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Unformatted text preview: ANSWERS TO TUTORIAL 3 – ACTSC 232, WINTER 2010 1. (a) The present value of the benefits is Z = ( 2000 v T 75 , T 75 ≤ 5 1200 v 5 , T 75 > 5 . (b) The APV is E [ Z ] = 2000 Z 5 e- . 03 t × 1 25 dt + 1200 e- . 03 × 5 × 20 25 = 1197 . 73 . (c) We have E [ Z 2 ] = 2000 2 Z 5 e- 2 × . 03 t × 1 25 dt + 1200 2 e- 2 × . 03 × 5 × 20 25 = 1544576 . 00 . Thus, V ar [ Z ] = E [ Z 2 ]- ( E [ Z ]) 2 = 110028 . 00. (d) We know that Z ∈ { 1032 . 86 } ∪ [1721 . 42 , 2000) . Thus, the probability is Pr { 1700 < Z < 1900 } = Pr { 1721 . 42 < 2000 v T 75 < 1900 } = 0 . 13161 . (e) The probability is Pr { 900 < Z < 1100 } = Pr { Z = 1200 v 5 } = 0 . 8 . 2. (a) A 1 50: 20 = 1 60 ∑ 19 k =0 v k +1 = 1 60 a 20 = 0 . 24796 . (b) A 50: 20 = A 1 50: 20 + v 20 20 p 50 = 0 . 61708 . (c) ( IA ) 50 = 1 60 59 X k =0 ( k + 1) v k +1 = 1 60 ( Ia ) 60 = 10 . 1788 . 3. (a) i. The present value is Z = ( 1000 v T 95 , T 95 ≤ 3 1000 v 3 , T 95 > 3 ii. The single benefit premium is 1000 ¯ A 95: 3 = 1000 i...
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