Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Akturaria |
| Periode Ujian | : | November 2015 |
| Nomor Soal | : | 29 |
SOAL
Berdasarkan soal nomor 27. Hitunglah nilai dari \(e_0^0\) dari fungsi survival tersebut (pembulatan terdekat)
- 81
- 95
- 105
- 121
- 140
| Diketahui | \(\begin{array}{*{20}{c}} {{S_0}\left( t \right) = \frac{{{{\left( {121 – t} \right)}^{\frac{1}{2}}}}}{{11}};}&{t \in \left[ {0,121} \right]} \end{array}\) |
| Rumus yang digunakan | \({}_t{p_x} = \frac{{{S_0}\left( {x + t} \right)}}{{{S_0}\left( x \right)}}\)
\(e_x^0 = \int\limits_0^\infty {{}_t{p_x}dt} \) |
| Proses pengerjaan | \({}_t{p_0} = \frac{{{S_0}\left( {0 + t} \right)}}{{{S_0}\left( 0 \right)}} = \frac{{{{\left( {121 – 0 – t} \right)}^{\frac{1}{2}}}}}{{11}} \cdot \frac{{11}}{{{{\left( {121 – 0} \right)}^{\frac{1}{2}}}}} = \frac{{{{\left( {121 – t} \right)}^{\frac{1}{2}}}}}{{11}}\) |
| \(e_0^0 = \int\limits_0^{121} {\left[ {\frac{{{{\left( {121 – t} \right)}^{\frac{1}{2}}}}}{{11}}} \right]dt} \)
\(e_0^0 = – \frac{2}{{3\left( {11} \right)}}\left[ {{{\left( {121 – t} \right)}^{\frac{3}{2}}}} \right]_0^{121}\)
\(e_0^0 = – \frac{2}{{3\left( {11} \right)}}\left[ {0 – {{121}^{\frac{3}{2}}}} \right]\)
\(e_0^0 = 80.666667\) |
| Jawaban | a. 81 |
