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| rev | Admin | (Created page with "'''Answer: D''' Let <math>Y_{i}</math> be the present value random variable of the payment to life <math>i</math>. <math>E\left[Y_{i}\right]=\ddot{a}_{x}=\frac{1-A_{x}}{d}=11.55 \quad \operatorname{Var}\left[Y_{i}\right]=\frac{{ }^{2} A_{x}-\left(A_{x}\right)^{2}}{d^{2}}=\frac{0.22-0.45^{2}}{(0.05 / 1.05)^{2}}=7.7175</math> Then <math>Y=\sum_{i=1}^{100} Y_{i}</math> is the present value of the aggregate payments. <math>E[Y]=100 E\left[Y_{i}\right]=1155</math> and <ma...") | Jan 19'24 at 0:09 | +799 |