Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
Matematika Aktuaria |
Periode Ujian |
: |
November 2015 |
Nomor Soal |
: |
9 |
SOAL
Diberikan suatu rate kematian sebagai berikut
\({q_{75}} = 0,01\)
\({q_{76}} = 0,02\)
\({q_{77}} = 0,04\)
\(i = 0,05\)
Hitunglah \({\ddot a_{75:\left. {\overline {\, 3 \,}}\! \right| }}\) (pembulatan terdekat)
- 1,85
- 2,34
- 2,82
- 3,43
- 3,77
Diketahui |
\({q_{75}} = 0,01\)
\({q_{76}} = 0,02\)
\({q_{77}} = 0,04\)
\(i = 0,05\) |
Rumus yang digunakan |
\({}_t{p_x} = \prod\limits_{k = 0}^{t – 1} {{p_{x + k}}} \)
\({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \sum\limits_{k = 0}^{n – 1} {{v^k}{}_k{p_x}} \) |
Proses pengerjaan |
\({{\ddot a}_{75:\left. {\overline {\, 3 \,}}\! \right| }} = \sum\limits_{k = 0}^2 {{v^k}{}_k{p_x}} = {v^0}{}_0{p_{75}} + {v^1}{}_1{p_{75}} + {v^2}{}_2{p_{75}}\)
\({{\ddot a}_{75:\left. {\overline {\, 3 \,}}\! \right| }} = {v^0}{}_0{p_{75}} + {v^1}{p_{75}} + {v^2}{p_{75}}{p_{76}}\)
\({{\ddot a}_{75:\left. {\overline {\, 3 \,}}\! \right| }} = 1 + \frac{{\left( {1 – 0.01} \right)}}{{1.05}} + \frac{{\left( {1 – 0.01} \right)\left( {1 – 0.02} \right)}}{{{{1.05}^2}}}\)
\({{\ddot a}_{75:\left. {\overline {\, 3 \,}}\! \right| }} = 2.822857\) |
Jawaban |
c. 2,82 |