Pembahasan Soal Ujian Profesi Aktuaris
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 |