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Q5-Answer-232-2010W

Q5-Answer-232-2010W - 10< T 50<-1 δ log[1(1 ¯ P(1-v...

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SOLUTIONS TO Quiz 5 – ACTSC 232, WINTER 2010 1. (a) By the equivalence principle, ¯ P ¯ a 80 = 5000 ¯ A 80 . Thus, ¯ P = 5000 δ ¯ A 80 / (1 - ¯ A 80 ), where ¯ A 80 = Z 35 0 e - 0 . 01 t 1 35 dt = 0 . 843748 . Hence, P = 270 . 00. (b) L = 5000 v T 80 - ¯ P ¯ a T 80 = (5000 + ¯ P δ ) v T 80 - ¯ P δ . Thus, 0 . 2 = Pr { L > 0 } = Pr ( v T 80 > ¯ P 5000 δ + ¯ P ) = Pr ( T 80 < 1 δ log ¯ P + 5000 δ ¯ P !) . Hence, 0 . 2 = 1 δ log ¯ P + 5000 δ ¯ P ! × 1 35 and P = 689 . 58. 2. (a) We have L = 8000 v K x +1 - P ¨ a K x +1 , 0 K x 9 , 8000 v K x +1 - P ¨ a 10 , 10 K x 24 , (3000 + P ¨ s 10 (1 + i ) 15 ) v 25 - P ¨ a 10 , K x 25 . (b) By the equivalence principle, P ¨ a x : 10 = 8000 A 1 x : 25 + (3000 + P ¨ s 10 (1 + i ) 15 ) 25 E x . We have ¨ a x : 10 = 9 X k =0 v k k p x = 8 . 586 , 25 E x = v 25 25 p x = 0 . 416862 , A x : 25 = 1 - d ¨ a x : 25 = 0 . 664281 , A 1 x : 25 = A x : 25 - 25 E x = 0 . 247419 , ¨ s 10 = (1 + i ) 10 - 1 d = 11 . 1812 . Thus, P = 1407 . 82. 3. (a) We have L = ( - ¯ P ¯ a T 50 , 0 < T 50 10 , v 10 ¯ a T 50 - 10 - ¯ P ¯ a 10 = ¯ a T 50 - ¯ a 10 - ¯ P ¯ a 10 = ¯ a T 50 - (1 + ¯ P a 10 , T 50 > 10 . (b) We have 0 . 85 = Pr { L < 0 } = Pr { 0 < T 50 10 } + Pr { ¯ a T 50 - (1 + ¯ P a 10 < 0 , T 50 > 10 } = 0 . 25 + Pr { 1 - v T 50 < (1 + ¯ P )(1 - v 10 ) , T 50 > 10 } , 1
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which implies that
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Unformatted text preview: { 10 < T 50 <-1 δ log[1-(1 + ¯ P )(1-v 10 )] } =-1 δ log[1-(1 + ¯ P )(1-v 10 )]-10 40 . Hence ¯ P = 1 . 467 . Alternatively, 0 . 25 = Pr { L ≥ } = Pr { ¯ a T 50-(1 + ¯ P )¯ a 10 ≥ , T 50 > 10 } , which gives ¯ P = 1 . 467 . 4. We have L = v K x +1-P ¨ a K x +1 = ± 1 + P d ² v K x +1-P d and V ar ( L ) = ± 1 + P d ² 2 V ar ( v K x +1 ) = ± 1 + P d ² 2 ± 2 A x-( A x ) 2 ² . By the equivalence principle, A x = P ¨ a x . Thus, P = A x ¨ a x = A x 1-A x d . Hence, V ar ( L ) = 2 A x-( A x ) 2 (1-A x ) 2 . 2...
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