Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | Juni 2016 |
| Nomor Soal | : | 19 |
SOAL
Jika \({T_x}\) dan \({T_y}\) adalah saling bebas, hitunglah nilai dari \({}_{\left. 2 \right|}{q_{xy}}\) (pembulatan terdekat) diberikan:
\({\begin{array}{*{20}{c}}{{q_x} = 0,08}&{{q_{x + 1}} = 0,09}&{{q_{x + 2}} = 0,10}\end{array}}\)
\({\begin{array}{*{20}{c}}{{q_y} = 0,10}&{{q_{y + 1}} = 0,15}&{{q_{y + 2}} = 0,20}\end{array}}\)
- 0,10
- 0,14
- 0,18
- 0,20
- 0,24
| Diketahui | Jika \({T_x}\) dan \({T_y}\) adalah saling bebas, hitunglah nilai dari \({}_{\left. 2 \right|}{q_{xy}}\) (pembulatan terdekat) diberikan: \({\begin{array}{*{20}{c}}{{q_x} = 0,08}&{{q_{x + 1}} = 0,09}&{{q_{x + 2}} = 0,10}\end{array}}\)
\({\begin{array}{*{20}{c}}{{q_y} = 0,10}&{{q_{y + 1}} = 0,15}&{{q_{y + 2}} = 0,20}\end{array}}\) |
| Rumus yang digunakan | - \({}_{\left. t \right|u}{q_{xy}} = {}_t{p_{xy}} – {}_{t + u}{p_{xy}}\)
- \({}_t{p_{xy}} = {}_t{p_x} \cdot {}_t{p_y}\)
- \({}_t{p_x} = \prod\limits_{k = 0}^{t – 1} {{p_{x + k}}} \)
- \({}_t{p_x} = 1 – {}_t{q_x}\)
|
| Proses pengerjaan | \({}_{\left. 2 \right|}{q_{xy}} = {}_2{p_{xy}} – {}_3{p_{xy}}\)
\({}_{\left. 2 \right|}{q_{xy}} = {}_2{p_x} \cdot {}_2{p_y} – {}_3{p_x} \cdot{}_3{p_y}\)
\({}_{\left. 2 \right|}{q_{xy}} = \left( {{p_x}} \right)\left( {{p_{x + 1}}} \right)\left( {{p_y}} \right)\left( {{p_{y + 1}}} \right) – \left( {{p_x}} \right)\left( {{p_{x + 1}}} \right)\left( {{p_{x + 2}}} \right)\left( {{p_y}} \right)\left( {{p_{y + 1}}} \right)\left( {{p_{y + 2}}} \right)\)
\({}_{\left. 2 \right|}{q_{xy}} = \left( {0.92} \right)\left( {0.91} \right)\left( {0.9} \right)\left( {0.85} \right) – \left( {0.92} \right)\left( {0.91} \right)\left( {0.9} \right)\left( {0.9} \right)\left( {0.85} \right)\left( {0.8} \right)\)
\({}_{\left. 2 \right|}{q_{xy}} = 0.17932\) |
| Jawaban | C. 0,18 |
