Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2014 |
| Nomor Soal |
: |
25 |
SOAL
Diketahui bahwa \(q_x^{\left( 1 \right)} = 0,20\) dan \(q_x^{\left( 2 \right)} = 0,10\). Kemudian penurunan (decrement) tersebut berdistribusi seragam di antara interval \(\left( {x,x + 1} \right)\) dalam konteks multiple decrement.
Diketahui pula persamaan berikut:
\({}_tp_x^{‘\left( j \right)} = {\left( {1 – t \cdot q_x^{\left( \tau \right)}} \right)^{\frac{{q_x^{\left( j \right)}}}{{q_x^{\left( \tau \right)}}}}}\) dan \(t = 1\)
Tentukan nilai \(q_x^{‘\left( 2 \right)}\)
- 0,8879
- 0,1121
- 1,8879
- 1,1121
- Tidak ada jawaban yang benar
| Diketahui |
- \(q_x^{\left( 1 \right)} = 0,20\) dan \(q_x^{\left( 2 \right)} = 0,10\).
- Kemudian penurunan (decrement) tersebut berdistribusi seragam di antara interval \(\left( {x,x + 1} \right)\) dalam konteks multiple decrement.
- Diketahui pula persamaan berikut: \({}_tp_x^{‘\left( j \right)} = {\left( {1 – t \cdot q_x^{\left( \tau \right)}} \right)^{\frac{{q_x^{\left( j \right)}}}{{q_x^{\left( \tau \right)}}}}}\) dan \(t = 1\)
|
| Rumus yang digunakan |
\({}_tp_x^{‘\left( j \right)} = {\left( {1 – t \cdot q_x^{\left( \tau \right)}} \right)^{\frac{{q_x^{\left( j \right)}}}{{q_x^{\left( \tau \right)}}}}}\)
\(q_x^{\left( \tau \right)} = \sum\limits_{i = 1}^n {q_x^{\left( i \right)}} \) |
| Proses pengerjaan |
\(q_x^{\left( \tau \right)} = \sum\limits_{i = 1}^n {q_x^{\left( i \right)}} = q_x^{\left( 1 \right)} + q_x^{\left( 2 \right)} = 0.2 + 0.1 = 0.3\) |
|
\(q_x^{‘\left( 2 \right)} = 1 – p_x^{‘\left( 2 \right)}\)
\(q_x^{‘\left( 2 \right)} = 1 – {\left( {1 – q_x^{\left( \tau \right)}} \right)^{\frac{{q_x^{\left( 2 \right)}}}{{q_x^{\left( \tau \right)}}}}}\)
\(q_x^{‘\left( 2 \right)} = 1 – {\left( {1 – 0.3} \right)^{\frac{{0.1}}{{0.3}}}}\)
\(q_x^{‘\left( 2 \right)} = 0.112096\) |
| Jawban |
b. 0,1121 |
