Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2015 |
| Nomor Soal |
: |
14 |
SOAL
Untuk suatu “triple decrement” tabel, diberikan:
- \({\mu _{x + t}^{\left( 1 \right)} = 0,3;}\) \({t > 0}\)
- \({\mu _{x + t}^{\left( 2 \right)} = 0,5;}\) \({t > 0}\)
- \({\mu _{x + t}^{\left( 3 \right)} = 0,7;}\) \({t > 0}\)
Hitunglah \(q_x^{\left( 2 \right)}\) (pembulatan terdekat)
- 0,26
- 0,30
- 0,33
- 0,36
- 0,39
| Diketahui |
Suatu “triple decrement” tabel, diberikan:
- \({\mu _{x + t}^{\left( 1 \right)} = 0,3;}\) \({t > 0}\)
- \({\mu _{x + t}^{\left( 2 \right)} = 0,5;}\) \({t > 0}\)
- \({\mu _{x + t}^{\left( 3 \right)} = 0,7;}\) \({t > 0}\)
|
| Rumus yang digunakan |
\({}_tq_x^{\left( j \right)} = \int\limits_0^t {{}_sp_x^{\left( \tau \right)}\mu _{x + s}^{\left( j \right)}ds} \)
\({}_tp_x^{\left( \tau \right)} = \exp \left[ { – \int\limits_0^t {\mu _{x + s}^{\left( \tau \right)}ds} } \right]\)
\(\mu _{x + t}^{\left( \tau \right)} = \sum\limits_{j = 1}^n {\mu _{x + t}^{\left( j \right)}} \) |
| Proses pengerjaan |
\(\mu _{x + t}^{\left( \tau \right)} = \sum\limits_{j = 1}^3 {\mu _{x + t}^{\left( j \right)}} = 0.3 + 0.5 + 0.7 = 1.5\) |
| \({}_tp_x^{\left( \tau \right)} = \exp \left[ { – \int\limits_0^t {\mu _{x + s}^{\left( \tau \right)}ds} } \right] = \exp \left[ { – \int\limits_0^t {1.5ds} } \right]\)
\({}_tp_x^{\left( \tau \right)} = \exp \left( { – 1.5t} \right)\) |
| \(q_x^{\left( 2 \right)} = \int\limits_0^1 {{}_sp_x^{\left( \tau \right)}\mu _{x + s}^{\left( 2 \right)}ds} = \int\limits_0^1 {\exp \left( { – 1.5s} \right)\left( {0.5} \right)ds} \)
\(q_x^{\left( 2 \right)} = – \frac{{0.5}}{{1.5}}\left[ {\exp \left( { – 1.5} \right) – 1} \right]\)
\(q_x^{\left( 2 \right)} = 0.258956\) |
| Jawaban |
a. 0,26 |
