Revision as of 21:38, 19 January 2024 by Admin (Created page with "'''Answer: B''' By the equivalence principle, <math display="block"> 4500 \bar{a}_{x: \overline{20}}=100,000 \bar{A}_{x: \overline{20}}^{1}+R \bar{a}_{x: \overline{20}} </math> where <math display="block"> \begin{aligned} & \bar{A}_{x: 20 \mid}^{1}=\frac{\mu}{\mu+\delta}\left(1-e^{-20(\mu+\delta)}\right)=\frac{0.04}{0.12}\left(1-e^{-20(0.12)}\right)=0.3031 \\ & \bar{a}_{x: 20 \mid}=\frac{1-e^{-20(\mu+\delta)}}{\mu+\delta}=\frac{1-e^{-20(0.12)}}{0.12}=7.5774 \end{...")
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Exercise


ABy Admin
Jan 19'24

Answer

Answer: B

By the equivalence principle,

[[math]] 4500 \bar{a}_{x: \overline{20}}=100,000 \bar{A}_{x: \overline{20}}^{1}+R \bar{a}_{x: \overline{20}} [[/math]]


where

[[math]] \begin{aligned} & \bar{A}_{x: 20 \mid}^{1}=\frac{\mu}{\mu+\delta}\left(1-e^{-20(\mu+\delta)}\right)=\frac{0.04}{0.12}\left(1-e^{-20(0.12)}\right)=0.3031 \\ & \bar{a}_{x: 20 \mid}=\frac{1-e^{-20(\mu+\delta)}}{\mu+\delta}=\frac{1-e^{-20(0.12)}}{0.12}=7.5774 \end{aligned} [[/math]]


Solving for [math]R[/math], we have

[[math]] R=4500-100,000\left(\frac{0.3031}{7.5774}\right)=500 [[/math]]

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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