Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Jika diketahui
- \(e_0^ \circ = 40\)
- \(S\left( x \right) = 1 – \frac{x}{\omega }{\rm{,}}\) atau \(0 \le x \le \omega \)
Hitunglah \(e_{20}^ \circ \)
- 30
- 36
- 40
- 42
- 50
| Diketahui | \(e_0^ \circ = 40\) \(S\left( x \right) = 1 – \frac{x}{\omega }{\rm{,}}\) untuk \(0 \le x \le \omega \) |
| Rumus yang digunakan | \({}_t{p_x} = \frac{{S\left( {x + t} \right)}}{{S\left( x \right)}}\) \(e_x^ \circ = \int\limits_0^\infty {{}_t{p_x}dt} \) |
| Proses Pengerjaan | \({}_t{p_x} = \frac{{S\left( {x + t} \right)}}{{S\left( x \right)}} = \frac{{\frac{{\omega – x – t}}{\omega }}}{{\frac{{\omega – x}}{\omega }}}\) \(= \frac{{\omega – x – t}}{{\omega – x}}\) |
| \(e_0^ \circ = \int\limits_0^\omega {\frac{{\omega – t}}{\omega }dt} \)
\(= \left. {\frac{{\omega t – \frac{1}{2}{t^2}}}{\omega }} \right|_0^\omega \)z \(= \omega – \frac{1}{2}\omega \) \(40 = \frac{1}{2}\omega \) \(\omega = 80\) | |
| \(e_{20}^ \circ = \int\limits_0^{60} {{}_t{p_{20}}dt} \) \(= \int\limits_0^{60} {\frac{{60 – t}}{{60}}dt} \) \(= \left. {\frac{{60t – \frac{1}{2}{t^2}}}{{60}}} \right|_0^{60}\) \(= 60 – 30\) \(= 30\) | |
| Jawaban | a. 30 |