Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Di bawah ini adalah tabel untuk double decrement \(l_{40}^{\left( \tau \right)} = 2.200\) dengan
| \(x\) | \(q_x^{\left( d \right)}\) | \(q_x^{\left( w \right)}\) | \(q_x^{‘\left( d \right)}\) | \(q_x^{‘\left( w \right)}\) |
| 40 | 0,24 | 0,12 | 0,26 | \(y\) |
| 41 | — | — | 0,20 | \(2y\) |
Hitunglah \(l_{42}^{\left( \tau \right)}\) (dibulatkan ke satuan terdekat)
- 803
- 822
- 840
- 860
- 880
| Diketahui | \(l_{40}^{\left( \tau \right)} = 2.200\) dengan
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| Rumus yang digunakan | \(q_x^{\left( d \right)} = q_x^{‘\left( d \right)}\left[ {1 – \frac{1}{2} \cdot q_x^{‘\left( w \right)}} \right]\) dan \(q_x^{\left( w \right)} = q_x^{‘\left( w \right)}\left[ {1 – \frac{1}{2} \cdot q_x^{‘\left( d \right)}} \right]\) \({}_tp_x^{\left( \tau \right)} = {}_tp_x^{‘\left( d \right)} \cdot {}_tp_x^{‘\left( w \right)}\) dengan \({}_tp_x^{‘\left( j \right)} = 1 – {}_tq_x^{‘\left( j \right)}\) \(l_{x + 1}^{\left( \tau \right)} = l_x^{\left( \tau \right)} \cdot p_x^{\left( \tau \right)}\) | |||||||||||||||
| Proses pengerjaan | \(q_{40}^{\left( w \right)} = q_{40}^{‘\left( w \right)}\left[ {1 – \frac{1}{2} \cdot q_{40}^{‘\left( d \right)}} \right]\)
\(0,12 = y\left[ {1 – \frac{{0,26}}{2} \cdot } \right]\)
\(y = \frac{{0,12}}{{0,87}} = 0,137931\) sehingga \(q_{40}^{‘\left( w \right)} = 0,137931\) dan \(q_{41}^{‘\left( w \right)} = 2y = 2\left( {\frac{4}{{29}}} \right) = 0,275862\) | |||||||||||||||
| \(l_{42}^{\left( \tau \right)} = l_{41}^{\left( \tau \right)} \cdot p_{41}^{\left( \tau \right)} = l_{40}^{\left( \tau \right)} \cdot p_{40}^{\left( \tau \right)} \cdot p_{41}^{\left( \tau \right)}\) \(= l_{40}^{\left( \tau \right)} \cdot p_{40}^{‘\left( d \right)} \cdot p_{40}^{‘\left( w \right)} \cdot p_{41}^{\left( d \right)} \cdot p_{41}^{\left( w \right)}\) \(= 2.200 \cdot \left( {1 – 0,26} \right) \cdot \left( {1 – \frac{4}{{29}}} \right) \cdot \left( {1 – 0,20} \right) \cdot \left( {1 – \frac{8}{{29}}} \right)\) \(= 813,0321046\) | ||||||||||||||||
| Jawaban | b. 822 |