Exam p samplesolutions

1. 62&,(7< 2) $&78$5,(6$68$/7 $78$5,$/ 62,(7 (;$0 3 352%$%,/,7 3 6$03/( (;$0 62/87,216 RSULJKW E WKH 6RFLHW RI $FWXDULHV DQG WKH DVXDOW $FWXDULDO 6RFLHW 6RPH RI WKH TXHVWLRQV LQ WKLV VWXG QRWH DUH WDNHQ IURP SDVW 62$$6 H[DPLQDWLRQV Page 1 of 67

2. 1. Solution: D Let G event that a viewer watched gymnastics B event that a viewer watched baseball S event that a viewer watched soccer Then we want to find Pr ª G ‰ B ‰ S

3. º 1 Pr G ‰ B ‰ S

4. c ¬ ¼ 1 ª Pr G

5. Pr B

6. Pr S

7. Pr G ˆ B

8. Pr G ˆ S

9. Pr B ˆ S

10. Pr G ˆ B ˆ S

11. º ¬ ¼ 1 0.28 0.29 0.19 0.14 0.10 0.12 0.08

12. 1 0.48 0.52 -------------------------------------------------------------------------------------------------------- 2. Solution: A Let R = event of referral to a specialist L = event of lab work We want to find P[RˆL] = P[R] + P[L] – P[R‰L] = P[R] + P[L] – 1 + P[~(R‰L)] = P[R] + P[L] – 1 + P[~Rˆ~L] = 0.30 + 0.40 – 1 + 0.35 = 0.05 . -------------------------------------------------------------------------------------------------------- 3. Solution: D First note P A ‰ B@ P A@ P B @ P A ˆ B @ P A ‰ B '@ P A@ P B '@ P A ˆ B '@ Then add these two equations to get P A ‰ B @ P A ‰ B '@ 2 P A@ P B @ P B '@

13. P A ˆ B @ P A ˆ B '@

14. 0.7 0.9 2 P A@ 1 P ª A ˆ B

15. ‰ A ˆ B '

16. º ¬ ¼ 1.6 2 P A@ 1 P A@ P A@ 0.6 Page 2 of 67

17. 4. Solution: A For i 1, 2, let Ri event that a red ball is drawn form urn i Bi event that a blue ball is drawn from urn i . Then if x is the number of blue balls in urn 2, 0.44 Pr[ R1 R2

18. * B1 B2

19. ] Pr[ R1 R2 ] Pr B1 B2 @ Pr R1 @ Pr R2 @ Pr B1 @ Pr B2 @ 4 § 16 · 6 § x · ¨ ¸ ¨ ¸ 10 © x 16 ¹ 10 © x 16 ¹ Therefore, 32 3x 3x 32 2.2 x 16 x 16 x 16 2.2 x 35.2 3 x 32 0.8 x 3.2 x 4 -------------------------------------------------------------------------------------------------------- 5. Solution: D Let N(C) denote the number of policyholders in classification C . Then N(Young ˆ Female ˆ Single) = N(Young ˆ Female) – N(Young ˆ Female ˆ Married) = N(Young) – N(Young ˆ Male) – [N(Young ˆ Married) – N(Young ˆ Married ˆ Male)] = 3000 – 1320 – (1400 – 600) = 880 . -------------------------------------------------------------------------------------------------------- 6. Solution: B Let H = event that a death is due to heart disease F = event that at least one parent suffered from heart disease Then based on the medical records, 210 102 108 P ªH ˆ F c º ¬ ¼ 937 937 937 312 625 P ªF c º ¬ ¼ 937 937 P ª H ˆ F º 108 625 108 ¬ c ¼ and P ª H | F c º ¬ ¼ 0.173 P ªF º c ¬ ¼ 937 937 625 Page 3 of 67

20. 7. Solution: D Let A event that a policyholder has an auto policy H event that a policyholder has a homeowners policy Then based on the information given, Pr A ˆ H

21. 0.15 Pr A ˆ H c

22. Pr A

23. Pr A ˆ H

24. 0.65 0.15 0.50 Pr Ac ˆ H

25. Pr H

26. Pr A ˆ H

27. 0.50 0.15 0.35 and the portion of policyholders that will renew at least one policy is given by 0.4 Pr A ˆ H c

28. 0.6 Pr Ac ˆ H

29. 0.8 Pr A ˆ H

30. 0.4

31. 0.5

32. 0.6

33. 0.35

34. 0.8

35. 0.15

36. 0.53 53%

37. -------------------------------------------------------------------------------------------------------- 100292 01B-9 8. Solution: D Let C = event that patient visits a chiropractor T = event that patient visits a physical therapist We are given that Pr C @ Pr T @ 0.14 Pr C T

38. 0.22 Pr C c T c

39. 0.12 Therefore, 0.88 1 Pr ªC c T c º ¬ ¼ Pr C * T @ Pr C @ Pr T @ Pr C T @ Pr T @ 0.14 Pr T @ 0.22 2 Pr T @ 0.08 or Pr T @ 0.88 0.08

40. 2 0.48 Page 4 of 67

41. 9. Solution: B Let M event that customer insures more than one car S event that customer insures a sports car Then applying DeMorgan’s Law, we may compute the desired probability as follows: Pr M c ˆ S c

42. Pr ª M ‰ S

43. º 1 Pr M ‰ S

44. 1 ¬ Pr M

45. Pr S

46. Pr M ˆ S

47. ¼ c ª º ¬ ¼ 1 Pr M

48. Pr S

49. Pr S M

50. Pr M

51. 1 0.70 0.20 0.15

52. 0.70

53. 0.205 -------------------------------------------------------------------------------------------------------- 10. Solution: C Consider the following events about a randomly selected auto insurance customer: A = customer insures more than one car B = customer insures a sports car We want to find the probability of the complement of A intersecting the complement of B (exactly one car, non-sports). But P ( Ac ˆ Bc) = 1 – P (A ‰ B) And, by the Additive Law, P ( A ‰ B ) = P ( A) + P ( B ) – P ( A ˆ B ). By the Multiplicative Law, P ( A ˆ B ) = P ( B _ A ) P (A) = 0.15 * 0.64 = 0.096 It follows that P ( A ‰ B ) = 0.64 + 0.20 – 0.096 = 0.744 and P (Ac ˆ Bc ) = 0.744 = 0.256 -------------------------------------------------------------------------------------------------------- 11. Solution: B Let C = Event that a policyholder buys collision coverage D = Event that a policyholder buys disability coverage Then we are given that P[C] = 2P[D] and P[C ˆ D] = 0.15 . By the independence of C and D, it therefore follows that 0.15 = P[C ˆ D] = P[C] P[D] = 2P[D] P[D] = 2(P[D])2 (P[D])2 = 0.15/2 = 0.075 P[D] = 0.075 and P[C] = 2P[D] = 2 0.075 Now the independence of C and D also implies the independence of CC and DC . As a result, we see that P[CC ˆ DC] = P[CC] P[DC] = (1 – P[C]) (1 – P[D]) = (1 – 2 0.075 ) (1 – 0.075 ) = 0.33 . Page 5 of 67

54. 12. Solution: E “Boxed” numbers in the table below were computed. High BP Low BP Norm BP Total Regular heartbeat 0.09 0.20 0.56 0.85 Irregular heartbeat 0.05 0.02 0.08 0.15 Total 0.14 0.22 0.64 1.00 From the table, we can see that 20% of patients have a regular heartbeat and low blood pressure. -------------------------------------------------------------------------------------------------------- 13. Solution: C The Venn diagram below summarizes the unconditional probabilities described in the problem. In addition, we are told that 1 P A ˆ B ˆ C@ x P A ˆ B ˆ C | A ˆ B@ 3 P A ˆ B@ x 0.12 It follows that 1 1 x x 0.12

55. x 0.04 3 3 2 x 0.04 3 x 0.06 Now we want to find P ª A ‰ B ‰ C

56. º c P ª A ‰ B ‰ C

57. | Ac º c ¬ ¼ ¬ ¼ P ªA ºc ¬ ¼ 1 P A ‰ B ‰ C@ 1 P A@ 1 3 0.10

58. 3 0.12

59. 0.06 1 0.10 2 0.12

60. 0.06 0.28 0.467 0.60 Page 6 of 67

61. 14. Solution: A k 1 11 1 1 1 §1· pk = pk 1 pk 2 ˜ ˜ pk 3 ... ¨ ¸ p0 kt0 5 55 5 5 5 ©5¹ f f k §1· p0 5 1= ¦ pk ¦ ¨ 5 ¸ p0 k 0© ¹ 1 4 p0 k 0 1 5 p0 = 4/5 . Therefore, P[N 1] = 1 – P[N d1] = 1 – (4/5 + 4/5 ˜ 1/5) = 1 – 24/25 = 1/25 = 0.04 . -------------------------------------------------------------------------------------------------------- 15. Solution: C A Venn diagram for this situation looks like: We want to find w 1 x y z

62. 1 1 5 We have x y , xz , yz 4 3 12 Adding these three equations gives 1 1 5 x y

63. x z

64. y z

65. 4 3 12 2 x y z

66. 1 1 x y z 2 1 1 w 1 x y z

67. 1 2 2 Alternatively the three equations can be solved to give x = 1/12, y = 1/6, z =1/4 § 1 1 1· 1 again leading to w 1 ¨ ¸ © 12 6 4 ¹ 2 Page 7 of 67

68. 16. Solution: D Let N1 and N 2 denote the number of claims during weeks one and two, respectively. Then since N1 and N 2 are independent, Pr N1 N 2 7@ ¦ Pr N1 n @ Pr N 2 7 n@ 7 n 0 § 1 ·§ 1 · ¦ 7 ¨ n 1 ¸¨ 8 n ¸ n 0 © 2 ¹© 2 ¹ 1 ¦ 7 n 0 9 2 8 1 1 29 2 6 64 -------------------------------------------------------------------------------------------------------- 17. Solution: D Let O Event of operating room charges E Event of emergency room charges Then 0.85 Pr O ‰ E

69. Pr O

70. Pr E

71. Pr O ˆ E

72. Pr O

73. Pr E

74. Pr O

75. Pr E

76. Independence

77. Since Pr E c

78. 0.25 1 Pr E

79. , it follows Pr E

80. 0.75 . So 0.85 Pr O

81. 0.75 Pr O

82. 0.75

83. Pr O

84. 1 0.75

85. 0.10 Pr O

86. 0.40 -------------------------------------------------------------------------------------------------------- 18. Solution: D Let X1 and X2 denote the measurement errors of the less and more accurate instruments, respectively. If N(P,V) denotes a normal random variable with mean P and standard deviation V, then we are given X1 is N(0, 0.0056h), X2 is N(0, 0.0044h) and X1, X2 are X1 X 2 0.00562 h 2 0.00442 h 2 independent. It follows that Y = is N (0, ) = N(0, 2 4 0.00356h) . Therefore, P[0.005h d Y d 0.005h] = P[Y d 0.005h] – P[Y d 0.005h] = P[Y d 0.005h] – P[Y t 0.005h] ª 0.005h º = 2P[Y d 0.005h] – 1 = 2P « Z d 1 = 2P[Z d 1.4] – 1 = 2(0.9192) – 1 = 0.84. ¬ 0.00356h »¼ Page 8 of 67

87. 19. Solution: B Apply Bayes’ Formula. Let A Event of an accident B1 Event the driver’s age is in the range 16-20 B2 Event the driver’s age is in the range 21-30 B3 Event the driver’s age is in the range 30-65 B4 Event the driver’s age is in the range 66-99 Then Pr A B1

88. Pr B1

89. Pr B1 A

90. Pr A B1

91. Pr B1

92. Pr A B2

93. Pr B2

94. Pr A B3

95. Pr B3

96. Pr A B4

97. Pr B4

98. 0.06

99. 0.08

100. 0.1584 0.06

101. 0.08

102. 0.03

103. 0.15

104. 0.02

105. 0.49

106. 0.04

107. 0.28

108. -------------------------------------------------------------------------------------------------------- 20. Solution: D Let S = Event of a standard policy F = Event of a preferred policy U = Event of an ultra-preferred policy D = Event that a policyholder dies Then P D | U @ P U @ P U | D @ P D | S @ P S @ P D | F @ P F @ P D | U @ P U @ 0.001

109. 0.10

110. 0.01

111. 0.50

112. 0.005

113. 0.40

114. 0.001

115. 0.10

116. 0.0141 -------------------------------------------------------------------------------------------------------- 21. Solution: B Apply Baye’s Formula: Pr ªSeri. Surv.º ¬ ¼ Pr ªSurv. Seri.º Pr Seri.@ ¬ ¼ Pr ªSurv. Crit.º Pr Crit.@ Pr ªSurv. Seri.º Pr Seri.@ Pr ªSurv. Stab.º Pr Stab.@ ¬ ¼ ¬ ¼ ¬ ¼ 0.9

117. 0.3

118. 0.29 0.6

119. 0.1

120. 0.9

121. 0.3

122. 0.99

123. 0.6

124. Page 9 of 67

125. 22. Solution: D Let H Event of a heavy smoker L Event of a light smoker N Event of a non-smoker D Event of a death within five-year period 1 ª º Now we are given that Pr ¬ D L ¼ 2 Pr ª D N º and Pr ª D L º ¬ ¼ ¬ ¼ Pr ª D H º 2 ¬ ¼ Therefore, upon applying Bayes’ Formula, we find that Pr ª D H º Pr H @ ¬ ¼ Pr ª H D º ¬ ¼ Pr ª D N º Pr N Pr ª D L º Pr L Pr ª D H º Pr H ¬ ¼ @ ¬ ¼ @ ¬ ¼ @ 2 Pr ª D L º 0.2

126. ¬ ¼ 0.4 0.42 Pr ª D L º 0.5

127. Pr ª D L º 0.3

128. 2 Pr ª D L º 0.2

129. 0.25 0.3 0.4 1 2 ¬ ¼ ¬ ¼ ¬ ¼ -------------------------------------------------------------------------------------------------------- 23. Solution: D Let C = Event of a collision T = Event of a teen driver Y = Event of a young adult driver M = Event of a midlife driver S = Event of a senior driver Then using Bayes’ Theorem, we see that P[C Y ]P[Y ] P[Y~C] = P[C T ]P[T ] P[C Y ]P[Y ] P[C M ]P[ M ] P[C S ]P[ S ] (0.08)(0.16) = = 0.22 . (0.15)(0.08) (0.08)(0.16) (0.04)(0.45) (0.05)(0.31) -------------------------------------------------------------------------------------------------------- 24. Solution: B Observe Pr 1 d N d 4@ ª1 1 1 1 º ª1 1 1 1 1º Pr ¬ N t 1 N d 4 º ª ¼ « 6 12 20 30 » « 2 6 12 20 30 » Pr N d 4@ ¬ ¼ ¬ ¼ 10 5 3 2 20 2 30 10 5 3 2 50 5 Page 10 of 67

130. 25. Solution: B Let Y = positive test result D = disease is present (and ~D = not D) Using Baye’s theorem: P[Y | D]P[ D] (0.95)(0.01) P[D|Y] = = 0.657 . P[Y | D]P[ D] P[Y |~ D]P[~ D] (0.95)(0.01) (0.005)(0.99) -------------------------------------------------------------------------------------------------------- 26. Solution: C Let: S = Event of a smoker C = Event of a circulation problem Then we are given that P[C] = 0.25 and P[S~C] = 2 P[S~CC] P[ S C ]P[C ] Now applying Bayes’ Theorem, we find that P[C~S] = P[ S C ]P[C ] P[ S C C ]( P[C C ]) 2 P[ S C C ]P[C ] 2(0.25) 2 2 = . 2 P[ S C ]P[C ] P[ S C ](1 P[C ]) C C 2(0.25) 0.75 23 5 -------------------------------------------------------------------------------------------------------- 27. Solution: D Use Baye’s Theorem with A = the event of an accident in one of the years 1997, 1998 or 1999. P[ A 1997]P[1997] P[1997|A] = P[ A 1997][ P[1997] P[ A 1998]P[1998] P[ A 1999]P[1999] (0.05)(0.16) = = 0.45 . (0.05)(0.16) (0.02)(0.18) (0.03)(0.20) -------------------------------------------------------------------------------------------------------- Page 11 of 67

131. 28. Solution: A Let C = Event that shipment came from Company X I1 = Event that one of the vaccine vials tested is ineffective P I1 | C @ P C @ Then by Bayes’ Formula, P C | I1 @ P I1 | C @ P C @ P ª I1 | C c º P ªC c º ¬ ¼ ¬ ¼ Now 1 P C @ 5 1 4 P ªC c º 1 P C @ 1 ¬ ¼ 5 5 P I1 | C @

132. 0.10

133. 0.90

134. 30 1 29 0.141 º

135. 0.02

136. 0.98

137. 29 P ª I1 | C c ¬ ¼ 30 1 0.334 Therefore, P C | I1 @ 0.141

138. 1/ 5

139. 0.096 0.141

140. 1/ 5

141. 0.334

142. 4 / 5

143. -------------------------------------------------------------------------------------------------------- 29. Solution: C Let T denote the number of days that elapse before a high-risk driver is involved in an accident. Then T is exponentially distributed with unknown parameter O . Now we are given that 50 ³ Oe Ot 0.3 = P[T d 50] = dt e O t 50 = 1 – e–50O 0 0 Therefore, e–50O = 0.7 or O = (1/50) ln(0.7) 80 ³ Oe Ot It follows that P[T d 80] = dt e O t 80 0 = 1 – e–80O 0 (80/50) ln(0.7) =1–e = 1 – (0.7)80/50 = 0.435 . -------------------------------------------------------------------------------------------------------- 30. Solution: D e O O 2 e O O 4 Let N be the number of claims filed. We are given P[N = 2] = 3 = 3 ˜ P[N 2! 4! = 4]24 O = 6 O 2 4 O2 = 4 Ÿ O = 2 Therefore, Var[N] = O = 2 . Page 12 of 67

144. 31. Solution: D Let X denote the number of employees that achieve the high performance level. Then X follows a binomial distribution with parameters n 20 and p 0.02 . Now we want to determine x such that Pr X ! x @ d 0.01 or, equivalently, 0.99 d Pr X d x @ ¦

145. 0.02

146. 0.98

147. x 20 k 20 k k 0 k The following table summarizes the selection process for x: x Pr X x @ Pr X d x @ 0.98

148. 0.668 20 0 0.668 20 0.02

149. 0.98

150. 19 1 0.272 0.940 190 0.02

151. 0.98

152. 2 18 2 0.053 0.993 Consequently, there is less than a 1% chance that more than two employees will achieve the high performance level. We conclude that we should choose the payment amount C such that 2C 120, 000 or C 60, 000 -------------------------------------------------------------------------------------------------------- 32. Solution: D Let X = number of low-risk drivers insured Y = number of moderate-risk drivers insured Z = number of high-risk drivers insured f(x, y, z) = probability function of X, Y, and Z Then f is a trinomial probability function, so Pr z t x 2@ f 0, 0, 4

153. f 1, 0,3

154. f 0,1,3

155. f 0, 2, 2

156. 4! 0.20

157. 4 0.50

158. 0.20

159. 4 0.30

160. 0.20

161. 0.30

162. 0.20

163. 4 3 3 2 2 2!2! 0.0488 Page 13 of 67

164. 33. Solution: B Note that § 1 · Pr X ! x @ 0.005 20 t

165. dt 20 ³ 0.005 ¨ 20t t 2 ¸ 20 x x © 2 ¹ § 1 · § 1 · 0.005 ¨ 400 200 20 x x 2 ¸ 0.005 ¨ 200 20 x x 2 ¸ © 2 ¹ © 2 ¹ where 0 x 20 . Therefore, Pr X ! 16@ 200 20 16

166. 1 16

167. 2 2 8 1 Pr ª X ! 16 X ! 8º ¬ ¼ Pr X ! 8@ 200 20 8

168. 1 8

169. 2 72 9 2 -------------------------------------------------------------------------------------------------------- 34. Solution: C We know the density has the form C 10 x

170. for 0 x 40 (equals zero otherwise). 2 40 First, determine the proportionality constant C from the condition ³ 0 f ( x)dx 1 : 40 C C 2 C 10 x

171. dx C (10 x) 1 40 2 1 ³ 0 0 10 50 25 C so C 25 2 , or 12.5 . Then, calculate the probability over the interval (0, 6): 1 6 §1 1· 12.5³ 10 x

172. dx 10 x

173. ¨ ¸ 12.5

174. 0.47 . 6 2 0 0 © 10 16 ¹ -------------------------------------------------------------------------------------------------------- 35. Solution: C Let the random variable T be the future lifetime of a 30-year-old. We know that the density of T has the form f (x) = C(10 + x)2 for 0 x 40 (and it is equal to zero otherwise). First, determine the proportionality constant C from the condition ³ 0 f ( x)dx 1: 40 40 2 1 = ³ f ( x)dx C (10 x) 1 |0 40 C 0 25 25 so that C = = 12.5. Then, calculate P(T 5) by integrating f (x) = 12.5 (10 + x)2 2 over the interval (0.5). Page 14 of 67

175. 36. Solution: B To determine k, note that 1 k k 1 = ³ k 1 y

176. dy 1 y

177. 4 5 1 0 5 0 5 k=5 We next need to find P[V 10,000] = P[100,000 Y 10,000] = P[Y 0.1] 1 ³ 5 1 y

178. dy 1 y

179. 4 5 1 = 0.1 = (0.9)5 = 0.59 and P[V 40,000] 0.1 1 ³ 5 1 y

180. dy 1 y

181. 4 5 1 = P[100,000 Y 40,000] = P[Y 0.4] = 0.4 = (0.6)5 = 0.078 . 0.4 It now follows that P[V 40,000~V 10,000] P[V ! 40, 000 ˆ V ! 10, 000] P[V ! 40, 000] 0.078 = = 0.132 . P[V ! 10, 000] P[V ! 10, 000] 0.590 ͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲͲ 37. Solution: D Let T denote printer lifetime. Then f(t) = ½ e–t/2, 0 d t d ¥ Note that 1 1 P[T d 1] = ³ e t / 2 dt e t / 2 1 = 1 – e–1/2 = 0.393 0 2 0 2 1 ³ 2e t / 2 e t / 2 2 P[1 d T d 2] = dt 1 = e –1/2 e –1 = 0.239 1 Next, denote refunds for the 100 printers sold by independent and identically distributed random variables Y1, . . . , Y100 where 200 with probability 0.393 ° Yi ®100 with probability 0.239 i = 1, . . . , 100 °0 with probability 0.368 ¯ Now E[Yi] = 200(0.393) + 100(0.239) = 102.56 100 Therefore, Expected Refunds = ¦ E Y @ = 100(102.56) = 10,256 . i 1 i Page 15 of 67

182. 38. Solution: A Let F denote the distribution function of f. Then F x

183. Pr X d x @ x x ³1 3t 4 dt t 3 1 1 x 3 Using this result, we see Pr ª X 2

184. ˆ X t 1.5

185. º ¬ ¼ Pr X 2@ Pr X d 1.5@ Pr X 2_ X t 1.5@ Pr X t 1.5@ Pr X t 1.5@ F 2

186. F 1.5

187. 1.5

188. 2

189. 3 3 3 §3· 1 ¨ ¸ 0.578 1 F 1.5

190. 1.5

191. 3 ©4¹ -------------------------------------------------------------------------------------------------------- 39. Solution: E Let X be the number of hurricanes over the 20-year period. The conditions of the problem give x is a binomial distribution with n = 20 and p = 0.05 . It follows that P[X 2] = (0.95)20(0.05)0 + 20(0.95)19(0.05) + 190(0.95)18(0.05)2 = 0.358 + 0.377 + 0.189 = 0.925 . -------------------------------------------------------------------------------------------------------- 40. Solution: B Denote the insurance payment by the random variable Y. Then 0 if 0 X d C Y ® ¯ X C if C X 1 Now we are given that 0.5 C 0.5 C 0.64 Pr Y 0.5

192. Pr 0 X 0.5 C

193. ³ 0.5 C

194. 2 2 2 x dx x 0 0 Therefore, solving for C, we find C r0.8 0.5 Finally, since 0 C 1 , we conclude that C 0.3 Page 16 of 67

195. 41. Solution: E Let X = number of group 1 participants that complete the study. Y = number of group 2 participants that complete the study. Now we are given that X and Y are independent. Therefore, ^ P ª X t 9

196. ˆ Y 9

197. º ‰ ª X 9

198. ˆ Y t 9

199. º ¬ ¼ ¬ ¼ ` P ª X t 9

200. ˆ Y 9

201. º P ª X 9

202. ˆ Y t 9

203. º ¬ ¼ ¬ ¼ 2 P ª X t 9

204. ˆ Y 9

205. º ¬ ¼ (due to symmetry) 2 P X t 9 @ P Y 9 @ 2 P X t 9@ P X 9@ (again due to symmetry) 2 P X t 9@ 1 P X t 9@

206. 2 ª 10

207. 0.2

208. 0.8

209. 10

210. 0.8

211. º ª1 10

212. 0.2

213. 0.8

214. 10

215. 0.8

216. º 9 10 9 10 ¬ 9 10 ¼¬ 9 10 ¼ 2 0.376@1 0.376@ 0.469 -------------------------------------------------------------------------------------------------------- 42. Solution: D Let IA = Event that Company A makes a claim IB = Event that Company B makes a claim XA = Expense paid to Company A if claims are made XB = Expense paid to Company B if claims are made Then we want to find ¬^ Pr ª I A ˆ I B º ‰ ª I A ˆ I B

217. ˆ X A X B

218. º C ¼ ¬ ¼ ` Pr ª I A ˆ I B º Pr ª I A ˆ I B

219. ˆ X A X B

220. º ¬ C ¼ ¬ ¼ Pr ª I A º Pr I B @ Pr I A @ Pr I B @ Pr X A X B @ ¬ ¼ C (independence) 0.60

221. 0.30

222. 0.40

223. 0.30

224. Pr X B X A t 0@ 0.18 0.12 Pr X B X A t 0@ Now X B X A is a linear combination of independent normal random variables. Therefore, X B X A is also a normal random variable with mean M E X B X A @ E X B @ E X A @ 9, 000 10, 000 1, 000 and standard deviation V Var X B

225. Var X A

226. 2000

227. 2000

228. 2 2 2000 2 It follows that Page 17 of 67

229. ª 1000 º Pr X B X A t 0@ Pr « Z t (Z is standard normal) ¬ 2000 2 » ¼ ª 1 º Pr « Z t ¬ 2 2»¼ ª 1 º 1 Pr « Z ¬ 2 2»¼ 1 Pr Z 0.354@ 1 0.638 0.362 Finally, ^ Pr ª I A ˆ I B º ‰ ª I A ˆ I B

230. ˆ X A X B

231. º ¬ C ¼ ¬ ¼ ` 0.18 0.12

232. 0.362

233. 0.223 -------------------------------------------------------------------------------------------------------- 43. Solution: D If a month with one or more accidents is regarded as success and k = the number of failures before the fourth success, then k follows a negative binomial distribution and the requested probability is 4 k 3 k § 3 · § 2 · 3 Pr k t 4@ 1 Pr k d 3@ 1 ¦ k

234. ¨ ¸ ¨ ¸ k 0 ©5¹ © 5¹ § 3· 4 ª 3 § 2 ·0 4 § 2 ·1 5 § 2 · 2 6 § 2 ·3 º 1 ¨ ¸ « 0

235. ¨ ¸ 1

236. ¨ ¸ 2

237. ¨ ¸ 3

238. ¨ ¸ » © 5¹ « ©5¹ ¬ ©5¹ ©5¹ © 5¹ » ¼ 4 § 3 · ª 8 8 32 º 1 ¨ ¸ «1 » © 5 ¹ ¬ 5 5 25 ¼ 0.2898 Alternatively the solution is 4 4 4 2 4 3 § 2· 4 § 2· 3 5 § 2· § 3· 6 § 2· § 3· ¨ ¸ 1

239. ¨ ¸ 2

240. ¨ ¸ ¨ ¸ 3

241. ¨ ¸ ¨ ¸ 0.2898 ©5¹ © 5¹ 5 © 5¹ © 5¹ © 5¹ © 5¹ which can be derived directly or by regarding the problem as a negative binomial distribution with i) success taken as a month with no accidents ii) k = the number of failures before the fourth success, and iii) calculating Pr k d 3@ Page 18 of 67

242. 44. Solution: C If k is the number of days of hospitalization, then the insurance payment g(k) is g(k) = 100k ^ for k 1, 2, 3 300 50 (k 3) for k 4, 5. 5 Thus, the expected payment is ¦ g (k ) p k 1 k 100 p1 200 p2 300 p3 350 p4 400 p5 = 1 100 u 5 200 u 4 300 u 3 350 u 2 400 u 1

243. =220 15 -------------------------------------------------------------------------------------------------------- 45. Solution: D 0 4 x2 4 x 2 x3 x3 8 64 56 28 Note that E X

244. 0 ³ 2 10 dx ³ 0 10 dx 30 2 30 0 30 30 30 15 -------------------------------------------------------------------------------------------------------- 46. Solution: D The density function of T is 1 t / 3 f t

245. e , 0 t f 3 Therefore, E X @ E ª max T , 2

246. º ¬ ¼ 2 2 t / 3 f t ³0 3 e dt ³ e t / 3 dt 2 3 f 2e t / 3 _ 0 te t / 3 _ f ³ e t / 3 dt 2 2 2 2 / 3 2 / 3 2e 2 2e 3e t / 3 _ f 2 2 3e 2 / 3 Page 19 of 67

247. 47. Solution: D Let T be the time from purchase until failure of the equipment. We are given that T is exponentially distributed with parameter O = 10 since 10 = E[T] = O . Next define the payment x for 0 d T d 1 °x ° P under the insurance contract by P ® for 1 T d 3 ° 2 °0 ¯ for T ! 3 We want to find x such that 1 3 1 x –t/10 x 1 –t/10 x 1000 = E[P] = ³ ³ 2 10 e dt = xe t /10 e dt + e t /10 3 1 0 10 1 0 2 = x e –1/10 + x – (x/2) e –3/10 + (x/2) e –1/10 = x(1 – ½ e –1/10 – ½ e–3/10) = 0.1772x . We conclude that x = 5644 . -------------------------------------------------------------------------------------------------------- 48. Solution: E Let X and Y denote the year the device fails and the benefit amount, respectively. Then the density function of X is given by f x

248. 0.6

249. 0.4

250. , x 1, 2,3... x 1 and 1000 5 x

251. if x 1, 2,3, 4 ° y ® °0 ¯ if x ! 4 It follows that E Y @ 4000 0.4

252. 3000 0.6

253. 0.4

254. 2000 0.6

255. 0.4

256. 1000 0.6

257. 0.4

258. 2 3 2694 -------------------------------------------------------------------------------------------------------- 49. Solution: D Define f ( X ) to be hospitalization payments made by the insurance policy. Then 100 X ° if X 1, 2,3 f (X ) ® ¯300 25 X 3

259. ° if X 4,5 and Page 20 of 67

260. 5 E ª f X

261. º ¬ ¼ ¦ f k

262. Pr X k 1 k@ §5· § 4· § 3· § 2· §1· 100 ¨ ¸ 200 ¨ ¸ 300 ¨ ¸ 325 ¨ ¸ 350 ¨ ¸ © 15 ¹ © 15 ¹ © 15 ¹ © 15 ¹ © 15 ¹ 1 640 100 160 180 130 70@ 213.33 3 3 -------------------------------------------------------------------------------------------------------- 50. Solution: C Let N be the number of major snowstorms per year, and let P be the amount paid to (3 / 2) n e 3/ 2 the company under the policy. Then Pr[N = n] = , n = 0, 1, 2, . . . and n! 0 for N 0 P ® . ¯10, 000( N 1) for N t 1 f (3 / 2) n e 3/ 2 Now observe that E[P] = ¦10, 000(n 1) n 1 n! f (3 / 2) n e 3/ 2 = 10,000 e–3/2 + n 0 ¦10, 000(n 1) n! = 10,000 e–3/2 + E[10,000 (N – 1)] = 10,000 e + E[10,000N] – E[10,000] = 10,000 e–3/2 + 10,000 (3/2) – 10,000 = 7,231 . –3/2 -------------------------------------------------------------------------------------------------------- 51. Solution: C Let Y denote the manufacturer’s retained annual losses. x for 0.6 x d 2 Then Y ® ¯2 for x ! 2 f 2 ª 2.5(0.6) 2.5 º ª 2.5(0.6) 2.5 º 2 2.5(0.6) 2.5 2(0.6) 2.5 ³ dx ³ 2 « ³ x 2.5 f and E[Y] = x« » » dx dx 0.6 ¬ x3.5 ¼ 2 ¬ x 3.5 ¼ 0.6 x 2.5 2 2.5(0.6) 2.5 2(0.6) 2.5 2.5(0.6) 2.5 2.5(0.6) 2.5 (0.6) 2.5 = 2 1.5 = 0.9343 . 1.5 x1.5 0.6 (2) 2.5 1.5(2)1.5 1.5(0.6)1.5 2 Page 21 of 67

263. 52. Solution: A Let us first determine K. Observe that § 1 1 1 1· § 60 30 20 15 12 · § 137 · 1 K ¨1 ¸ K ¨ ¸ K¨ ¸ © 2 3 4 5¹ © 60 ¹ © 60 ¹ 60 K 137 It then follows that Pr N n @ Pr ª N n Insured Suffers a Loss º Pr Insured Suffers a Loss @ ¬ ¼ 60 3 0.05

264. , N 1,...,5 137 N 137 N Now because of the deductible of 2, the net annual premium P E X @ where 0 , if N d 2 X ® ¯N 2 , if N ! 2 Then, 3 § 1 · 3 ª 3 º ª º EX @ ¦ N 2

265. 137 N 1

266. ¨ 137 ¸ 2 «137 4

267. » 3 «137 5

268. » 5 P 0.0314 N 3 © ¹ ¬ ¼ ¬ ¼ -------------------------------------------------------------------------------------------------------- 53. Solution: D y for 1 y d 10 Let W denote claim payments. Then W ® ¯10 for y t 10 10 f 2 2 2 10 10 f It follows that E[W] = ³ y 3 dy ³ 10 3 dy 2 = 2 – 2/10 + 1/10 = 1.9 . 1 y 10 y y1 y 10 Page 22 of 67

269. 54. Solution: B Let Y denote the claim payment made by the insurance company. Then 0 with probability 0.94 ° Y ®Max 0, x 1

270. with probability 0.04 °14 ¯ with probability 0.02 and E Y @ 0.94

271. 0

272. 0.04

273. 0.5003

274. ³1 x 1

275. e x / 2 dx 0.02

276. 14

277. 15 0.020012

278. ª ³1 xe x / 2 dx ³ e x / 2 dx º 0.28 15 15 « ¬ 1 » ¼ 0.28 0.020012

279. ª 2 xe x / 2 _ 15 2³ e x / 2 dx ³ e x / 2 dx º 15 15 « ¬ 1 1 1 » ¼ 0.28 0.020012

280. ª 30e 7.5 2e 0.5 ³ e x / 2 dx º 15 « ¬ 1 » ¼ 0.28 0.020012

281. ª 30e 2e 2e _ 1 º ¬ 7.5 0.5 x / 2 15 ¼ 0.28 0.020012

282. 30e 7.5 2e 0.5 2e7.5 2e0.5

283. 0.28 0.020012

284. 32e 7.5 4e 0.5

285. 0.28 0.020012

286. 2.408

287. 0.328 (in thousands) It follows that the expected claim payment is 328 . -------------------------------------------------------------------------------------------------------- 55. Solution: C k The pdf of x is given by f(x) = , 0 x f . To find k, note (1 x) 4 f k k 1 f k 1=³ (1 x)4 dx 0 3 (1 x)3 0 3 k=3 f 3x It then follows that E[x] = ³ (1 x) 0 4 dx and substituting u = 1 + x, du = dx, we see f f f 3(u 1) ª u 2 u 3 º ª1 1º E[x] = ³ du = 3 ³ (u 3 u 4 )du 3« » 3 « » = 3/2 – 1 = ½ . 1 u4 1 ¬ 2 3 ¼1 ¬ 2 3¼ Page 23 of 67

288. 56. Solution: C Let Y represent the payment made to the policyholder for a loss subject to a deductible D. 0 for 0 d X d D That is Y ® ¯ x D for D X d 1 Then since E[X] = 500, we want to choose D so that 1 ( x D) 2 1000 (1000 D) 2 1000 1 1 500 ³ ( x D)dx 4 D 1000 1000 2 D 2000 (1000 – D)2 = 2000/4 ˜ 500 = 5002 1000 – D = r 500 D = 500 (or D = 1500 which is extraneous). -------------------------------------------------------------------------------------------------------- 57. Solution: B 1 We are given that Mx(t) = for the claim size X in a certain class of accidents. (1 2500t ) 4 (4)(2500) 10, 000 First, compute Mxc(t) = (1 2500t ) 5 (1 2500t )5 (10, 000)(5)(2500) 125, 000, 000 Mxs(t) = (1 2500t )6 (1 2500t )6 Then E[X] = Mxc (0) = 10,000 E[X2] = Mxs (0) = 125,000,000 Var[X] = E[X2] – {E[X]}2 = 125,000,000 – (10,000)2 = 25,000,000 Var[ X ] = 5,000 . -------------------------------------------------------------------------------------------------------- 58. Solution: E Let XJ, XK, and XL represent annual losses for cities J, K, and L, respectively. Then X = XJ + XK + XL and due to independence M(t) = E ªe xt º E ªe J K L

289. º E ªe xJ t º E ª e xK t º E ªe xLt º x x x t ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ ¬ ¼ = MJ(t) MK(t) ML(t) = (1 – 2t)–3 (1 – 2t)–2.5 (1 – 2t)–4.5 = (1 – 2t)–10 Therefore, Mc(t) = 20(1 – 2t)–11 Ms(t) = 440(1 – 2t)–12 Msc(t) = 10,560(1 – 2t)–13 E[X3] = Msc(0) = 10,560 Page 24 of 67

290. 59. Solution: B The distribution function of X is given by 2.5 x 2.5 200

291. 200

292. 200

293. 2.5 2.5 F x

294. x ³ 200 t 3.5 dt t 2.5 1 x 2.5 , x ! 200 200 th Therefore, the p percentile x p of X is given by 200

295. 2.5 F xp

296. p 1 2.5 100 xp 200

297. 2.5 1 0.01 p 2.5 xp 200 1 0.01 p

298. 25 xp 200 xp 1 0.01 p

299. 25 200 200 It follows that x 70 x 30 93.06 0.30

300. 0.70

301. 25 25 -------------------------------------------------------------------------------------------------------- 60. Solution: E Let X and Y denote the annual cost of maintaining and repairing a car before and after the 20% tax, respectively. Then Y = 1.2X and Var[Y] = Var[1.2X] = (1.2)2 Var[X] = (1.2)2(260) = 374 . -------------------------------------------------------------------------------------------------------- z 61. Solution: A f The first quartile, Q1, is found by ¾ = f(x) dx . That is, ¾ = (200/Q1)2.5 or Q1 Q1 = 200 (4/3)0.4 = 224.4 . Similarly, the third quartile, Q3, is given by Q3 = 200 (4)0.4 = 348.2 . The interquartile range is the difference Q3 – Q1 . Page 25 of 67

302. 62. Solution: C First note that the density function of X is given by 1 °2 if x 1 ° ° f x

303. ® x 1 if 1 x 2 ° ° °0 ¯ otherwise Then 2 1 §1 3 1 2· EX

304. 1 ³ x x 1

305. dx 1

307. 1 ³ x 2 x 1

308. dx 1

311. ª E X

312. º 23 16 5 2 E X 2 ¬ ¼ ¨ ¸ 12 © 3 ¹ 12 9 36 -------------------------------------------------------------------------------------------------------- 63. Solution: C X if 0 d X d 4 Note Y ® ¯4 if 4 X d 5 Therefore, 41 54 1 2 4 4 5 E Y @ ³ xdx ³ dx x _ 0 x_ 4 0 5 4 5 10 5 16 20 16 8 4 12 10 5 5 5 5 5 41 5 16 1 3 4 16 5 E ªY 2 º ³ x 2 dx ³ ¬ ¼ 05 dx x _ 0 x_ 4 4 5 15 5 64 80 64 64 16 64 48 112 15 5 5 15 5 15 15 15 2 112 § 12 · Var Y @ E ªY 2 º E Y @

314. 64. Solution: A Let X denote claim size. Then E[X] = [20(0.15) + 30(0.10) + 40(0.05) + 50(0.20) + 60(0.10) + 70(0.10) + 80(0.30)] = (3 + 3 + 2 + 10 + 6 + 7 + 24) = 55 E[X2] = 400(0.15) + 900(0.10) + 1600(0.05) + 2500(0.20) + 3600(0.10) + 4900(0.10) + 6400(0.30) = 60 + 90 + 80 + 500 + 360 + 490 + 1920 = 3500 Var[X] = E[X2] – (E[X])2 = 3500 – 3025 = 475 and Var[ X ] = 21.79 . Now the range of claims within one standard deviation of the mean is given by [55.00 – 21.79, 55.00 + 21.79] = [33.21, 76.79] Therefore, the proportion of claims within one standard deviation is 0.05 + 0.20 + 0.10 + 0.10 = 0.45 . -------------------------------------------------------------------------------------------------------- 65. Solution: B Let X and Y denote repair cost and insurance payment, respectively, in the event the auto is damaged. Then 0 if x d 250 Y ® ¯ x 250 if x ! 250 and 1 1 12502 E Y @ ³ x 250

315. dx x 250

316. 1500 1500 2 250 521 250 1500 3000 3000 1 1 12503 x 250

317. dx x 250

318. 1500 1500 E ¬Y 2 º ³ 2 3 ª ¼ 250 434, 028 250 1500 4500 4500 Var Y @ E ªY 2 º ^ E Y @` 434, 028 521

319. 2 2 ¬ ¼ Var Y @ 403 -------------------------------------------------------------------------------------------------------- 66. Solution: E Let X1, X2, X3, and X4 denote the four independent bids with common distribution function F. Then if we define Y = max (X1, X2, X3, X4), the distribution function G of Y is given by G y

320. Pr Y d y @ Pr ª X 1 d y

321. ˆ X 2 d y

322. ˆ X 3 d y

323. ˆ X 4 d y

324. º ¬ ¼ Pr X 1 d y @ Pr X 2 d y @ Pr X 3 d y @ Pr X 4 d y @ ª F y

325. º 4 ¬ ¼ 1 3 5 1 sinS y

326. , 4 d yd 16 2 2 It then follows that the density function g of Y is given by Page 27 of 67

327. g y

328. G ' y

329. 1 1 sinS y

330. S cosS y

331. 3 4 S 3 5 cosS y 1 sinS y

332. 3 , d yd 4 2 2 Finally, E Y @ yg y

333. dy 5/ 2 ³ 3/ 2 S ycosS y 1 sinS y

334. dy 5/ 2 ³ 3 3/ 2 4 -------------------------------------------------------------------------------------------------------- 67. Solution: B The amount of money the insurance company will have to pay is defined by the random variable 1000 x if x 2 Y ® ¯2000 if x t 2 where x is a Poisson random variable with mean 0.6 . The probability function for X is e 0.6 0.6

335. k p x

336. k 0,1, 2,3 and k! k f 0.6 E Y @ 0 1000 0.6

337. e 2000e ¦ k 2 0.6 0.6 k! § f 0.6 k · 1000 0.6

338. e 0.6 2000 ¨ e 0.6 ¦ k 0 e0.6 0.6

340. k f 2000e 0.6 ¦ k 0 k! 2000e 0.6 1000 0.6

341. e 0.6 2000 2000e 0.6 600e0.6 573 0.6k e0.6 ¦ k f 1000

342. 0.6

343. e0.6 2000

344. 2 2 E ªY 2 º ¬ ¼ 2 k! 0.6k 2000

345. e0.6 ¦ k f 2000

346. e 0.6 ª 2000

347. 1000

348. º 0.6

349. e 0.6 2 2 2 2 0 k! ¬ ¼ 2000

350. 2000

351. e 0.6 ª 2000

352. 1000

353. º 0.6

354. e 0.6 2 2 2 2 ¬ ¼ 816,893 Var Y @ E ªY 2 º ^ E Y @` 816,893 573

355. 2 2 ¬ ¼ 488,564 Var Y @ 699 Page 28 of 67

356. 68. Solution: C Note that X has an exponential distribution. Therefore, c = 0.004 . Now let Y denote the x for x 250 claim benefits paid. Then Y ® and we want to find m such that 0.50 ¯250 for x t 250 m m = ³ 0.004e 0.004 x dx e 0.004 x 0 = 1 – e–0.004m 0 This condition implies e–0.004m = 0.5 Ÿ m = 250 ln 2 = 173.29 . -------------------------------------------------------------------------------------------------------- 69. Solution: D The distribution function of an exponential random variable T with parameter T is given by F t

357. 1 e t T , t ! 0 Since we are told that T has a median of four hours, we may determine T as follows: 1 F 4

358. 1 e4 T 2 1 e 4 T 2 4 ln 2

359. T 4 T ln 2

360. 5ln 2

361. Therefore, Pr T t 5

362. 1 F 5

363. e 5 T e 4 25 4 0.42 -------------------------------------------------------------------------------------------------------- 70. Solution: E Let X denote actual losses incurred. We are given that X follows an exponential distribution with mean 300, and we are asked to find the 95th percentile of all claims that exceed 100 . Consequently, we want to find p95 such that Pr[100 x p95 ] F ( p95 ) F (100) 0.95 = where F(x) is the distribution function of X . P[ X ! 100] 1 F (100) Now F(x) = 1 – e–x/300 . 1 e p95 / 300 (1 e 100 / 300 ) e 1/ 3 e p95 / 300 Therefore, 0.95 = 1 e1/ 3e p95 / 300 1 (1 e 100 / 300 ) e 1/ 3 e p95 / 300 = 0.05 e –1/3 p95 = 300 ln(0.05 e–1/3) = 999 Page 29 of 67

364. 71. Solution: A The distribution function of Y is given by G y

365. Pr T 2 d y

366. Pr T d y F y

368. 1 4 y for y ! 4 . Differentiate to obtain the density function g y

369. 4 y 2 Alternate solution: Differentiate F t

370. to obtain f t

371. 8t 3 and set y t 2 . Then t y and g y

372. f t y

374. dt dy f y

375. dt y

376. d §1 · 8 y 3 2 ¨ y 1 2 ¸ ©2 ¹ 4 y 2 -------------------------------------------------------------------------------------------------------- 72. Solution: E We are given that R is uniform on the interval 0.04, 0.08

377. and V 10, 000e R Therefore, the distribution function of V is given by F v

378. Pr V d v @ Pr ª10, 000e R d v º Pr ª R d ln v

379. ln 10, 000

380. º ¬ ¼ ¬ ¼ ln v

381. ln 10,000

382. 1 ln v

383. ln 10,000

384. 1 25ln v

385. 25ln 10, 000

386. 1 0.04 ³ 0.04 dr r 0.04 0.04 ª § v · º 25 « ln ¨ ¸ 0.04 » ¬ © 10, 000 ¹ ¼ -------------------------------------------------------------------------------------------------------- 73. Solution: E ª

387. º Y

388. 10 10 8 F y

389. Pr Y d y @ Pr ª10 X 0.8 d y º 8 ¬ ¼ Pr « X d Y » 1 e 10 ¬ 10 ¼ 1 1 § Y · 4 Y 10

390. 5 4 Therefore, f y

391. Fc y

393. 74. Solution: E First note R = 10/T . Then ª10 º ª 10 º § 10 · FR(r) = P[R d r] = P « d r » P «T t » 1 FT ¨ ¸ . Differentiating with respect to ¬T ¼ ¬ r¼ © r ¹ § § 10 · · §d · § 10 · r fR(r) = FcR(r) = d/dr ¨1 FT ¨ ¸ ¸ ¨ FT t

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