November 2001 - Course 1 SOA Solutions

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15. 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 ⎪ X i = ⎨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 ⎣ i ⎦ ( ) ( ) ( ) ( ) ( ) ( ) 2 [ Xi] E⎡X ⎤ i { E[ Xi] } ( ) E⎡X ⎤ = 0 0.6 + 1 0.1 + 2 0.3 = 1.3 , and 2 2 Var = ⎣ ⎦ − = 1.3 − 0.7 = 0.81 Since X1,..., X 100 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 X + ... + E X = 100 0.7 = 70 [ ] [ ] [ ] ( ) 1 100 and variance Var S = Var X + ... + Var X = 100 0.81 = 81 [ ] [ ] [ ] ( ) 1 100 Consequently, ⎡S −70 90.5 −70⎤ Pr[ S ≤ 90.5] = Pr ⎢ ≤ ⎣ 9 9 ⎥ ⎦ = Pr[ Z ≤2.28] = 0.99 16. C Observe Pr 4 < S < 8 = Pr ⎡ ⎣4 < S < 8 N = 1⎤ ⎦Pr N = 1 + Pr ⎡ ⎣4 < S < 8 N > 1⎤ ⎦Pr N > 1 1 − 8 ( ) ( ) 4 − 1 − 5 5 1 2 −1 = e − e + e −e * 3 6 = 0.122 *Uses that if X has an exponential distribution with mean µ [ ] [ ] [ ] b 1 1 − − −t µ −t µ µ Pr ( a≤ X ≤ b) = Pr ( X ≥a) −Pr ( X ≥ b) = ∫ e dt− e dt = e −e µ µ ∫µ ∞ a ∞ b a Course 1 Solutions 9 November 2001
17. B Note that 20 [ > ] = ( − ) ∫ Pr X x 0.005 20 t dt x ⎛ 1 2⎞ 20 = 0.005⎜20t− t ⎟ x ⎝ 2 ⎠ ⎛ 1 2 ⎞ = 0.005⎜400 −200 − 20x + x ⎟ ⎝ 2 ⎠ ⎛ 1 2 ⎞ = 0.005⎜200 − 20x+ x ⎟ ⎝ 2 ⎠ where 0< x < 20 . Therefore, [ X > ] [ X ] ( ) 1 2( ) 2 ( ) 1 ( ) Pr 16 200 − 20 16 + 16 8 1 Pr ⎡ ⎣X > 16 X > 8⎤ ⎦ = = = = Pr > 8 200 − 20 8 + 8 72 9 2 2 18. D Note that ⎧2 x for x< 0 and x> 1 ⎪ f′ ( x) = ⎨ 1 ⎪ for 0 < x < 1 ⎩2 x Furthermore, lim lim 0 = x − f′ ( x) ≠ ( ) 0 x + f′ x =+∞ → →0 and lim lim 1 2 = f ( x ) f ( x + ′ ≠ − ′ ) = x→1 x→1 2 so f is not differentiable at x = 0 or x = 1, although f is differentiable everywhere else. By contrast, f is continuous everywhere since lim f ( x lim ) f ( x + = − ) = 0 x→0 x→0 and lim f x lim = f x = ( ) ( ) + − 1 x→1 x→1 Course 1 Solutions 10 November 2001
Page 1 and 2: Course 1 Solutions November 2001 Ex
Page 3 and 4: 3. A Since 3 4 2 5 S = 175L A , we
Page 5 and 6: 7. A Cov C , C = Cov X + Y, X + 1.2
Page 7 and 8: 8. D The function F() t , 1≤t ≤
Page 9: 12. E Observe that x f′ x = f′
Page 13 and 14: 21. D Let N1 and N 2 denote the num
Page 15 and 16: 24. D 1 1 4 2 100 x y must be maxim
Page 17 and 18: 27. D Let X denote the number of em
Page 19 and 20: 29. B Let X and Y denote repair cos
Page 21 and 22: 31. E The general solution of the d
Page 23 and 24: 35. E If X is the random variable r
Page 25: 39. D We are given that P x =− x

November 2001 - Course 1 SOA Solutions

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