Given the life table below, and a constant force of interest = 5%, compute the following. (Note that Z means a present value random variable for a single insurance payout as appropriate to the notation, like in the lectures)

a) A5 b) A! 15:4 c) A5:1 (that is, 4E5) 1 C 0 1 4 5 6 7 8 9 10 11 12 1, 100.000 97,000 94,000 89,000 82,000 74,000 65,000 55,000 43,000 34,000 22,000 10,000 0 a) A5 b) A! 15:4 c) A5:1 (that is, 4E5) 1 C 0 1 4 5 6 7 8 9 10 11 12 1, 100.000 97,000 94,000 89,000 82,000 74,000 65,000 55,000 43,000 34,000 22,000 10,000 0