EXAM P SAMPLE SOLUTIONS

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33. Solution: B Note that 20 ⎛ 1 2⎞ 20 Pr[ X > x] = ∫ 0.005( 20 − t) dt = 0.005 20 x ⎜ t − t ⎟ x ⎝ 2 ⎠ ⎛ 1 2⎞ ⎛ 1 2 = 0.005⎜400 −200 − 20x + x ⎟ = 0.005⎜200 − 20x+ x ⎝ 2 ⎠ ⎝ 2 ⎞ ⎟ ⎠ where 0 < x < 20 . Therefore, 2 Pr[ X > 16] 200 − 20( 16) + 1 ( 16) 8 1 Pr ⎡ 2 ⎣X > 16 X > 8⎤ ⎦ = = = = 2 Pr[ X > 8] 200 − 20( 8) + 1 ( 8) 72 9 2 -------------------------------------------------------------------------------------------------------- 34. Solution: C C x − We know the density has the form + for 0< x < 40 (equals zero otherwise). ( ) 2 10 First, determine the proportionality constant C from the condition ∫ 40 0 f ( xdx= ) 1 40 40 −2 −1 C C 2 1= ∫ C( 10 + x) dx=− C(10 + x) = − = C 0 0 10 50 25 so C = 25 2 , or 12.5 . Then, calculate the probability over the interval (0, 6): 6 −2 −1 6 ⎛ 1 1 ⎞ 12.5∫ ( 10 + x) dx=− ( 10 + x) = ( 12.5) 0.47 0 ⎜ − ⎟ = . 0 ⎝10 16 ⎠ -------------------------------------------------------------------------------------------------------- 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 40 ∫ 0 f ( x) dx= 1: 40 −1 40 2 1 = ∫ f ( xdx ) =− C(10 + x) | 0 = 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 55 :
36. Solution: B To determine k, note that 1 4 k 5 1 1 = ∫ k( 1− y) dy =− ( 1− y) 5 0 k = 5 0 k = 5 We next need to find P[V > 10,000] = P[100,000 Y > 10,000] = P[Y > 0.1] 1 ∫ 0.1 ( − y) dy =−( − y) 4 5 1 = 51 1 = (0.9) 5 = 0.59 and P[V > 40,000] = P[100,000 Y > 40,000] = P[Y > 0.4] = 0.1 1 ∫ 0.4 ( − y) dy =−( − y) 4 5 1 51 1 It now follows that P[V > 40,000⏐V > 10,000] PV [ > 40,000 ∩ V > 10,000] PV [ > 40,000] 0.078 = = = = 0.132 . PV [ > 10,000] PV [ > 10,000] 0.590 -------------------------------------------------------------------------------------------------------- 37. Solution: D Let T denote printer lifetime. Then f(t) = ½ e –t/2 , 0 ≤ t ≤ ∞ Note that 1 1 −t/2 −t/21–1/2 P[T ≤ 1] = ∫ e dt = e = 1 – e = 0.393 2 0 0 2 0.4 = (0.6) 5 = 0.078 . 1 −t/2 −t/22 P[1 ≤ T ≤ 2] = ∫ e dt = e 1 2 1 –1/2 –1 = e − e = 0.239 Next, denote refunds for the 100 printers sold by independent and identically distributed random variables Y 1, . . . , Y100 where ⎧200 ⎪ Yi = ⎨100 ⎪⎩ 0 with probability 0.393 with probability 0.239 with probability 0.368 i = 1, . . . , 100 Now E[Y ] = 200(0.393) + 100(0.239) = 102.56 i Therefore, Expected Refunds = 100 ∑ E [ Yi ] = 100(102.56) = 10,256 . i= 1 Page 15 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: 31. Solution: D Let X denote the nu
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 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

EXAM P SAMPLE SOLUTIONS

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