Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Berdasarkan soal nomor 14, hitunglah \(Var\left[ T \right]\) untuk \(x = 30\)
- 457,77
- 465,88
- 469,32
- 476,11
- 489,67
| Diketahui | \({}_t{p_x} = {\left( {\frac{{120 – x – t}}{{120 – x}}} \right)^{\frac{1}{6}}}\) \(\mathop {{e_{30}}}\limits^ \circ = 77,142857\) |
| Rumus yang digunakan | \(Var\left[ T \right] = 2\int\limits_0^\infty {t \cdot {}_t{p_x}dt} – \mathop {e_x^2}\limits^ \circ \) |
| Proses pengerjaan | \(Var\left[ T \right] = 2\int\limits_0^{90} {t \cdot {}_t{p_{30}}dt} – \mathop {e_{30}^2}\limits^ \circ \) \(= 2\int\limits_0^{90} {t \cdot {{\left( {\frac{{90 – t}}{{90}}} \right)}^{\frac{1}{6}}}dt} – {\left( {77,142857} \right)^2}\) \(= 2{\left( {\frac{1}{{90}}} \right)^{\frac{1}{6}}}\int\limits_0^{90} {t \cdot {{\left( {90 – t} \right)}^{\frac{1}{6}}}dt} – 5951,020386{\rm{ }}\) misal \(90 – t = {u^6}\) maka \(– dt = 6{u^5}\) \(= – 2{\left( {\frac{1}{{90}}} \right)^{\frac{1}{6}}}\int\limits_{{{90}^{\frac{1}{6}}}}^0 {\left( {90 – {u^6}} \right) \cdot 6{u^6}} du – 5951,020386\) \(= – 2{\left( {\frac{1}{{90}}} \right)^{\frac{1}{6}}}\int\limits_{{{90}^{\frac{1}{6}}}}^0 {\left( {540{u^6} – 6{u^{12}}} \right)} du – 5951,020386\) \(= – 2{\left( {\frac{1}{{90}}} \right)^{\frac{1}{6}}}\left[ { – \frac{{540 \cdot {{\left( {{{90}^{\frac{1}{6}}}} \right)}^7}}}{7} + \frac{{6 \cdot {{\left( {{{90}^{\frac{1}{6}}}} \right)}^{13}}}}{{13}}} \right] – 5951,020386\) \(= \frac{{1080 \cdot {{90}^{\frac{6}{6}}}}}{7} – \frac{{12 \cdot {{90}^{\frac{{12}}{6}}}}}{{13}} – 5951,020386\) \(= 457,7708228\) |
| Jawaban | a. 457,77 |