EXAM P SAMPLE SOLUTIONS

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2 ∞ ∞ 9 ⎛ 1 ⎞ 9 ⎛ 2 1 ⎞ E [ Y ] = ∫ 1 dy 1 dy 1 3 ⎜ − 3 ⎟ = − + 1 3 3 6 y y ∫ ⎜ ⎟ ⎝ ⎠ y ⎝ y y ⎠ ∞ ⎛ 9 18 9 ⎞ ⎡ 9 18 9 ⎤ = ∫1 ⎜ − + dy 3 6 9 ⎟ = 2 5 8 y y y ⎢− + − 2y 5y 8y ⎥ ⎝ ⎠ ⎣ ⎦ ⎡1 2 1⎤ = 9 ⎢ − + = 2.025 (in thousands) ⎣2 5 8⎥ ⎦ -------------------------------------------------------------------------------------------------------- 77. Solution: D 2 21 ( x ydxdy ) 1 1 8 + Prob. = 1− ∫∫ = 0.625 Note c { } ( X ≤ ) U( Y ≤ ) = ( X > ) I ( Y > ) Pr ⎡⎣ 1 1 ⎤⎦ Pr ⎡⎣ 1 1 ⎤⎦ (De Morgan's Law) = 1− Pr⎡⎣( X > 1) I ( Y > 1) ⎤⎦ 2 21 = 1− ∫ ( ) 1∫ x+ y dxdy 1 8 1 21 2 2 = 1− ( ) 1 8∫ x+ y dy 1 2 1 2 = 1− ⎡ 16 ∫ 1 ⎣ 2 + 2 − 2 + 1 ⎤ ⎦ 1 = 1− ⎡ 48 ⎣ 3 + 2 − 3 + 1 ⎤ ⎦ 2 1 1 = 1− 48 64−27− 2 + 18 30 = 1− = = 0.625 48 48 ( y ) ( y ) dy ( y ) ( y ) ( 7 8) -------------------------------------------------------------------------------------------------------- 78. Solution: B That the device fails within the first hour means the joint density function must be integrated over the shaded region shown below. This evaluation is more easily performed by integrating over the unshaded region and subtracting from 1. Page 32 of 55 ∞ 1
( X < ) ∪ ( Y < ) Pr ⎡⎣ 1 1 ⎤⎦ 2 3 3 3 3 3 x+ y x + 2xy 1 = 1− ∫∫ dx dy = 1− dy 1 ( 9 6y1 2y) dy 27 ∫ = − + − − 54 54 ∫ 1 1 1 1 1 3 1 3 1 2 1 32 11 ( ) ( ) ( ) 54 ∫ y dy y y 1 54 1 54 54 27 = 1− 8+ 4 = 1− 8 + 2 = 1− 24+ 18−8− 2 = 1− = = 0.41 -------------------------------------------------------------------------------------------------------- 79. Solution: E The domain of s and t is pictured below. Note that the shaded region is the portion of the domain of s and t over which the device fails sometime during the first half hour. Therefore, ⎡⎛ 1⎞ ⎛ 1⎞⎤ 1/2 1 1 1/2 Pr ⎢⎜S ≤ ⎟∪⎜T ≤ ⎟ = f ( s, t) dsdt+ f ( s, t) d 2 2 ⎥ ∫0 ∫1/2 ∫0∫ sdt 0 ⎣⎝ ⎠ ⎝ ⎠⎦ (where the first integral covers A and the second integral covers B). -------------------------------------------------------------------------------------------------------- 80. Solution: C By the central limit theorem, the total contributions are approximately normally nμ = 2025 3125 = 6,328,125 and standard deviation distributed with mean ( )( ) σ n = 250 2025 = 11,250 . From the tables, the 90 th percentile for a standard normal random variable is 1.282 . Letting p be the 90 th percentile for total contributions, p−nμ = 1.282, and so p= nμ + 1.282σ n = 6,328,125 + ( 1.282)( 11, 250) = 6,342,548 . σ n Page 33 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: 74. Solution: E First note R = 10/T
Page 35 and 36: 40 − 3n −2≥ n To find such an
Page 37 and 38: -----------------------------------
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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