EXAM P SAMPLE SOLUTIONS

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-------------------------------------------------------------------------------------------------------- 85. Solution: B Denote the policy premium by P . Since x is exponential with parameter 1000, it follows from what we are given that E[X] = 1000, Var[X] = 1,000,000, Var[ X ] = 1000 and P = 100 + E[X] = 1,100 . Now if 100 policies are sold, then Total Premium Collected = 100(1,100) = 110,000 Moreover, if we denote total claims by S, and assume the claims of each policy are independent of the others then E[S] = 100 E[X] = (100)(1000) and Var[S] = 100 Var[X] = (100)(1,000,000) . It follows from the Central Limit Theorem that S is approximately normally distributed with mean 100,000 and standard deviation = 10,000 . Therefore, ⎡ 110,000 −100,000⎤ P[S ≥ 110,000] = 1 – P[S ≤ 110,000] = 1 – P⎢Z ≤ 10,000 ⎥ = 1 – P[Z ≤ 1] = 1 ⎣ ⎦ – 0.841 ≈ 0.159 . -------------------------------------------------------------------------------------------------------- 86. Solution: E Let X1,..., X 100 denote the number of pensions that will be provided to each new recruit. Now under the assumptions given, ⎧0 with probability 1− 0.4 = 0.6 ⎪ Xi = ⎨1 with probability ( 0.4)( 0.25) = 0.1 ⎪ ⎩2 with probability ( 0.4)( 0.75) = 0.3 for i = 1,...,100 . Therefore, E X = 0 0.6 + 1 0.1 + 2 0.3 = 0.7 , [ ] ( )( ) ( )( ) ( )( ) i 2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) 2 ⎡ ⎤ = + + = E X ⎣ i ⎦ 0 0.6 1 0.1 2 0.3 1.3 , and 2 [ Xi] E⎡X ⎤ i { E[ Xi] } ( ) 2 2 Var = ⎣ ⎦ − = 1.3− 0.7 = 0.81 Since X1,..., X100 are assumed by the consulting actuary to be independent, the Central Limit Theorem then implies that S = X1 + ... + X100 is approximately normally distributed with mean E[ S] = E[ X1] + ... + E[ X100] = 100( 0.7) = 70 and variance Var[ S] = Var [ X1] + ... + Var[ X100 ] = 100( 0.81) = 81 Consequently, ⎡S−70 90.5 −70⎤ Pr[ S ≤ 90.5] = Pr ⎢ ≤ ⎣ 9 9 ⎥ ⎦ = Pr[ Z ≤2.28] = 0.99 Page 36 of 55
-------------------------------------------------------------------------------------------------------- 87. Solution: D Let X denote the difference between true and reported age. We are given X is uniformly distributed on (−2.5,2.5) . That is, X has pdf f(x) = 1/5, −2.5 < x < 2.5 . It follows that μ = E[X] = 0 x 2.5 2 3 3 x x 2.5 2(2.5) σx ∫ dx = −2.5 = 5 15 15 −2.5 2 = Var[X] = E[X 2 ] = =2.083 σx =1.443 Now X 48 , the difference between the means of the true and rounded ages, has a distribution that is approximately normal with mean 0 and standard deviation 1.443 48 = 0.2083 . Therefore, ⎡ 1 1 ⎤ ⎡ −0.25 0.25 ⎤ P ⎢ − ≤ X48 ≤ = P ≤ Z ≤ ⎣ 4 4⎥ ⎦ ⎢ ⎣0.2083 0.2083⎥ = P[−1.2 ≤ Z ≤ 1.2] = P[Z ≤ 1.2] – P[Z ≤ – ⎦ 1.2] = P[Z ≤ 1.2] – 1 + P[Z ≤ 1.2] = 2P[Z ≤ 1.2] – 1 = 2(0.8849) – 1 = 0.77 . -------------------------------------------------------------------------------------------------------- 88. Solution: C Let X denote the waiting time for a first claim from a good driver, and let Y denote the waiting time for a first claim from a bad driver. The problem statement implies that the respective distribution functions for X and Y are F x −x /6 = 1 − e , x> 0 and ( ) ( ) − y /3 G y = 1 − e , y > 0 Therefore, Pr ⎡⎣ X ≤3∩ Y ≤ 2 ⎤⎦ = Pr X ≤3PrY ≤ 2 ( ) ( ) [ ] [ ] ( ) ( ) ( )( ) = F 3 G 2 = 1−e 1− e = 1−e − e + e − −1/2 −2/3−2/3−1/2 7/6 Page 37 of 55
Page 1 and 2: SOCIETY OF ACTUARIES/CASUALTY ACTUA
Page 3 and 4: 4. Solution: A For i = 1, 2, let Ri
Page 5 and 6: 9. Solution: B Let M = event that c
Page 7 and 8: 14. Solution: A 1 11 1 1 1 ⎛1⎞
Page 9 and 10: 19. Solution: B Apply Bayes’ Form
Page 11 and 12: 25. Solution: B Let Y = positive te
Page 13 and 14: 31. Solution: D Let X denote the nu
Page 15 and 16: 36. Solution: B To determine k, not
Page 17 and 18: 41. Solution: E Let X = number of g
Page 19 and 20: 44. Solution: C If k is the number
Page 21 and 22: 5 ( ) = ∑ ( ) Pr[ = ] E⎡⎣f X
Page 23 and 24: 54. Solution: B Let Y denote the cl
Page 25 and 26: 59. Solution: B The distribution fu
Page 27 and 28: 64. Solution: A Let X denote claim
Page 29 and 30: 68. Solution: C Note that X has an
Page 31 and 32: 74. Solution: E First note R = 10/T
Page 33 and 34: ( X < ) ∪ ( Y < ) Pr ⎡⎣ 1 1
Page 35: 40 − 3n −2≥ n To find such an
Page 39 and 40: 92. Solution: B Let X and Y denote
Page 41 and 42: Consequently, 1 2 1 2 ( , ) 34 f t
Page 43 and 44: 99. Solution: C We use the relation
Page 45 and 46: 105. Solution: A The calculation re
Page 47 and 48: 2 [ X Y] E⎡( X Y) ⎤ ( E[ X Y] )
Page 49 and 50: 111. Solution: E 3 f Pr ⎡ ⎣1< Y
Page 51 and 52: 116. Solution: D Denote the number
Page 53 and 54: 120. Solution: A We are given that
Page 55: 124. Solution: A − x −2 y Becau

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