Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | April 2019 |
| Nomor Soal | : | 17 |
SOAL
Diberikan informasi sebagai berikut:
- \({T_x}\) dan \({T_y}\) saling lepas (independent)
| \(k\) | \({q_{x + k}}\) | \({q_{y + k}}\) |
| 0 | 0,08 | 0,10 |
| 1 | 0,09 | 0,15 |
| 2 | 0,10 | 0,20 |
Hitunglah \({}_{\left. 2 \right|}{q_{xy}}\) (gunakan pembulatan terdekat)
- 0,179
- 0,192
- 0,205
- 0,218
- 0,231
| Diketahui | Diberikan informasi sebagai berikut:- \({T_x}\) dan \({T_y}\) saling lepas (independent)
| \(k\) | \({q_{x + k}}\) | \({q_{y + k}}\) | | 0 | 0,08 | 0,10 | | 1 | 0,09 | 0,15 | | 2 | 0,10 | 0,20 |
|
| Rumus yang digunakan | \({}_{\left. t \right|u}{q_{xy}} = {}_t{p_{xy}} \cdot {}_u{q_{x + t:y + t}} = {}_t{p_{xy}} – {}_{t + u}{p_{xy}}\)
\({}_t{p_{xy}} = {}_t{p_x} \cdot {}_t{p_y}\)
\({}_t{p_x} = \prod\limits_{k = 0}^{t – 1} {{p_{x + k}}} \)
\({}_t{p_x} = 1 – {}_t{q_x}\) |
| Proses pengerjaan | \({}_{\left. 2 \right|}{q_{xy}} = {}_2{p_{xy}} – {}_3{p_{xy}}\)
\({}_{\left. 2 \right|}{q_{xy}} = {}_2{p_x} \cdot {}_2{p_y} – {}_3{p_x} \cdot {}_3{p_y}\)
\({}_{\left. 2 \right|}{q_{xy}} = \left[ {\left( {{p_x}} \right)\left( {{p_{x + 1}}} \right)} \right]\left[ {\left( {{p_y}} \right)\left( {{p_{y + 1}}} \right)} \right] – \left[ {\left( {{p_x}} \right)\left( {{p_{x + 1}}} \right)\left( {{p_{x + 2}}} \right)} \right]\left[ {\left( {{p_y}} \right)\left( {{p_{y + 1}}} \right)\left( {{p_{y + 2}}} \right)} \right]\)
\({}_{\left. 2 \right|}{q_{xy}} = \left[ {\left( {0.92} \right)\left( {0.91} \right)} \right]\left[ {\left( {0.9} \right)\left( {0.85} \right)} \right] – \left[ {\left( {0.92} \right)\left( {0.91} \right)\left( {0.9} \right)} \right]\left[ {\left( {0.9} \right)\left( {0.85} \right)\left( {0.8} \right)} \right]\)
\({}_{\left. 2 \right|}{q_{xy}} = 0.1793\) |
| Jawaban | a. 0,179 |
