Revision as of 22:21, 18 January 2024 by Admin (Created page with "You are given: (i) <math>\quad A_{35}=0.188</math> (ii) <math>A_{65}=0.498</math> (iii) <math>{ }_{30} p_{35}=0.883</math> (iv) <math>\quad i=0.04</math> Calculate <math>1000 \ddot{a}_{35: 30}^{(2)}</math> using the two-term Woolhouse approximation. <ul class="mw-excansopts"><li> 17,060</li><li> 17,310</li><li> 17,380</li><li> 17,490</li><li> 17,530</li></ul> {{soacopyright|2024}}")
Jan 18'24
Exercise
You are given:
(i) [math]\quad A_{35}=0.188[/math]
(ii) [math]A_{65}=0.498[/math]
(iii) [math]{ }_{30} p_{35}=0.883[/math]
(iv) [math]\quad i=0.04[/math]
Calculate [math]1000 \ddot{a}_{35: 30}^{(2)}[/math] using the two-term Woolhouse approximation.
- 17,060
- 17,310
- 17,380
- 17,490
- 17,530
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Jan 19'24
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.