Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Berikut diberikan tabel double decrement
- \(\mu _{x + 0.5}^{(1)} = 0.02\)
- \(q_x^{(2)} = 0.01\)
- setiap decrement seragam
Hitunglah nilai \(p_x^{(1)}\).
- 0,975
- 0,9803
- 0,9831
- 0,9860
- 0,9901
| Diketahui |
dan setiap decrement menyebar seragam (UDD) |
| Rumus yang digunakan | Untuk kasus UUD maka \(\mu _{x + t}^{(j)} = \mu _x^{(j)}(t)\) \(\mu _x^{(j)}(t) = \frac{{q_x^{(j)}}}{{_tp_x^{(\tau )}}} = \frac{{q_x^{(j)}}}{{1 – t{\rm{ }}q_x^{(\tau )}}}\) \(_s{q_{x + t}} = \frac{{s{q_x}}}{{1 – t{q_x}}}\) |
| Proses Pengerjaan | Dengan mengaplikasikan rumus di atas diperoleh dipunyai \(q_x^{(2)} = 0.01,\) maka \(q_x^{(\tau )} = q_x^{(1)} + q_x^{(2)} = 0.01 + q_x^{(1)}\) \(\mu _{x + 0.5}^{(1)} = \mu _x^{(1)}(0.5) = \frac{{q_x^{(1)}}}{{_{0.5}p_x^{(\tau )}}} = \frac{{q_x^{(j)}}}{{1 – 0.5{\rm{ }}q_x^{(\tau )}}} = 0.02\) Sehingga \(\frac{{q_x^{(1)}}}{{1 – 0.5{\rm{ }}q_x^{(\tau )}}} = 0.02\) \(\Leftrightarrow \frac{{q_x^{(1)}}}{{1 – 0.5{\rm{ }}(0.01 + q_x^{(1)})}} = 0.02\) \(\Leftrightarrow q_x^{(1)} = 0.02(1 – 0.5{\rm{ }}(0.01 + q_x^{(1)}))\) \(\Leftrightarrow q_x^{(1)} = 0.02(1 – 0.005{\rm{ }} + 0.5q_x^{(1)})\) \(\Leftrightarrow q_x^{(1)} = 0.02 – 0.0001{\rm{ }} + 0.01q_x^{(1)}\) \(\Leftrightarrow 1.01q_x^{(1)} = 0.02 – 0.0001{\rm{ }}\) \(\Leftrightarrow q_x^{(1)} = 0.1970297\)maka |
| Jawaban | b. 0,9803 |