ACTSC-232-1081-Midterm2_solutions

# ACTSC-232-1081-Midterm2_solutions - ACTSC 232 WINTER 2008...

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ACTSC 232 – WINTER 2008 MID-TERM EXAM #2 – March 10 2008 SUGGESTED SOLUTIONS NAME: I.D.: Aids: Calculator Time: 50 minutes Examiner: Johnny Li Question Maximum Mark 1 5 2 9 3 18 4 8 Total 40 1

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1. You are given the following select-and-ultimate life table: [ x ] l [ x ] l [ x ]+1 l x +2 x + 2 [70] 22507 22200 21722 72 [71] 21500 21188 20696 73 [72] 20443 20126 19624 74 [73] 19339 19019 18508 75 [74] 18192 17871 17355 76 Compute the following: (a) The probability that a life age 71 dies between ages 75 and 76 given that the life was selected at age 70. [2] 4 | 1 q [70]+1 = l [70]+5 - l [70]+6 l [70]+1 = l 75 - l 76 l [70]+1 = 18508 - 17355 22200 = 0 . 05194 . (b) 0 . 5 p [70] , assuming UDD. [3] 0 . 5 p [70] = l [70]+0 . 5 l [70] = 1 2 ( l [70] + l [70]+1 ) l [70] = 1 2 (22507 + 22200) 22507 = 0 . 99318 . 2
2. You are given that for S T 0 ( t ) = 1 - t ω , t < ω . (a) Find an expression for μ x . [3] μ x = - S 0 T 0 ( x ) S T 0 ( x ) = 1 ω ω - x ω = 1 ω - x . (b) Assuming that the force of interest

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## This note was uploaded on 03/20/2011 for the course ACTSC 232 taught by Professor Matthewtill during the Summer '08 term at Waterloo.

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ACTSC-232-1081-Midterm2_solutions - ACTSC 232 WINTER 2008...

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