Revision as of 00:14, 19 January 2024 by Admin (Created page with "'''Answer: C''' <math display="block"> \begin{aligned} \ddot{a}_{35: 30}^{(2)} & \approx \ddot{a}_{35: \overline{30}}-\frac{(m-1)}{2 m}\left(1-v^{30}{ }_{30} p_{35}\right) \\ \ddot{a}_{35: \overline{30}} & =\frac{1-A_{35: 30}}{d}=\frac{1-A_{35: 30}^{1}-{ }_{30} E_{35}}{d} \\ & =\frac{1-\left(A_{35}-{ }_{30} E_{35} \times A_{65}\right)-{ }_{30} E_{35}}{d} \end{aligned} </math> Since <math>{ }_{30} E_{35}=v^{30}{ }_{30} p_{35}=0.2722</math>, then <math display="block...")
Exercise
Jan 19'24
Answer
Answer: C
[[math]]
\begin{aligned}
\ddot{a}_{35: 30}^{(2)} & \approx \ddot{a}_{35: \overline{30}}-\frac{(m-1)}{2 m}\left(1-v^{30}{ }_{30} p_{35}\right) \\
\ddot{a}_{35: \overline{30}} & =\frac{1-A_{35: 30}}{d}=\frac{1-A_{35: 30}^{1}-{ }_{30} E_{35}}{d} \\
& =\frac{1-\left(A_{35}-{ }_{30} E_{35} \times A_{65}\right)-{ }_{30} E_{35}}{d}
\end{aligned}
[[/math]]
Since [math]{ }_{30} E_{35}=v^{30}{ }_{30} p_{35}=0.2722[/math], then
[[math]]
\begin{aligned}
& \ddot{a}_{35: 30}=\frac{1-\left(A_{35}-v^{30}{ }_{30} p_{35} \times A_{65}\right)-v^{30}{ }_{30} p_{35}}{d} \\
&=\frac{1-(0.188-(0.2722)(0.498))-0.2722}{(0.04 / 1.04)} \\
&=17.5592 \\
& \ddot{a}_{35: 30 \mid}^{(2)} \approx 17.5592-\frac{1}{4}(1-0.2722)=17.38 \\
& 1000 \ddot{a}_{35: 30}^{(2)} \approx 1000 \times 17.38=17,380
\end{aligned}
[[/math]]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.