Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Metoda Statistika |
| Periode Ujian |
: |
November 2017 |
| Nomor Soal |
: |
11 |
SOAL
Pada sebuah model double decrement, diperoleh informasi sebagai berikut:
\(l_x^{\left( \tau \right)} = 100\)
\(l_{x + 3}^{\left( \tau \right)} = 50\)
\(_3q_x^{\left( 1 \right)} = 0,07\)
\(_{\left. 2 \right|}q_x^{\left( 2 \right)} = 0,08\)
Hitunglah \(_2q_x^{\left( 2 \right)}\)
- 0,15
- 0,20
- 0,25
- 0,30
- 0,35
| Diketahui |
\(l_x^{\left( \tau \right)} = 100\)
\(l_{x + 3}^{\left( \tau \right)} = 50\)
\(_3q_x^{\left( 1 \right)} = 0,07\)
\(_{\left. 2 \right|}q_x^{\left( 2 \right)} = 0,08\) |
| Rumus yang digunakan |
\(l_{x + 1}^{\left( \tau \right)} = l_x^{\left( \tau \right)} – d_x^{\left( 1 \right)} – d_x^{\left( 2 \right)}\)
\({}_tq_x^{\left( j \right)} = \frac{{d_x^{\left( j \right)} + d_{x + 1}^{\left( j \right)} + \cdots + d_{x + t – 1}^{\left( j \right)}}}{{l_x^{\left( \tau \right)}}}\)
\(_{\left. t \right|u}q_x^{\left( j \right)} = \frac{{{}_ud_{x + t}^{\left( j \right)}}}{{l_x^{\left( \tau \right)}}}\) |
| Proses pengerjaan |
\({}_3q_x^{\left( 1 \right)} = \frac{{d_x^{\left( 1 \right)} + d_{x + 1}^{\left( 1 \right)} + d_{x + 2}^{\left( 1 \right)}}}{{l_x^{\left( \tau \right)}}}\)
\(0,07 = \frac{{d_x^{\left( 1 \right)} + d_{x + 1}^{\left( 1 \right)} + d_{x + 2}^{\left( 1 \right)}}}{{100}}\)
\(d_x^{\left( 1 \right)} + d_{x + 1}^{\left( 1 \right)} + d_{x + 2}^{\left( 1 \right)} = 7\) |
| \(_{\left. 2 \right|}q_x^{\left( 2 \right)} = \frac{{d_{x + 2}^{\left( 2 \right)}}}{{l_x^{\left( \tau \right)}}}\)
\(0,08 = \frac{{d_{x + 2}^{\left( 2 \right)}}}{{100}}\)
\(d_{x + 2}^{\left( 2 \right)} = 8\) |
| \(l_{x + 3}^{\left( \tau \right)} = l_x^{\left( \tau \right)} – d_x^{\left( 1 \right)} – d_x^{\left( 2 \right)} – d_{x + 1}^{\left( 1 \right)} – d_{x + 1}^{\left( 2 \right)} – d_{x + 2}^{\left( 1 \right)} – d_{x + 2}^{\left( 2 \right)}\)
\(l_{x + 3}^{\left( \tau \right)} = l_x^{\left( \tau \right)} – \left( {d_x^{\left( 1 \right)} + d_{x + 1}^{\left( 1 \right)} + d_{x + 2}^{\left( 1 \right)}} \right) – \left( {d_x^{\left( 2 \right)} + d_{x + 1}^{\left( 2 \right)}} \right) – d_{x + 2}^{\left( 2 \right)}\)
\(d_x^{\left( 2 \right)} + d_{x + 1}^{\left( 2 \right)} = 100 – 7 – 8 – 50\)
\(= 35\) |
| \(_2q_x^{\left( 2 \right)} = \frac{{d_x^{\left( 2 \right)} + d_{x + 1}^{\left( 2 \right)}}}{{l_x^{\left( \tau \right)}}}\)
\(= \frac{{35}}{{100}}\)
\(= 0,35\) |
| Jawaban |
e. 0,35 |
