a P A B ) ( ) 246 b P A B = = ) ( ' ) 514 68 246 0.547 c P A B ) ( ' ') 514 940 0.881 940 940 0.453 940 = = = + + = = + + = d P A B . e) P( AB ) = P A B ( ) P ( B ) / / . = 246 940 = 314 940 0 783 f) P(BA ) = P B A ( ) P ( A ) / / . = 246 940 = 358 940 0 687 2-120. Let E denote a read error and let S,O,P denote skewed, off-center, and proper alignments, respectively. Then, a) P(E) = P(E|S) P(S) + P(E|O) P (O) + P(E|P) P(P) = 0.01(0.10) + 0.02(0.05) + 0.001(0.85) = 0.00285 b) P(S|E) = P ES P S ( ) ( ) P E ( ) . ( . ) . = 0 01 0 10 = . 0 00285 0 351 2-121. Let A denote the event that the ith row operates. Then, . i P(A1 ) = 0.98, P(A2 ) = (0.99)(0.99) = 0.9801, P(A3 ) = 0.9801, P(A4 ) = 0.98 The probability the circuit does not operate is 7 '4 '3 '2 '1 P(A )P(A )P(A )P(A ) = (0.02)(0.0199)(0.0199)(0.02) = 1.58 10 2-122. a) (0.4)(0.1) + (0.3)(0.1) +(0.2)(0.2) + (0.4)(0.1) = 0.15 b) P(4 or more|provided) = (0.4)(0.1)/0.15 = 0.267 Mind-Expanding Exercises 2-123. Let E denote a read error and let S, O, B, P denote skewed, off-center, both, and proper alignments, respectively. P(E) = P(E|S)P(S) + P(E|O)P(O) + P(E|B)P(B) + P(E|P)P(P) = 0.01(0.10) + 0.02(0.05) + 0.06(0.01) + 0.001(0.84) = 0.00344 2-124. Let n denote the number of washers selected. a) The probability that all are less than the target is 0.4n , by independence. n 0 4 . n 1 0.4 2 0.16 3 0.064 Therefore, n = 3 b) The requested probability is the complement of the probability requested in part a. Therefore, n = 3 2-21
• 22. 2-125. Let x denote the number of kits produced. Revenue at each demand 0 50 100 200 0 x 50 -5x 100x 100x 100x Mean profit = 100x(0.95)-5x(0.05)-20x 50 x 100 -5x 100(50)-5(x-50) 100x 100x Mean profit = [100(50)-5(x-50)](0.4) + 100x(0.55)-5x(0.05)-20x 100 x 200 -5x 100(50)-5(x-50) 100(100)-5(x-100) 100x Mean profit = [100(50)-5(x-50)](0.4) + [100(100)-5(x-100)](0.3) + 100x(0.25) - 5x(0.05) - 20x Mean Profit Maximum Profit 0 x 50 74.75 x $ 3737.50 at x=50 50 x 100 32.75 x + 2100 $ 5375 at x=100 100 x 200 1.25 x + 5250 $ 5500 at x=200 Therefore, profit is maximized at 200 kits. However, the difference in profit over 100 kits is small. 2-126. Let E denote the probability that none of the bolts are identified as incorrectly torqued. The requested probability is P(E'). Let X denote the number of bolts in the sample that are incorrect. Then, P(E) = P(E|X=0)P(X=0) + P(E|X=1) P(X=1) + P(E|X=2) P(X=2) + P(E|X=3) P(X=3) + P(E|X=4)P(X=4) and P(X=0) = (15/20)(14/19)(13/18)(12/17) = 0.2817. The remaining probability for x can be determined from the counting methods in Appendix B-1. Then, ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) P X ( ) P X P X ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! . ( ) ! ! ! ! ! ! ! ! ! . ( ) ! ! ! ! ! ! ! ! ! . 5 15 = = = = = 20 15 = = = = 20 15 = = = = 1 5 4 1 15 3 12 20 4 16 5 15 4 16 4 3 12 20 0 4696 2 5 3 2 15 2 13 20 4 16 0 2167 3 5 3 2 15 1 14 20 4 16 0 0309 1 3 4 2 5 2 4 3 5 1 20 4 P(X=4) = (5/20)(4/19)(3/18)(2/17) = 0.0010 and P(E|X=0) = 1, P(E|X=1) = 0.05, P(E|X=2) = 0 05 = 0 . 0025 , P(E|X=3) = 0.053 = 1.25 104 , P(E|X=4) = 0.054 = 6.25 106 . 2. Then, P E ( ) ( . ) . ( . ) . ( . ) . ( . ) 1 0 2817 0 05 0 4696 0 0025 0 2167 125 10 0 0309 6 25 10 0 0010 0 306 = + + + + = . (. ) . 4 6 and P(E') = 0.694 2-127. P A B P A B P A B ( ' ') ([ ' ']') ( ) 1 1 1 1 1 1 = = P A P B P A B P A P B P A P B P A P B [ ( ) ( ) ( )] ( ) ( ) ( ) ( ) = + = + = = [ ( )][ ( )] ( ') ( ') P A P B 2-22
• 23. 2-128. The total sample size is ka + a + kb + b = (k + 1)a + (k +1)b. P A k a b k a k b P B ka a k a k b and P A B ka k a k b ka k a b Then P A P B k a b ka a k a k b k a b k a k a b ka k a b P A B ( ) ( ) ( ) ( ) , ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) , ( ) ( ) ( )( ) [( ) ( ) ] ( )( ) ( ) ( ) ( )( ) ( ) = + + + + = + + + + = + + + = + + = + + + + + = + + + + = + + = 1 1 1 1 1 1 1 1 1 1 2 1 2 2 1 Section 2-1.4 on CD S2-1. From the multiplication rule, the answer is 5 3 4 2 = 120 S2-2. From the multiplication rule, 343=36 S2-3. From the multiplication rule, 3434=144 S2-4. From equation S2-1, the answer is 10! = 3628800 S2-5. From the multiplication rule and equation S2-1, the answer is 5!5! = 14400 S2-6. From equation S2-3, 7 ! 35 ! ! 3 4 = sequences are possible 140 140! 5 = = S2-7. a) From equation S2-4, the number of samples of size five is ( ) 416965528 5!135! 130 130! 4 = = b) There are 10 ways of selecting one nonconforming chip and there are ( ) 11358880 4!126! ways of selecting four conforming chips. Therefore, the number of samples that contain exactly one nonconforming chip is 10 (130 ) 113588800 4 = c) The number of samples that contain at least one nonconforming chip is the total number of samples ( 140 ) 5 minus the number of samples that contain no nonconforming chips ( 130 ) 5 . That is ( 140 ) - ( ) 5 5 = 130 140 5 135 130 5 125 ! 130721752 ! ! ! = ! ! S2-8. a) If the chips are of different types, then every arrangement of 5 locations selected from the 12 results in a 12 12 different layout. Therefore, P5 = ! = 95040 7 ! layouts are possible. b) If the chips are of the same type, then every subset of 5 locations chosen from the 12 results in a different layout. Therefore, ( 12 ) = 12 ! 5 = 792 ! ! 5 7 layouts are possible. 2-23
• 24. 7! = sequences are possible. S2-9. a) 21 2!5! 7! = sequences are possible. b) 2520 1!1!1!1!1!2! c) 6! = 720 sequences are possible. S2-10. a) Every arrangement of 7 locations selected from the 12 comprises a different design. 12 12 P7 = ! = 3991680 5 ! designs are possible. 12! = designs are b) Every subset of 7 locations selected from the 12 comprises a new design. 792 5!7! possible. c) First the three locations for the first component are selected in ( ) 220 12 12! 3 = = ways. Then, the four 3!9! 9 9! 4 = = locations for the second component are selected from the nine remaining locations in ( ) 126 4!5! ways. From the multiplication rule, the number of designs is 220 126 = 27720 S2-11. a) From the multiplication rule, 103= 1000 prefixes are possible b) From the multiplication rule, 8 2 10 = 160 are possible c) Every arrangement of three digits selected from the 10 digits results in a possible prefix. P3 10 = 10 ! = 720 7 ! prefixes are possible. S2-12. a) From the multiplication rule, 28 = 256 bytes are possible b) From the multiplication rule, 27 = 128 bytes are possible 24 24! 4 = = The number of samples in which exactly S2-13. a) The total number of samples possible is ( ) 10626. 4!20! 18! 18 6! 3 one tank has high viscosity is ( )( ) 4896 6 1 = = . Therefore, the probability is 3!15! 1!5! 0.461 4896 = 10626 18 18! 4 = = Therefore, the b) The number of samples that contain no tank with high viscosity is ( ) 3060. 4!14! 3060 = . requested probability is 1 0.712 10626 14! 4! 14 6! 2 c) The number of samples that meet the requirements is ( )( )( ) 2184 6 1 = = . 2!12! 1!3! 1!5! 4 1 2184 = Therefore, the probability is 0.206 10626 2-24
• 25. S2-14. a) The total number of samples is ( ) 3 12 12 = ! = 220 3 9 ! ! . The number of samples that result in one 10! 10 2! 2 nonconforming part is ( )( ) 90. 2 1 = = Therefore, the requested probability is 2!8! 1!1! 90/220 = 0.409. b) The number of samples with no nonconforming part is ( ) 120. 10 10! 3 = = The probability of at least one 3!7! 120 = . nonconforming part is 1 0.455 220 5 4 = S2-15. a) The probability that both parts are defective is 0.0082 49 50 50 50! 2 b) The total number of samples is ( ) 50 49 2 2!48! = = . The number of samples with two defective 5 5! 2 parts is ( ) 5 4 2 2!3! 5 4 5 4 2 = = . Therefore, the probability is 0.0082 50 49 50 49 2 = = . 2-25
• 26. CHAPTER 3 Section 3-1 3-1. The range of X is {0,1,2,...,1000} 3-2 The range of X is {0,1,2,...,50} 3-3. The range of X is {0,1,2,...,99999} 3-4 The range of X is {0,1,2,3,4,5} 3-5. The range of X is { . Because 490 parts are conforming, a nonconforming part must be 1,2,...,491 selected in 491 selections. } 3-6 The range of X is { } . Although the range actually obtained from lots typically might not exceed 10%. 0,1,2,...,100 3-7. The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is {0,1,2,...} 3-8 The range of X is conveniently modeled as all nonnegative integers. That is, the range of X is {0,1,2,...} 3-9. The range of X is {0,1,2,...,15} 3-10 The possible totals for two orders are 1/8 + 1/8 = 1/4, 1/8 + 1/4 = 3/8, 1/8 + 3/8 = 1/2, 1/4 + 1/4 = 1/2, 1/4 + 3/8 = 5/8, 3/8 + 3/8 = 6/8. Therefore the range of X is 1 4 3 8 1 2 5 8 6 8 , , , , 3-11 The range of X is {0,1,2,K,10000} 3-12 The range of X is {0,1,2,...,5000} Section 3-2 3-13. f P X (0) = ( = 0) = 1/ 6 + 1/ 6 = 1/ 3 f P X (1.5) ( 1.5) 1/ 3 (2) = 1/ 6 (3) = 1/ 6 = = = X X X X f f 3-14 a) P(X=1.5) = 1/3 b) P(0.5< X < 2.7) = P(X=1.5) + P(X=2) = 1/6 + 1/3 = 1/2 c) P(X > 3) = 0 d) P ( 0 X < 2) = P(X=0) + P(X=1.5) = 1/3 + 1/3 = 2/3 e) P(X=0 or X=2) = 1/3 + 1/6 = 1/2 3-15. All probabilities are greater than or equal to zero and sum to one. 3-1
• 27. a) P(X 2)=1/8 + 2/8 + 2/8 + 2/8 + 1/8 = 1 b) P(X > - 2) = 2/8 + 2/8 + 2/8 + 1/8 = 7/8 c) P(-1 X 1) = 2/8 + 2/8 + 2/8 =6/8 = 3/4 /2 d) P(X -1 or X=2) = 1/8 + 2/8 +1/8 = 4/8 =1 -16 All probabilities are greater than or equal to zero and sum to one. 14=0.4286 X=3)=1 -17. Probabilities are nonnegative and sum to one. 3/25 = 4/25 /25 -18 Probabilities are nonnegative and sum to one. ] = 63/64 /4) = 1/4 -19. P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 -20 P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 -21. P(X = 0) = 0.02 = 8 x 10 2)]=0.0012 -22 X = number of wafers that pass .096 -23 P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1 -24 X = number of components that meet specifications 0.02) = 0.068 -25. X = number of components that meet specifications 8)(0.01)+(0.05)(0.02)(0.99) = 0.00167 3 a) P(X 1)=P(X=1)=0.5714 b) P(X>1)= 1-P(X=1)=1-0.57 c) P(21)= P(X=1)+ P(X=2)+P( 3 a) P(X = 4) = 9/25 b) P(X 1) = 1/25 + c) P(2 X < 4) = 5/25 + 7/25 = 12 d) P(X > 10) = 1 3 a) P(X = 2) = 3/4(1/4)2 = 3/64 b) P(X 2) = 3/4[1+1/4+(1/4)2 c) P(X > 2) = 1 P(X 2) = 1/64 d) P(X 1) = 1 P(X 0) = 1 (3 3 3 3 3 -6 P(X = 1) = 3[0.98(0.02)(0.0 P(X = 2) = 3[0.98(0.98)(0.02)]=0.0576 P(X = 3) = 0.983 = 0.9412 3 P(X=0) = (0.2)3 = 0.008 P(X=1) = 3(0.2)2(0.8) = 0 P(X=2) = 3(0.2)(0.8)2 = 0.384 P(X=3) = (0.8)3 = 0.512 3 3 P(X=0) = (0.05)(0.02) = 0.001 P(X=1) = (0.05)(0.98) + (0.95)( P(X=2) = (0.95)(0.98) = 0.931 3 P(X=0) = (0.05)(0.02)(0.01) = 0.00001 P(X=1) = (0.95)(0.02)(0.01) + (0.05)(0.9 P(X=2) = (0.95)(0.98)(0.01) + (0.95)(0.02)(0.99) + (0.05)(0.98)(0.99) = 0.07663 P(X=3) = (0.95)(0.98)(0.99) = 0.92169 3-2
• 28. Section 3-3 0, 0 1/ 3 0 1.5 < x < x 2 / 3 1.5 < 2 3-26 where 5 / 6 2 3 < = x x x F x 1 3 ( ) f P X (0) = ( = 0) = 1/ 6 + 1/ 6 = 1/ 3 f P X (1.5) ( 1.5) 1/ 3 (2) = 1/ 6 (3) = 1/ 6 = = = X X X X f f 3-27. where 0, 2 1/8 2 1 3/8 1 0 5/8 0 1 7 /8 1 2 < x < x < x < < = x x x F x 1 2 ( ) ( 2) = 1/ 8 ( 1) = 2 / 8 (0) = 2 / 8 (1) = 2 / 8 (2) = 1/ 8 X X X X X f f f f f a) P(X 1.25) = 7/8 b) P(X 2.2) = 1 c) P(-1.1 < X 1) = 7/8 1/8 = 3/4 d) P(X > 0) = 1 P(X 0) = 1 5/8 = 3/8 0, < 0 x 1/ 25 0 < 1 x 4 / 25 1 < 2 3-28 where 9 / 25 2 3 16 / 25 3 4 x < < = x x x F x 1 4 ( ) (0) = 1/ 25 (1) = 3 / 25 (2) = 5 / 25 (3) = 7 / 25 (4) = 9 / 25 X X X X X f f f f f a) P(X < 1.5) = 4/25 b) P(X 3) = 16/25 c) P(X > 2) = 1 P(X 2) = 1 9/25 = 16/25 d) P(1 < X 2) = P(X 2) P(X 1) = 9/25 4/25 = 5/25 = 1/5 3-29. 0, < 1 x 0.1, 1 < 5 0.7, 5 < 10 = x x x F x 1, 10 ( ) where P(X = 10 million) = 0.3, P(X = 5 million) = 0.6, P(X = 1 million) = 0.1 3-3
• 29. 3-30 0, 10 0.2, 10 25 0.5, 25 50 < x < < = x x x F x 1, 50 ( ) where P(X = 50 million) = 0.5, P(X = 25 million) = 0.3, P(X = 10 million) = 0.2 3-31. where 0, 0 0.008, 0 1 0.104, 1 2 0.488, 2 3 < x < x < < = x x x F x 1, 3 ( ) . 3 (0) = 0.2 = 0.008, (1) = 3(0.2)(0.2)(0.8) = 0.096, (2) = 3(0.2)(0.8)(0.8) = 0.384, 3 (3) = (0.8) = 0.512, f f f f 3-32 0, 0.5 0.1, 0.5 5 0.4, 5 15 < < < = x x x x F x 1, 15 ( ) where P(X = 15 million) = 0.6, P(X = 5 million) = 0.3, P(X = -0.5 million) = 0.1 3-33. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.5, f(3) = 0.5 a) P(X 3) = 1 b) P(X 2) = 0.5 c) P(1 X 2) = P(X=1) = 0.5 d) P(X>2) = 1 P(X2) = 0.5 3-34 The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(1) = 0.7, f(4) = 0.2, f(7) = 0.1 a) P(X 4) = 0.9 b) P(X > 7) = 0 c) P(X 5) = 0.9 d) P(X>4) = 0.1 e) P(X2) = 0.7 3-35. The sum of the probabilities is 1 and all probabilities are greater than or equal to zero; pmf: f(-10) = 0.25, f(30) = 0.5, f(50) = 0.25 a) P(X50) = 1 b) P(X40) = 0.75 c) P(40 X 60) = P(X=50)=0.25 d) P(X = P X = P X ( 1.53) ( 2) 1 ( 1) [( ) ( ) ] 1 0.01 0.99 0.01 0.99 0.0169 = 20 0 20 + 20 1 19 = 0 1 b) X is binomial with n = 20 and p = 0.04 ( 1) 1 ( 1) 1 [( )0.04 0.96 ( )0.04 0.96 ] 0.1897 = 20 0 20 + 20 1 19 = 0 1 P X > = P X c) Let Y denote the number of times X exceeds 1 in the next five samples. Then, Y is binomial with n = 5 and p = 0.190 from part b. ( 1) 1 ( 0) 1 [(5 )0.19000.8105 ] 0.651 0 P Y = P Y = = = The probability is 0.651 that at least one sample from the next five will contain more than one defective. 3-67. Let X denote the passengers with tickets that do not show up for the flight. Then, X is binomial with n = 125 and p = 0.1. 125 125 + a P X P X = + ( ) ( ) ( ) ( ) ( ) ) ( 5) 1 ( 4) 125 125 + 0.9961 0 125 1 124 2 123 = + ) ( 5) 1 ( 5) 0.9886 0.1 0.9 125 4 0.1 0.9 3 0.1 0.9 2 0.1 0.9 1 0.1 0.9 0 1 3 122 4 121 > = = = b P X P X 3-68 Let X denote the number of defective components among those stocked. ( 100 ) 0 100 0 ( ) ( ) ( ) a P X ). ( = 0) = 0.02 0.98 = 0.133 b P X .) ( 2) = 0.02 0.98 + 0.02 0.98 + 0.02 0.98 = 0.666 ). ( 5) 0.981 102 2 100 2 102 1 101 1 102 0 102 0 = c P X 3-11
• 37. 3-69. Let X denote the number of questions answered correctly. Then, X is binomial with n = 25 and p = 0.25. 25 25 + + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0.25 (0.75) 0.2137 20 5 21 4 22 3 25 + + 23 2 24 1 25 0 10 25 + + 0 25 1 24 2 23 25 4 25 25 + 25 25 + 0.25 0.75 3 25 25 0.25 0.75 2 0.25 0.75 1 0.25 0.75 0 a P X ) ( 5) 0.25 0.75 9.677 10 25 0.25 0.75 24 0.25 0.75 23 0.25 0.75 22 0.25 0.75 21 0.25 0.75 20 ) ( 20) 3 22 4 21 = + < = = = b P X 3-70 Let X denote the number of mornings the light is green. ( ) ( ) 5 1 4 1 a P X ) ( = 1) = 0.2 0.8 = 0.410 20 4 16 4 ) ( = 4) = 0.2 0.8 = 0.218 c P X P X ) ( > 4) = 1 ( 4) = 1 0.630 = 0.370 b P X Section 3-7 3-71. a.) P(X = 1) = (1 0.5)0 0.5 = 0.5 b.) P(X = 4) = (1 0.5)3 0.5 = 0.54 = 0.0625 c.) P(X = 8) = (1 0.5)7 0.5 = 0.58 = 0.0039 d.) P(X 2) = P(X = 1) + P(X = 2) = (1 0.5)0 0.5 + (1 0.5)10.5 = 0.5 + 0.52 = 0.75 e.) P(X > 2) = 1 P(X 2) = 1 0.75 = 0.25 3-72 E(X) = 2.5 = 1/p giving p = 0.4 a.) P(X = 1) = (1 0.4)0 0.4 = 0.4 b.) P(X = 4) = (1 0.4)30.4 = 0.0864 c.) P(X = 5) = (1 0.5)4 0.5 = 0.05184 d.) P(X 3) = P(X = 1) + P(X = 2) + P(X = 3) = (1 0.4)0 0.4 + (1 0.4)10.4 + (1 0.4)2 0.4 = 0.7840 e.) P(X > 3) = 1 P(X 3) = 1 0.7840 = 0.2160 3-12
• 38. 3-73. Let X denote the number of trials to obtain the first successful alignment. Then X is a geometric random variable with p = 0.8 a) P(X = 4) = (1 0.8)30.8 = 0.230.8 = 0.0064 b) P(X 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = (1 0.8)0 0.8 + (1 0.8)10.8 + (1 0.8)2 0.8 + (1 0.8)30.8 = 0.8 + 0.2(0.8) + 0.22 (0.8) + 0.230.8 = 0.9984 c) P(X 4) = 1 P(X 3) = 1[P(X = 1) + P(X = 2) + P(X = 3)] = 1[(1 0.8)0 0.8 + (1 0.8)10.8 + (1 0.8)2 0.8] = 1[0.8 + 0.2(0.8) + 0.22 (0.8)] = 1 0.992 = 0.008 3-74 Let X denote the number of people who carry the gene. Then X is a negative binomial random variable with r=2 and p = 0.1 a) P(X 4) = 1 P(X < 4) = 1[P(X = 2) + P(X = 3)] 2 + 1 2 1 2 0 = + = (1 0.1) 0.1 1 (0.01 0.018) 0.972 1 1 (1 0.1) 0.1 1 = b) E(X ) = r / p = 2 / 0.1 = 20 3-75. Let X denote the number of calls needed to obtain a connection. Then, X is a geometric random variable with p = 0.02 a) P(X = 10) = (1 0.02)9 0.02 = 0.9890.02 = 0.0167 b) P(X > 5) = 1 P(X 4) = 1[P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)] = 1[0.02 + 0.98(0.02) + 0.982 (0.02) + 0.983 (0.02)] = 1 0.0776 = 0.9224 c) E(X) = 1/0.02 = 50 3-76 Let X denote the number of mornings needed to obtain a green light. Then X is a geometric random variable with p = 0.20. a) P(X = 4) = (1-0.2)30.2= 0.1024 b) By independence, (0.8)10 = 0.1074. (Also, P(X > 10) = 0.1074) 3-77 p = 0.005 , r = 8 a.) P(X = 8) = 0.0058 = 3.91x1019 b). 200 = E(X ) = 1 = days 0.005 c) Mean number of days until all 8 computers fail. Now we use p=3.91x10-19 18 ( ) 1 x = E Y = = days or 7.01 x1015 years 91 2.56 10 x 3.91 10 3-78 Let Y denote the number of samples needed to exceed 1 in Exercise 3-66. Then Y has a geometric distribution with p = 0.0169. a) P(Y = 10) = (1 0.0169)9(0.0169) = 0.0145 b) Y is a geometric random variable with p = 0.1897 from Exercise 3-66. P(Y = 10) = (1 0.1897)9(0.1897) = 0.0286 c) E(Y) = 1/0.1897 = 5.27 3-13
• 39. 3-79. Let X denote the number of trials to obtain the first success. a) E(X) = 1/0.2 = 5 b) Because of the lack of memory property, the expected value is still 5. x 1 3-80 Negative binomial random variable: f(x; p, r) = p x r pr . r (1 ) 1 When r = 1, this reduces to f(x; p, r) = (1p)x-1p, which is the pdf of a geometric random variable. Also, E(X) = r/p and V(X) = [r(1p)]/p2 reduce to E(X) = 1/p and V(X) = (1p)/p2, respectively. 3-81. a) E(X) = 4/0.2 = 20 19 16 4 = b) P(X=20) = (0.80) 0.2 0.0436 3 18 15 4 = c) P(X=19) = (0.80) 0.2 0.0459 3 20 17 4 = d) P(X=21) = (0.80) 0.2 0.0411 3 e) The most likely value for X should be near X. By trying several cases, the most likely value is x = 19. 3-82 Let X denote the number of attempts needed to obtain a calibration that conforms to specifications. Then, X is geometric with p = 0.6. P(X 3) = P(X=1) + P(X=2) + P(X=3) = 0.6 + 0.4(0.6) + 0.42(0.6) = 0.936. 3-83. Let X denote the number of fills needed to detect three underweight packages. Then X is a negative binomial random variable with p = 0.001 and r = 3. a) E(X) = 3/0.001 = 3000 b) V(X) = [3(0.999)/0.0012] = 2997000. Therefore, X = 1731.18 3-84 Let X denote the number of transactions until all computers have failed. Then, X is negative binomial random variable with p = 10-8 and r = 3. a) E(X) = 3 x 108 b) V(X) = [3(110-80]/(10-16) = 3.0 x 1016 3-14
• 40. 3-85 Let X denote a geometric random variable with parameter p. Let q = 1-p. q p d ( ) E X = x p p p xq p d 1 1 0 p 2 2 1 p (1 ) V X x p p px x p ( ) = ( ) (1 ) = 2 + (1 ) p x q xq q 2 = + = + 2 2 2 p x q p x q = [ ] [ ] [ ] [ ] 1 1 2 3 1 p q q q 2 3 ... = d + + + dq p p q q q (1 2 3 ...) 2 2 = d + + + dq p 1 3 2 1 q p pq q p q 2 (1 ) (1 ) = = + q p p 2 2 2 (1 ) 2 1 1 1 2 1 2 1 1 2 1 1 1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 d 2(1 ) 1 (1 ) 1 1 ( ) (1 ) 2 2 2 p p p p p p q dq q dq q p p dq p x x p p x x x x p x x x x x x p x x p x x x x x x = = + = = = = = = = = = = = = = = = = = Section 3-8 3-86 X has a hypergeometric distribution N=100, n=4, K=20 a.) ( )( ) ( ) 0.4191 ( 1) 20(82160) 100 3921225 4 80 3 20 P X = = 1 = = b.) P(X = 6) = 0 , the sample size is only 4 c.) ( )( ) ( ) 0.001236 ( 4) 4845(1) 100 3921225 4 80 0 20 P X = = 4 = = E X np n K ( ) = = = 4 20 = d.) 0.8 100 N 0.6206 4(0.2)(0.8) 96 = V X np p N n ( ) ( 1 ) = 1 99 = N 3-15
• 41. 3-87. a) ( 4 )( ) 1 (4 16 15 14) / 6 ( ) = 0.4623 (20 19 18 17) / 24 ( 1) 20 4 16 3 P X = = = b) ( )( ) ( ) 0.00021 ( 4) 1 20 (20 19 18 17) / 24 4 16 0 44 = P X = = = c) P X P X P X P X ( 2) = ( = 0) + ( = 1) + ( = 2) ( )( ) ( ) ( )( ) ( ) ( )( ) ( ) 0.9866 16 3 = + + 4 16 15 14 16 4 16 15 14 13 20 19 18 17 24 16 2 6 16 15 2 6 24 20 4 42 20 4 4 1 20 4 4 0 = = + + d) E(X) = 4(4/20) = 0.8 V(X) = 4(0.2)(0.8)(16/19) = 0.539 3-88 N=10, n=3 and K=4 0 1 2 3 0.5 0.4 0.3 0.2 0.1 0.0 x P(x) 3-16
• 42. 3-89. where 0, 0 1/ 6, 0 1 2 / 3, 1 2 29 / 30, 2 3 < x < x < x < = x x F x 1, 3 ( ) 0 .0333 10 3 4 6 0 4 3 f f 0 .3, (3 ) 10 10 3 6 6 1 4 4 2 ( 2 ) 0 .5 , 10 3 6 2 1 0 .1667 , (1) 3 3 0 ( 0 ) = = = = = = = = f f 3-90 Let X denote the number of unacceptable washers in the sample of 10. ( )( ) ( ) 70! 70 10 .) ( 0) 10!60! 65 64 63 62 61 75! a P X b P X P X .) ( 1) = 1 ( = 0) = 0.5214 ( )( ) ( ) 5!70! 70 9 .) ( 1) 9!61! 5 65 64 63 62 10 75! c P X .) ( ) 10(5/ 75) 2 / 3 0.3923 75 74 73 72 71 0.4786 75 74 73 72 71 10!65! 75 10 5 1 10!65! 75 10 5 0 = = = = = = = = = = = = d E X 3-91. Let X denote the number of men who carry the marker on the male chromosome for an increased risk for high blood pressure. N=800, K=240 n=10 a) n=10 ( )( ) ( ) ( 240! )( 560! ) 560 9 240 P ( X = 1) = 1 = 1!239! 9!551! = 0.1201 800 800! 10 10!790! b) n=10 P(X > 1) = 1 P(X 1) = 1[P(X = 0) + P(X = 1)] ( )( ) ( ) ( 240! )( 560! ) 560 10 240 P ( X = 0) = 0 = 0!240! 10!550! = 0.0276 800 800! 10 10!790! P(X > 1) = 1 P(X 1) = 1[0.0276 + 0.1201] = 0.8523 3-17
• 43. 3-92 . Let X denote the number of cards in the sample that are defective. a) P X P X ( 1) = 1 ( = 0) ( )( ) ( ) 120! 120 20 20 0 P X ( = 0) = = 20!100! = 0.0356 140! 20!120! 140 20 ( 1) = 1 0.0356 = 0.9644 P X b) P X P X = = ( )( ) ( ) 135! 135 20 ( 0) 20!115! 135!120! = = = = = 140! P X ( 1) 1 0.4571 0.5429 0.4571 115!140! ( 1) 1 ( 0) 20!120! 140 20 5 0 = = P X 3-93. Let X denote the number of blades in the sample that are dull. a) P X P X = = ( )( ) ( ) 38! ( 0) 5!33! 38!43! = = = = = 48! P X P X ( 1) 1 ( 0) 0.7069 0.2931 48!33! ( 1) 1 ( 0) 5!43! 48 5 38 5 10 0 = = = P X b) Let Y denote the number of days needed to replace the assembly. P(Y = 3) = 0.29312 (0.7069) = 0.0607 c) On the first day, ( )( 46 ) 46! ( 5 ) 0.8005 ( 0) 46!43! 48!41! 5!41! 48! 5!43! 48 5 2 P X = = 0 = = = On the second day, ( )( 42 ) 42! ( 5 ) 0.4968 ( 0) 42!43! 48!37! 5!37! 48! 5!43! 48 5 60 P X = = = = = On the third day, P(X = 0) = 0.2931 from part a. Therefore, P(Y = 3) = 0.8005(0.4968)(1-0.2931) = 0.2811. 3-94 Let X denote the count of the numbers in the state's sample that match those in the player's sample. Then, X has a hypergeometric distribution with N = 40, n = 6, and K = 6. a) ( )( 34 ) 1 ( 0 ) 7 ( 6) 40! P X = = = 40 6 66 2.61 10 6!34! = b) ( )( ) ( ) ( ) 5 40 6 34 1 40 6 6 ( 5) 5 6 34 = 5.3110 P X = = = c) ( )( ) ( 4) ( ) 0.00219 40 6 34 2 6 P X = = 4 = d) Let Y denote the number of weeks needed to match all six numbers. Then, Y has a geometric distribution with p = 1 and E(Y) = 1/p =3,838,380 weeks. This is more than 738 centuries! 3,838,380 3-18
• 44. 3-95. a) For Exercise 3-86, the finite population correction is 96/99. For Exercise 3-87, the finite population correction is 16/19. Because the finite population correction for Exercise 3-86 is closer to one, the binomial approximation to the distribution of X should be better in Exercise 3-86. b) Assuming X has a binomial distribution with n = 4 and p = 0.2, ( ) 4 1 3 1 P X ( = 1) = 0.2 0.8 = 0.4096 ( 4) ( )0.2 0.8 0.0016 0 4 44 = = = P X The results from the binomial approximation are close to the probabilities obtained in Exercise 3-86. c) Assume X has a binomial distribution with n = 4 and p = 0.2. Consequently, P(X = 1) and P(X = 4) are the same as computed in part b. of this exercise. This binomial approximation is not as close to the true answer as the results obtained in part b. of this exercise. 3-96 a.) From Exercise 3-92, X is approximately binomial with n = 20 and p = 20/140 = 1/7. ( 1) 1 ( 0) ( )( ) 0 ( 6 ) 20 1 0.0458 0.9542 P X = P X = = 20 = = 0 7 finite population correction is 120/139=0.8633 b) From Exercise 3-92, X is approximately binomial with n = 20 and p = 5/140 =1/28 7 1 ( 1) 1 ( 0) ( )( ) 0 ( 27 ) 20 1 0.4832 0.5168 P X = P X = = 20 1 = = 0 28 28 finite population correction is 120/139=0.8633 Section 3-9 3-97. a) P(X ) e e = = = = . ! 0 4 0 0 0183 4 0 4 b) P(X 2) = P(X = 0) + P(X = 1) + P(X = 2) 441 4 2 1 e 4 e e = + + = 4 2 0 2381 ! ! . c) P(X ) e = = = . ! 4 4 4 01954 4 4 d) P(X ) e = = = . ! 8 4 8 0 0298 4 8 3-98 a) P(X = 0) = e0.4 = 0.6703 0 . 4 0 . 4 2 b) P(X 2 ) e e ( 04 . ) e ! ( . ) ! . . = + + = 1 04 2 0 4 0 9921 c) P(X 4 ) e ( 04 . ) ! . . = = = 4 0 000715 0 4 4 d) P(X ) e ( . ) 8 0 4 ! . . = = = 8 109 10 0 4 8 8 3-19
• 45. 3-99. . Therefore, = ln(0.05) = 2.996. P X e ( ) . = = = 0 005 Consequently, E(X) = V(X) = 2.996. 3-100 a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with = 10. P(X ) e = = = . ! 5 10 5 00378 10 5 . 10 10 2 10 3 b) P(X 3 ) e e 10 e e = + + + = . ! ! ! 1 10 2 10 3 10 0 0103 c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with = 20. P(Y ) e = = = . ! 15 20 15 0 0516 20 15 d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with = 5. P(W ) e = = = . ! 5 5 5 01755 5 5 3-101. a) Let X denote the number of flaws in one square meter of cloth. Then, X is a Poisson random variable 0.1 2 ( 2) (0.1) with = 0.1. = = = 0.0045 2! e P X b) Let Y denote the number of flaws in 10 square meters of cloth. Then, Y is a Poisson random variable e 1 1 1 with = 1. P ( Y = 1) = = e 1 = 0.3679 1! c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable with = 2. P(W = 0) = e2 = 0.1353 d) P(Y 2) = 1 P(Y 1) = 1 P(Y = 0) P(Y = 1) = 1 e 1 e 1 = 0.2642 3-102 a) E(X ) = = 0.2 errors per test area 0.2 0.2 2 (0.2) P ( X 2) e e 0.2 e = 0.2 + + = b.) 0.9989 2! 1! 99.89% of test areas 3-103. a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with = 10. P(X = 0) = e10 = 4.54105 b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a Poisson random variable with = 1. P(Y 1) = 1 P(Y = 0) = 1 e1 = 0.6321 c) The assumptions of a Poisson process require that the probability of a count is constant for all intervals. If the probability of a count depends on traffic load and the load varies, then the assumptions of a Poisson process are not valid. Separate Poisson random variables might be appropriate for the heavy and light load sections of the highway. 3-20
• 46. 3-104 a.) E(X ) = = 0.01failures per 100 samples. Let Y= the number of failures per day E(Y) = E(5X ) = 5E(X ) = 5 = 0.05 failures per day. b.)Let W= the number of failures in 500 participants, now =0.05 and P(W = 0) = e0.05 = 0.9512 3-105. a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random variable with = 0.5. P(X = 0) = e0.5 = 0.6065 b) Let Y denote the number of cars with no flaws, ) 10 ( 0 10 = (0.6065) (0.3935) 0.0067 10 10 P Y = = c) Let W denote the number of cars with surface flaws. Because the number of flaws has a Poisson distribution, the occurrences of surface flaws in cars are independent events with constant probability. From part a., the probability a car contains surface flaws is 10.6065 = 0.3935. Consequently, W is binomial with n = 10 and p = 0.3935. 10 0 10 = (0.3935) (0.6065) 0.0067 0 10 1 9 = (0.3935) (0.6065) 0.0437 1 P W ( 0) = = P W ( 1) = = ( 1) = 0.0067 + 0.00437 = 0.00504 P W 3-106 a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with = 0.16. P(X = 0) = e0.16 = 0.8521 b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with = 0.48. P(Y 1) = 1 P(Y = 0) = 1 e48 = 0.3812 Supplemental Exercises 3-107. Let X denote the number of totes in the sample that do not conform to purity requirements. Then, X has a hypergeometric distribution with N = 15, n = 3, and K = 2. 0.3714 1 13!12! 10!15! 13 15 3 3 2 0 ( 1) 1 ( 0) 1 = = P X = P X = = 3-108 Let X denote the number of calls that are answered in 30 seconds or less. Then, X is a binomial random variable with p = 0.75. a) P(X = 9) = (0.75) (0.25) 0.1877 10 9 1 9 = b) P(X 16) = P(X=16) +P(X=17) + P(X=18) + P(X=19) + P(X=20) 20 + + 16 4 17 3 18 2 (0.75) (0.25) 0.4148 20 20 (0.75) (0.25) + 20 19 (0.75) (0.25) 18 (0.75) (0.25) 20 17 (0.75) (0.25) 20 16 19 1 20 0 = + = c) E(X) = 20(0.75) = 15 3-21
• 47. 3-109. Let Y denote the number of calls needed to obtain an answer in less than 30 seconds. a) P(Y = 4) = (1 0.75)30.75 = 0.2530.75 = 0.0117 b) E(Y) = 1/p = 1/0.75 = 4/3 3-110 Let W denote the number of calls needed to obtain two answers in less than 30 seconds. Then, W has a negative binomial distribution with p = 0.75. a) P(W=6) = 5 1 0 25 4 0 75 2 0 0110 ( . ) ( . ) = . b) E(W) = r/p = 2/0.75 = 8/3 5 5 e P X ( 5) 5 3-111. a) Let X denote the number of messages sent in one hour. = = = 0.1755 5! b) Let Y denote the number of messages sent in 1.5 hours. Then, Y is a Poisson random variable with 7.5 10 ( 10) (7.5) =7.5. = = = 0.0858 10! e P Y c) Let W denote the number of messages sent in one-half hour. Then, W is a Poisson random variable with = 2.5. P(W < 2) = P(W = 0) + P(W = 1) = 0.2873 3-112 X is a negative binomial with r=4 and p=0.0001 E(X ) = r / p = 4 / 0.0001 = 40000 requests 3-113. X Poisson( = 0.01), X Poisson( = 1) 0.9810 1 1 1 2 1 3 (1) P Y e e e e = 1 + + + = 3! (1) 2! ( 3) (1) 1! 3-114 Let X denote the number of individuals that recover in one week. Assume the individuals are independent. Then, X is a binomial random variable with n = 20 and p = 0.1. P(X 4) = 1 P(X 3) = 1 0.8670 = 0.1330. 3-115 a.) P(X=1) = 0 , P(X=2) = 0.0025, P(X=3) = 0.01, P(X=4) = 0.03, P(X=5) = 0.065 P(X=6) = 0.13, P(X=7) = 0.18, P(X=8) = 0.2225, P(X=9) = 0.2, P(X=10) = 0.16 b.) P(X=1) = 0.0025, P(X=1.5) = 0.01, P(X=2) = 0.03, P(X=2.5) = 0.065, P(X=3) = 0.13 P(X=3.5) = 0.18, P(X=4) = 0.2225, P(X=4.5) = 0.2, P(X=5) = 0.16 3-116 Let X denote the number of assemblies needed to obtain 5 defectives. Then, X is a negative binomial random variable with p = 0.01 and r=5. a) E(X) = r/p = 500. b) V(X) =(5* 0.99/0.012 = 49500 and X = 222.49 3-117. If n assemblies are checked, then let X denote the number of defective assemblies. If P(X 1) 0.95, then P(X=0) 0.05. Now, P(X=0) = and 0.99 n n n n 0.05. Therefore, 0 = (0.01) (0.99) 99 0 (ln(0.99)) ln(0.05) = 298.07 ln(0.05) ln(0.95) n n This would require n = 299. 3-22
• 48. 3-118 Require f(1) + f(2) + f(3) + f(4) = 1. Therefore, c(1+2+3+4) = 1. Therefore, c = 0.1. 3-119. Let X denote the number of products that fail during the warranty period. Assume the units are independent. Then, X is a binomial random variable with n = 500 and p = 0.02. a) P(X = 0) = (0.02)0 (0.98)500 = 4.1 x 10 0 500 -5 b) E(X) = 500(0.02) = 10 c) P(X >2) = 1 P(X 2) = 0.9995 3-120 (0) = (0.1)(0.7) + (0.3)(0.3) = 0.16 X f (1) = (0.1)(0.7) + (0.4)(0.3) = 0.19 X f (2) = (0.2)(0.7) + (0.2)(0.3) = 0.20 X f (3) = (0.4)(0.7) + (0.1)(0.3) = 0.31 X f (4) = (0.2)(0.7) + (0)(0.3) = 0.14 X f 3-121. a) P(X 3) = 0.2 + 0.4 = 0.6 b) P(X > 2.5) = 0.4 + 0.3 + 0.1 = 0.8 c) P(2.7 < X < 5.1) = 0.4 + 0.3 = 0.7 d) E(X) = 2(0.2) + 3(0.4) + 5(0.3) + 8(0.1) = 3.9 e) V(X) = 22(0.2) + 32(0.4) + 52(0.3) + 82(0.1) (3.9)2 = 3.09 3-122 x 2 5.7 6.5 8.5 f(x) 0.2 0.3 0.3 0.2 3-123. Let X denote the number of bolts in the sample from supplier 1 and let Y denote the number of bolts in the sample from supplier 2. Then, x is a hypergeometric random variable with N = 100, n = 4, and K = 30. Also, Y is a hypergeometric random variable with N = 100, n = 4, and K = 70. a) P(X=4 or Y=4) = P(X = 4) + P(Y = 4) 70 30 100 = 0.2408 70 30 100 4 4 0 4 0 4 = + 70 30 70 30 + = b) P[(X=3 and Y=1) or (Y=3 and X = 1)]= 0.4913 100 4 3 1 1 3 = 3-124 Let X denote the number of errors in a sector. Then, X is a Poisson random variable with = 0.32768. a) P(X>1) = 1 P(X1) = 1 e-0.32768 e-0.32768(0.32768) = 0.0433 b) Let Y denote the number of sectors until an error is found. Then, Y is a geometric random variable and P = P(X 1) = 1 P(X=0) = 1 e-0.32768 = 0.2794 E(Y) = 1/p = 3.58 3-23
• 49. 3-125. Let X denote the number of orders placed in a week in a city of 800,000 people. Then X is a Poisson random variable with = 0.25(8) = 2. a) P(X 3) = 1 P(X 2) = 1 [e-2 + e-2(2) + (e-222)/2!] = 1 0.6767 = 0.3233. b) Let Y denote the number of orders in 2 weeks. Then, Y is a Poisson random variable with = 4, and P(Y>2) =1- P(Y 2) = e-4 + (e-441)/1!+ (e-442)/2! =1 - [0.01832+0.07326+0.1465] = 0.7619. 3-126 a.) hypergeometric random variable with N = 500, n = 5, and K = 125 0.2357 f E X 6.0164 10 2.5524 11 375 125 500 5 5 0 = (0) = = E 0.3971 f E X 125(8.10855 8) 2.5525 11 375 125 500 5 4 1 = (1) = = E 0.2647 7750(8718875) 2.5524 11 375 125 500 5 3 2 = (2) = = E f X 0.0873 317750(70125) 2.5524 11 375 125 500 5 2 3 = (3) = = E f X 0.01424 9691375(375) 2.5524 11 375 125 500 5 1 4 = (4) = = E f X 0.00092 f E X 2.3453 8 2.5524 11 375 125 500 5 0 5 = (5) = = E b.) x 0 1 2 3 4 5 6 7 8 9 10 f(x) 0.0546 0.1866 0.2837 0.2528 0.1463 0.0574 0.0155 0.0028 0.0003 0.0000 0.0000 3-24
• 50. 3-127. Let X denote the number of totes in the sample that exceed the moisture content. Then X is a binomial random variable with n = 30. We are to determine p. 30 If P(X 1) = 0.9, then P(X = 0) = 0.1. Then ( ) 0 (1 ) 30 = 0.1 , giving 30ln(1p)=ln(0.1), 0 which results in p = 0.0739. p p 3-128 Let t denote an interval of time in hours and let X denote the number of messages that arrive in time t. Then, X is a Poisson random variable with = 10t. Then, P(X=0) = 0.9 and e-10t = 0.9, resulting in t = 0.0105 hours = 0.63 seconds 3-129. a) Let X denote the number of flaws in 50 panels. Then, X is a Poisson random variable with = 50(0.02) = 1. P(X = 0) = e-1 = 0.3679. b) Let Y denote the number of flaws in one panel, then P(Y 1) = 1 P(Y=0) = 1 e-0.02 = 0.0198. Let W denote the number of panels that need to be inspected before a flaw is found. Then W is a geometric random variable with p = 0.0198 and E(W) = 1/0.0198 = 50.51 panels. c) P(Y 1) = 1 P(Y = 0) = 1 e0.02 = 0.0198 Let V denote the number of panels with 1 or more flaws. Then V is a binomial random variable with n=50 and p=0.0198 50 + 0 50 0.01981 (0.9802)49 ( 2 ) 1 50 0.0198 (.9802) 0 P V = 50 2 48 = 0.0198 (0.9802) 0.9234 2 + Mind Expanding Exercises 3-130. Let X follow a hypergeometric distribution with parameters K, n, and N. To solve this problem, we can find the general expectation: E(Xk) = = ( = ) = n i ikP X i 0 N K n i = n i k N n K i i 0 Using the relationships N K i and 1 K K = 1 i i = N n 1 1 n N n we can substitute into E(XK): 3-25
• 51. E(Xk) = = = n i ikP X i 0 ( ) N K n i 1 N K N K K N N K [( 1) ] N K 1 1 1 1 ( 1) 1 1 1 1 1 0 1 0 1 0 = = = nK nK = + = + = = k n j k n i k n i k E Z N N n n j j K j N n n i i n i n i i Now, Z is also a hypergeometric random variable with parameters n 1, N 1, and K 1. To find the mean of X, E(X), set k = 1: E(X) = E Z nK N N nK = [( + 1)11] = If we let p = K/N, then E(X) = np. In order to find the variance of X using the formula V(X) = E(X2) [E(X)}2, the E(X2) must be found. Substituting k = 2 into E(Xk) we get E Z nK N = + = + n K E Z E nK ( ) (1) ( 1)( 1) + [ ] nK = + = 1 1 ( 2 ) [( 1)2 1 ] ( 1) N N N E Z N E X nK ( 1)( 1) 2 nK 1 ( n 1)( K 1) 1 + n K + nK nK nK Therefore, V(X) = = N N N N N N 1 1 N n If we let p = K/N, the variance reduces to V(X) = (1 ) 1 np p N 3-131. Show that using an infinite sum. = = 1 (1 ) 1 1 i p i p To begin, , by definition of an infinite sum this can be rewritten = as = = 1 1 1 (1 ) 1 (1 ) i i i p i p p p 1 1 (1 ) (1 ) 1 p 1 = = = = p p p p i p i 3-26
• 52. 3-132 E X a a b b a ( ) = [( + ( + 1) + ... + ]( + 1) = = 2 2 + + = ( ) ( ) [ ] + + + + + i b a i b a b a i b ( b 1)(2 b 1) a a a b a b b a a b a b a ( 1) 1 12 ( 1) ( 1) ( )( 1) ( ) ( 1)( ) + ( ) ( 1) ( 1) 1 ( 1)( ) 4 2 ( 1) (2 1) 6 6 1 4 1 ( ) 2 ( 1) 2 ( 1) 2 ( 1) 2 2 ( 1) 2 2 2 2 2 2 1 1 1 + = + + + + + + + = + + + = + = + = + + = + + + = = = = = b a b a b a b a V X b a b a b a b a b a b a b a b a b b a a b a i i b i a b i a b i a b a b i a i 3-133 Let X denote the number of nonconforming products in the sample. Then, X is approximately binomial with p = 0.01 and n is to be determined. If P(X 1) 0.90 , then P(X = 0) 0.10 . Now, P(X = 0) = (n )p0 (1 p)n (1 p)n (1 p)n 0.10 0 = . Consequently, , and n 229.11 . Therefore, n = 230 is required ln 0.10 = ln(1 ) p 3-134 If the lot size is small, 10% of the lot might be insufficient to detect nonconforming product. For example, if the lot size is 10, then a sample of size one has a probability of only 0.2 of detecting a nonconforming product in a lot that is 20% nonconforming. If the lot size is large, 10% of the lot might be a larger sample size than is practical or necessary. For example, if the lot size is 5000, then a sample of 500 is required. Furthermore, the binomial approximation to the hypergeometric distribution can be used to show the following. If 5% of the lot of size 5000 is nonconforming, then the probability of zero nonconforming product in the sample is approximately 7 1012 . Using a sample of 100, the same probability is still only 0.0059. The sample of size 500 might be much larger than is needed. 3-27
• 53. 3-135 Let X denote the number of panels with flaws. Then, X is a binomial random variable with n =100 and p is the probability of one or more flaws in a panel. That is, p = 1 e0.1 = 0.095. P X P X P X P X P X P X P X ( < 5) = ( 4) = ( = 0) + ( = 1) + ( = 2) + ( = 3) + ( = 4) ( ) ( ) ( ) ( ) ( ) 0.034 p p p p p p (1 ) (1 ) (1 ) = + + 100 4 96 4 100 3 97 3 (1 ) (1 ) 100 2 98 2 100 1 99 1 100 0 100 0 + + = p p p p 3-136 Let X denote the number of rolls produced. Revenue at each demand 0 1000 2000 3000 0 x 1000 0.05x 0.3x 0.3x 0.3x mean profit = 0.05x(0.3) + 0.3x(0.7) - 0.1x 1000 x 2000 0.05x 0.3(1000) + 0.05(x-1000) 0.3x 0.3x mean profit = 0.05x(0.3) + [0.3(1000) + 0.05(x-1000)](0.2) + 0.3x(0.5) - 0.1x 2000 x 3000 0.05x 0.3(1000) + 0.05(x-1000) 0.3(2000) + 0.05(x-2000) 0.3x mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000) + 0.05(x-2000)](0.3) + 0.3x(0.2) - 0.1x 3000 x 0.05x 0.3(1000) + 0.05(x-1000) 0.3(2000) + 0.05(x-2000) 0.3(3000)+ 0.05(x-3000) mean profit = 0.05x(0.3) + [0.3(1000)+0.05(x-1000)](0.2) + [0.3(2000)+0.05(x-2000)]0.3 + [0.3(3000)+0.05(x- 3000)]0.2 - 0.1x Profit Max. profit 0 x 1000 0.125 x $ 125 at x = 1000 1000 x 2000 0.075 x + 50 $ 200 at x = 2000 2000 x 3000 200 $200 at x = 3000 3000 x -0.05 x + 350 3-137 Let X denote the number of acceptable components. Then, X has a binomial distribution with p = 0.98 and n is to be determined such that P(X 100) 0.95 . n P(X 100) 102 0.666 103 0.848 104 0.942 105 0.981 Therefore, 105 components are needed. 3-28
• 54. CHAPTER 4 Section 4-2 < = = = = P X e xdx e x e 4-1. a) (1 ) ( ) 1 0.3679 1 1 b) (1 2.5) ( ) 2.5 1 2.5 0.2858 < < = = = = P X e xdx e x e e 1 2.5 1 3 = = = P X exdx c) ( 3) 0 3 d) ( 4) ( ) 1 4 0.9817 4 < = = = = P X e xdx e x e 0 4 0 = = = = P X e xdx e x e e) (3 ) ( ) 3 0.0498 3 3 4-2. a) ( < ) = = ( ) = = 0.10 x P x X e xdx e e . x x x Then,x=ln(0.10) = 2.3 x = = = = x x x P X x e xdx e e . Then, x = ln(0.9) = 0.1054 b) ( ) ( ) 1 0.10 0 0 P X < = x dx = x = 4 3 , because f (x) = 0 X for x < 3. 4-3 a) 0.4375 16 8 16 ( 4) 4 2 2 3 4 2 3 = P X > = x dx = x = 5 3.5 because f (x) = 0 X for x > 5. b) , 0.7969 16 8 16 ( 3.5) 5 2 2 3.5 5 2 3.5 = 5 4 c) 0.5625 16 8 16 (4 5) 5 2 2 4 5 2 4 = < < = = = P X x dx x 4.5 3 d) 0.7031 16 8 16 ( 4.5) 4.5 2 2 3 4.5 2 3 = < = = = P X x dx x P X > + P X < = x dx + x dx = x + x = 5 4.5 3.5 3 . e) 0.5 16 16 8 8 16 16 ( 4.5) ( 3.5) 3.5 2 2 2 2 3 5 2 4.5 3.5 2 3 5 4.5 = + 4-1
• 55. < = ( 4) = = P X e x dx e x , because f (x) = 0 X for x < 4. This can also be obtained from the fact that f (x) is a probability density function for 4 < x. X b) (2 5) 1 1 0.6321 5 4-4 a) (1 ) 1 4 ( 4) 4 P X = e ( x 4) dx = e ( x 4) = e = 4 c) P(5 < X ) = 1 P(X 5) . From part b., P(X 5) = 0.6321 . Therefore, P(5 < X ) = 0.3679 . d) (8 12) 12 4 8 0.0180 5 4 < < = ( 4) = = = P X e x dx e x e e 8 ( 4) 12 8 x < = ( 4) = = = x x x P X x e x dx e e . Then, x = 4 ln(0.10) = 6.303 e) ( ) 1 ( 4) 0.90 4 ( 4) 4 4-5 a) P(0 < X ) = 0.5 , by symmetry. b) (0.5 ) 1.5 0.5 0.5 0.0625 0.4375 1 P < X = x2dx = x = = 0.5 3 1 0.5 c) ( 0.5 0.5) 1.5 0.5 0.125 0.5 P X = x 2 dx = x 3 = 0.5 d) P(X < 2) = 0 e) P(X < 0 or X > 0.5) = 1 f) ( ) 1.5 0.5 3 1 0.5 0.5 3 0.05 0.5 0.5 1 P x < X = x2dx = x = x = x x Then, x = 0.9655 x 1000 x P X e dx e e ( 3000) 3 > = = = = 4-6. a) 0.05 1000 1000 3000 3000 x P X e dx e e e x (1000 2000) 1 2 b) 0.233 1000 2000 1000 2000 1000 1000 1000 < < = = = = x P X e dx e e x ( 1000) 1 c) 1 0.6321 1000 1000 0 1000 0 1000 1000 < = = = = x x x x 1000 x P X x e dx e e ( ) /1000 < = = = = d) 1 0.10 1000 1000 0 0 . Then, ex/1000 = 0.9 , and x = 1000 ln 0.9 = 105.36. 4-2
• 56. 4-7 a) ( 50) 2.0 2 0.5 50.25 P X > = dx = x = 50 50.25 50 50.25 ( ) 0.90 2.0 2 100.5 2 50.25 > = = = = Then, 2x = 99.6 and x = 49.8. b) P X x dx x x x x 4-8. a) ( 74.8) 1.25 1.25 0.25 74.8 P X < = dx = x = 74.6 b) P(X < 74.8 or X > 75.2) = P(X < 74.8) + P(X > 75.2) because the two events are mutually exclusive. The result is 0.25 + 0.25 = 0.50. c) (74.7 75.3) 1.25 1.25 75.3 1.25(0.6) 0.750 74.8 74.6 P < X < = dx = x = = 74.7 75.3 74.7 4-9 a) P(X < 2.25 or X > 2.75) = P(X < 2.25) + P(X > 2.75) because the two events are mutually exclusive. Then, P(X < 2.25) = 0 and 2.8 dx = = P(X > 2.75) = 2 2(0.05) 0.10 . 2.75 b) If the probability density function is centered at 2.5 meters, then fX(x) = 2 for 2.3 < x < 2.8 and all rods will meet specifications. x ( ) 1 2 1 and x2 4-10. Because the integral f x dx is not changed whether or not any of the endpoints x x are included in the integral, all the probabilities listed are equal. Section 4-3 4-11. a) P(X 6) = 1 (6) = 0 X P X F 4-12. a) ( 1.8) ( 1.8) (1.8) X P X < = P X = F because X is a continuous random variable. Then, (1.8) = 0.25(1.8) + 0.5 = 0.95 X F b) P(X > 1.5) = 1 P(X 1.5) = 1.125 = 0.875 c) P(X < -2) = 0 d) (1 < 0 ( ) e x 4-14. Now, f (x) = x / 8 for 3 < x < 5 and 16 9 x x 8 16 ( ) 2 3 2 3 F x = xdx = x = x X for 0 < x. Then, x 0, 3 x x ,3 < 5 < = 1, 5 16 9 ( ) 2 x F x X x ( ) (4 ) x x 1 x X F x = e dx = e = e 4-15. Now, f (x) = e(4x) for 4 < x and (4 ) 4 (4 ) 4 x for 4 < x. Then, x 0, 4 F x X x ( ) e (4 ) x = 1 , > 4 4-16. Now, e x /1000 f x 1000 ( ) = for 0 < x and /1000 x ( ) 1/1000 /1000 x x 1 x F x = e x dx = e /1000) = e X 4 4 for 0 < x. Then, x 0, 0 F x X x ( ) e /1000 x = 1 , > 0 P(X>3000) = 1- P(X 3000) = 1- F(3000) = e-3000/1000 = 0.5 x F x = dx = x 4-17. Now, f(x) = 1.25 for 74.6 < x < 75.4 and ( ) 1.25 1.25 93.25 74.6 for 74.6 < x < 75.4. Then, 0, < 74.6 x x 1.25 93.25, 74.6 < 75.4 = x x F x 1, 75.4 ( ) P(X > 75) = 1 P(X 75) = 1 F(75) = 1 0.5 = 0.5 because X is a continuous random variable. 4-4
• 58. 4-18 f (x) = 2e2x , x > 0 4-19. 0.2, 0 4 < < = 0.04, 4 < 9 ( ) x x f x 4-20. 0.25, 2 1 < < = 0.5, 1 < 1.5 ( ) x x f x X x = = = for 0 < x < 2. Then, ( ) 0.5 0.25 2 F x xdx x x 4-21. 0.5 2 2 0 0 x 0, 0 ( ) 2 0.25 , 0 2 < x x < = x x F x 1, 2 Section 4-4 4-22. 2 2 ( ) 0.25 0.25 4 0 4 0 2 = = = E X xdx x 4 3 2 3 2 3 = = V X x dx x ( ) 0.25( 2) 0.25 ( 2) 3 4 0 4 0 3 2 = + = 4-23. 2.6667 3 ( ) 0.125 0.125 4 0 4 0 3 = 2 = = E X x dx x = 2 = 3 16 + V ( X ) 0.125 x ( x ) dx 0.125 ( x x x ) dx 4 0 2 8 0.125( 1 ) 0.88889 2 64 9 16 3 3 4 64 9 4 0 4 0 2 3 3 4 3 = + = x x x 4-5
• 59. E X x dx x 4-24. 0 4 ( ) 1.5 1.5 1 1 1 1 4 = 3 = = x 3 2 4 V X x x dx x dx ( ) = 1.5 ( 0) = 1.5 0.6 5 1.5 1 1 5 1 1 1 1 = = 5 3 4-25. 4.083 24 8 24 ( ) 5 3 3 3 5 3 3 = = = = E X x xdx x V X x x dx x x 16.6709 x dx 5 3 2 = = + 0.3264 5 3 16.6709 2 8.166 3 1 8 4 8 8.166 8 8 8 ( ) ( 4.083) 5 3 4 3 2 3 2 = = + x x x 4-26. ( ) 2 50 50.25 E X = xdx = x = 49.75 2 50.25 49.75 50.25 = = 2 + 50.25 2 V X x dx x x dx ( ) 2( 50) 2 ( 100 2500) 49.75 49.75 3 2 2( 100 2500 ) = + 0.0208 50.25 3 2 49.75 = x x x 4-27. a.) 600 ln 109.39 ( ) 600 120 = dx = x = 2 100 120 100 x E X x 120 120 ( ) ( 109.39) 600 600 1 2(109.39) V X x x x dx dx = = + 120 100 100 2 1 (109.39) 2 100 2 x x x x 600( 218.78ln 109.39 ) 33.19 2 2 = = b.) Average cost per part = $0.50*109.39 = $54.70 = 3 = = E X x x dx x 4-28. ( ) 2 2 2 1 1 1 4-6
• 60. = 4-29. a) . E(X ) x10e 10( x 5)dx 5 Using integration by parts with u = x and dv = 10e10( x5)dx , we obtain 5.1 = 10( 5) + = = 10 ( ) 5 5 10( 5) 5 10( 5) 5 x E X xe x e x dx e = V(X ) (x 5.1)210e 10( x 5)dx Now, . Using the integration by parts with and 5 u = (x 5.1)2 dv = 10e10( x5) , we obtain = + V(X ) (x 5.1)2 e 10( x 5) 2 (x 5.1)e x dx . 5 10( 5) 5 From the definition of E(X) the integral above is recognized to equal 0. Therefore, V(X ) = (5 5.1)2 = 0.01. b) ( > 5.1) = 10 10( 5) = = 10(5.1 5) = 0.3679 P X e x dx e x e 5.1 10( 5) 5.1 4-30. a) E X x dx x = = = V X x dx x ( ) ( 1205) 0.1 0.1( 1205) , ( ) 2.887 8.333 3 ( ) 0.1 0.05 1205 1210 1200 1210 3 1200 2 1210 1200 2 1210 1200 = = = = = Therefore V X x b) Clearly, centering the process at the center of the specifications results in the greatest proportion of cables within specifications. (1195 1205) (1200 1205) 0.1 0.1 0.5 1205 P < X < = P < X < = dx = x = 1200 1205 1200 Section 4-5 4-31. a) E(X) = (5.5+1.5)/2 = 3.5, 1.333, 1.333 1.155 ( ) (5.5 1.5) 12 2 = = = = x V X and . b) ( 2.5) 0.25 0.25 0.25 2.5 P X < = dx = x = 1.5 2.5 1.5 4-7
• 61. 4-32. a) E(X) = (-1+1)/2 = 0, 1/ 3, 0.577 ( ) (1 ( 1)) 12 2 = = = x V X and x ( ) 0.5 0.5(2 ) 2 b) P x < X < x = dt = t = x = x x x x 1 Therefore, x should equal 0.90. 4-33. a) f(x)= 2.0 for 49.75 < x < 50.25. E(X) = (50.25 + 49.75)/2 = 50.0, 0.0208, 0.144 ( ) (50.25 49.75) 12 2 = = = x V X and . x b) F ( x ) = 2.0 dx for 49.75 < x < 50.25. Therefore, 49.75 0, 49.75 x x 2 99.5, 49.75 < 50.25 < = x x F x 1, 50.25 ( ) c) P(X < 50.1) = F(50.1) = 2(50.1) 99.5 = 0.7 4-34. a) The distribution of X is f(x) = 10 for 0.95 < x < 1.05. Now, 0, 0.95 10 9.5, 0.95 1.05 x x < < = x x F x X 1, 1.05 ( ) b) P ( X >1.02) =1 P ( X 1.02) =1 F = X (1.02) 0.3 c) If P(X > x)=0.90, then 1 F(X) = 0.90 and F(X) = 0.10. Therefore, 10x - 9.5 = 0.10 and x = 0.96. (1.05 0.95)2 d) E(X) = (1.05 + 0.95)/2 = 1.00 and V(X) = 0.00083 12 = 4-8
• 62. ( ) (1.5 + 2.2) = 4-35 1.85min 2 E X = 2 2 0.0408min ( ) (2.2 1.5) = 12 V X = ( 2) 1 2 P X < = dx dx x b) (1/ 0.7) (1/ 0.7) (1/ 0.7)(0.5) 0.7143 (2.2 1.5) 1.5 2 1.5 2 1.5 = = = = x x ( ) 1 = = c.) x = for 1.5 < x < 2.2. Therefore, F X dx dx x 1.5 1.5 1.5 (1/ 0.7) (1/ 0.7) (2.2 1.5) 0, < 1.5 x x (1/ 0.7) 2.14, 1.5 < 2.2 = x x F x 1, 2.2 ( ) 4-36 f (x) = 0.04 for 50< x = = = 70 60 60 50 P(X 60) 0.04dx 0.04x 0.4 b) < = = = 50 + ( ) 75 50 = E X = seconds c) 62.5 2 52.0833 ( ) (75 50) 12 2 = V X = seconds2 4-37. a) The distribution of X is f(x) = 100 for 0.2050 < x < 0.2150. Therefore, 0, 0.2050 100 20.50, 0.2050 0.2150 x < < = x x x F x 1, 0.2150 ( ) b) P(X > 0.2125) = 1 F(0.2125) = 1[100(0.2125) 20.50] = 0.25 c) If P(X > x)=0.10, then 1 F(X) = 0.10 and F(X) = 0.90. Therefore, 100x - 20.50 = 0.90 and x = 0.2140. d) E(X) = (0.2050 + 0.2150)/2 = 0.2100 m and (0.2150 0.2050) 2 V(X) = = 6 2 8.33 10 m 12 4-9
• 63. 40 P X > = dx = x = 4-38. a) ( 35) 0.1 0.1 0.5 35 40 35 40 b) P(X > x)=0.90 and P X x dt x . ( > ) = 0.1 = 0.1(40 x ) Now, 0.1(40-x) = 0.90 and x = 31 (40 30)2 c) E(X) = (30 + 40)/2 = 35 and V(X) = 8.33 12 = Section 4-6 4-39. a) P(Z 2.15) = p(Z < 2.15) = 0.98422 e) P(2.34 < Z < 1.76) = P(Z 2.34) = 0.95116 4-40. a) P(1 < Z < 1) = P(Z < 1) P(Z > 1) = 0.84134 (1 0.84134) = 0.68268 b) P(2 < Z < 2) = P(Z < 2) [1 P(Z < 2)] = 0.9545 c) P(3 < Z < 3) = P(Z < 3) [1 P(Z < 3)] = 0.9973 d) P(Z > 3) = 1 P(Z < 3) = 0.00135 e) P(0 < Z < 1) = P(Z < 1) P(Z < 0) = 0.84134 0.5 = 0.34134 4-41 a) P(Z < 1.28) = 0.90 b) P(Z < 0) = 0.5 c) If P(Z > z) = 0.1, then P(Z < z) = 0.90 and z = 1.28 d) If P(Z > z) = 0.9, then P(Z < z) = 0.10 and z = 1.28 e) P(1.24 < Z < z) = P(Z < z) P(Z < 1.24) = P(Z < z) 0.10749. Therefore, P(Z < z) = 0.8 + 0.10749 = 0.90749 and z = 1.33 4-42. a) Because of the symmetry of the normal distribution, the area in each tail of the distribution must equal 0.025. Therefore the value in Table II that corresponds to 0.975 is 1.96. Thus, z = 1.96. b) Find the value in Table II corresponding to 0.995. z = 2.58. c) Find the value in Table II corresponding to 0.84. z = 1.0 d) Find the value in Table II corresponding to 0.99865. z = 3.0. 4-10
• 64. 4-43. a) P(X < 13) = P(Z < (1310)/2) = P(Z < 1.5) = 0.93319 b) P(X > 9) = 1 P(X < 9) = 1 P(Z < (910)/2) = 1 P(Z < 0.5) = 0.69146. c) P(6 < X < 14) = P 6 10 Z 2 14 10 2 < < = P(2 < Z < 2) = P(Z < 2) P(Z < 2)] = 0.9545. d) P(2 < X < 4) = P 2 10 Z 2 4 10 2 < < = P(4 < Z < 3) = P(Z < 3) P(Z < 4) = 0.00132 e) P(2 < X < 8) = P(X < 8) P(X < 2) = P Z < P Z < 8 10 2 2 10 2 = P(Z < 1) P(Z < 6) = 0.15866. P Z x 10 = 0.5. Therefore, 2 4-44. a) P(X > x) = > 2 x10 = 0 and x = 10. P Z x 10 = 1 b) P(X > x) = > 2 P Z x 10 < 2 = 0.95. Therefore, P Z < x 10 2 = 0.05 and x 10 2 = 1.64. Consequently, x = 6.72. P(Z 0) P Z x 10 P x 10 < Z < = c) P(x < X < 10) = 0 2 < < 2 P Z x 10 = 0.2. = 0.5 < 2 Therefore, P Z < x 10 2 = 0.3 and x 10 2 = 0.52. Consequently, x = 8.96. d) P(10 x < X < 10 + x) = P(x/2 < Z < x/2)= 0.95. Therefore, x/2 = 1.96 and x = 3.92 e) P(10 x < X < 10 + x) = P(x/2 < Z < x/2) = 0.99. Therefore, x/2 = 2.58 and x = 5.16 4-11
• 65. P Z 11 5 4-45. a) P(X < 11) = < 4 = P(Z < 1.5) = 0.93319 P Z 0 5 b) P(X > 0) = > 4 = P(Z > 1.25) = 1 P(Z < 1.25) = 0.89435 7 5 P 3 5 Z c) P(3 < X < 7) = < < 4 4 = P(0.5 < Z < 0.5) = P(Z < 0.5) P(Z < 0.5) = 0.38292 9 5 P 2 5 Z d) P(2 < X < 9) = < < 4 4 = P(1.75 < Z < 1) = P(Z < 1) P(Z < 1.75)] = 0.80128 P 2 5 Z 8 5 =P(0.75 < Z < 0.75) = P(Z < 0.75) P(Z < 0.75) = 0.54674 e) P(2 < X < 8) = < < 4 4 P Z x 5 = 0.5. 4-46. a) P(X > x) = > 4 Therefore, x = 5. b) P(X > x) = P Z x 5 = 0.95. > 4 P Z x 5 = 0.05 Therefore, < 4 Therefore, x 5 4 = 1.64, and x = 1.56. P x 5 Z = 0.2. < < 1 c) P(x < X < 9) = 4 Therefore, P(Z < 1) P(Z < x 5 4 )= 0.2 where P(Z < 1) = 0.84134. Thus P(Z < x 5 4 ) = 0.64134. Consequently, x 5 4 = 0.36 and x = 6.44. 4-12
• 66. P 3 5 Z x = 0.95. d) P(3 < X < x) = < < 4 5 4 P Z x 5 P(Z < 0.5) = 0.95 and Therefore, < 4 P Z x 5 0.30854 = 0.95. < 4 Consequently, P Z x 5 = 1.25854. Because a probability can not be greater than one, there is no solution for x. In fact, P(3 < X) = P(0.5 < Z) = 0.69146. Therefore, even if x is set to infinity the probability requested cannot equal 0.95. e) P(5 x < X x) = > 100 x6000 Therefore, 100 = 1.65 and x = 5835. P Z 40 35 4-48. a) P(X < 40) = < 2 = P(Z < 2.5) = 0.99379 P Z 30 35 b) P(X < 30) = < 2 = P(Z < 2.5) = 0.00621 0.621%are scrapped 4-13
• 67. P Z 0.62 0.5 4-49. a) P(X > 0.62) = > 0.05 = P(Z > 2.4) = 1 P(Z 12.6) = P Z> . . . 12 6 12 4 01 = P(Z > 2) = 0.02275. Therefore, the proportion of cans scrapped is 0.00135 + 0.02275 = 0.0241, or 2.41% c) P(12.4 x < X < 12.4 + x) = 0.99. P x x Therefore, < Z < = 0.99 0.1 0.1 P Z x < = 0.995 and x = 0.1(2.58) = 0.258. Consequently, 0.1 The limits are ( 12.142, 12.658). P Z 45 65 = P(Z < -3) = 0.00135 4-51. a) P(X 1) = 1- P(Z < 1) = 1 - 0.841345= 0.158655 > b) P(X > 65) = 5 P Z x 60 = 0.99. c) P(X < x) = < 5 x60 Therefore, 5 = 2.33 and x = 72 4-14
• 68. 4-52. a) If P(X > 12) = 0.999, then P Z> 12 01 . = 0.999. Therefore, 12 . = 3.09 and = 12.309. 0 1 b) If P(X > 12) = 0.999, then P Z> 12 005 . = 0.999. Therefore, 12 . = -3.09 and = 12.1545. 0 05 P Z 0.5 0.4 4-53. a) P(X > 0.5) = > 0.05 = P(Z > 2) = 1 0.97725 = 0.02275 0.5 0.4 P 0.4 0.4 Z b) P(0.4 < X < 0.5) = < < 0.05 0.05 = P(0 < Z < 2) = P(Z < 2) P(Z < 0) = 0.47725 P Z x 0.4 = 0.90. c) P(X > x) = 0.90, then > 0.05 x0.4 Therefore, 0.05 = 1.28 and x = 0.336. P Z 70 60 4-54 a) P(X > 70) = > 4 = P Z < 1 ( 2.5) = = 1 0.99379 0.00621 P Z 58 60 b) P(X < 58) = < 4 = P Z < ( 0.5) 0.308538 = c) 1,000,000 bytes*8 bits/byte = 8,000,000 bits 133.33seconds 8,000,000 bits = , 60 000 bits/sec 4-15
• 69. a) P(X > 90.3) + P(X < 89.7) P Z 89.7 90.2 P Z 90.3 90.2 + = > 0.1 < 0.1 = P(Z > 1) + P(Z < 5) = 1 P(Z < 1) + P(Z < 5) =1 0.84134 +0 = 0.15866. Therefore, the answer is 0.15866. b) The process mean should be set at the center of the specifications; that is, at = 90.0. P 89.7 90 Z 90.3 90 c) P(89.7 < X < 90.3) = < < 0.1 0.1 = P(3 < Z < 3) = 0.9973. The yield is 100*0.9973 = 99.73% 90.3 90 P 89.7 90 Z = P(3 < Z < 3) =