Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Untuk suatu “independent lives” \(\left( x \right)\) dan \(\left( y \right)\):
- \({q_x} = 0,05\)
- \({q_y} = 0,10\)
- Kematian berdistribusi seragam pada setiap tahun usia
Hitunglah \({}_{0,75}{q_{xy}}\) (pembulatan terdekat)
- 0,1088
- 0,1097
- 0,1106
- 0,1116
- 0,1125
| Diketahui | Suatu “independent lives” \(\left( x \right)\) dan \(\left( y \right)\):
|
| Rumus yang digunakan | Asumsi usia pecahan berdistribusi seragam \({}_s{q_x} = s \cdot {q_x}\) \({}_t{p_{xy}} = {}_t{p_x} \cdot {}_t{p_y}\) \({}_t{q_{xy}} = 1 – {}_t{p_{xy}}\) |
| Proses pengerjaan | \({}_{0.75}{q_x} = 0.75 \cdot {q_x} = 0.75\left( {0.05} \right) = 0.0375\) \({}_{0.75}{q_y} = 0.75 \cdot {q_y} = 0.75\left( {0.10} \right) = 0.075\) |
| \({}_{0.75}{p_{xy}} = {}_{0.75}{p_x} \cdot {}_{0.75}{p_y}\) \({}_{0.75}{p_{xy}} = \left( {1 – {}_{0.75}{q_x}} \right)\left( {1 – {}_{0.75}{q_y}} \right)\) \({}_{0.75}{p_{xy}} = \left( {1 – 0.0375} \right)\left( {1 – 0.075} \right) = 0.8903125\) | |
| \({}_{0.75}{q_{xy}} = 1 – {}_{0.75}{p_{xy}} = 1 – 0.8903125\) \({}_{0.75}{q_{xy}} = 0.1096875\) | |
| Jawaban | b. 0,1097 |