Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Misalkan \(X\) adalah variabel acak untuk umur pada saat kematian dengan
\({\mu _x} = \frac{1}{{3\left( {100 – x} \right)}}{\rm{,}}\) untuk \(0 \le x \le 100\)
Hitunglah \(e_{19}^ \circ \), yaitu rata-rata pengharapan hidup untuk seseorang yang berusia 19.
- 42,67
- 58,89
- 60,75
- 67,67
- 71,75
| Diketahui | \({\mu _x} = \frac{1}{{3\left( {100 – x} \right)}}{\rm{,}}\) untuk \(0 \le x \le 100\) |
| Rumus yang digunakan | \(_t{p_x} = \exp \left[ { – \int\limits_x^{x + t} {\mu \left( y \right)dy} } \right]\) \(e_x^ \circ = \int\limits_0^\infty {_t{p_x}dt} \) |
| Proses pengerjaan | \(_t{p_{19}} = \exp \left[ { – \int\limits_{19}^{19 + t} {\frac{1}{{3\left( {100 – y} \right)}}dy} } \right]\) \(= \exp \left[ { – \frac{1}{3}\int\limits_{19}^{19 + t} {\frac{1}{{\left( {100 – y} \right)}}dy} } \right],{\rm{ }}\) misal \(u = 100 – y\) maka \(du = – dy\) \(= \exp \left[ { – \frac{1}{3}\int\limits_{81}^{81 – t} { – \frac{1}{u}du} } \right]\) \(= \exp \left[ {\frac{{\ln \left( {81 – t} \right) – \ln \left( {81} \right)}}{3}} \right]\) \(= \frac{{{{\left( {81 – t} \right)}^{\frac{1}{3}}}}}{{{{81}^{\frac{1}{3}}}}}\) |
| \(e_{19}^ \circ = \int\limits_0^{81} {\frac{{{{\left( {81 – t} \right)}^{\frac{1}{3}}}}}{{{{81}^{\frac{1}{3}}}}}dt} \) \(= \frac{1}{{{{81}^{\frac{1}{3}}}}}\int\limits_0^{81} {{{\left( {81 – t} \right)}^{\frac{1}{3}}}dt} ,{\rm{ }}\) misalkan \({u^3} = 81 – t\) maka \(3{u^2}du = – dt\) \(= \frac{1}{{{{81}^{\frac{1}{3}}}}}\int\limits_{{{81}^{\frac{1}{3}}}}^0 { – u \cdot } 3{u^2}du\) \(= \frac{1}{{{{81}^{\frac{1}{3}}}}}\left( {\frac{{3 \cdot {{\left( {{{81}^{\frac{1}{3}}}} \right)}^4}}}{4}} \right)\) \(= \frac{{3 \cdot {{81}^{\frac{3}{3}}}}}{4}\) \(= 60,75\) | |
| Jawaban | c. 60,75 |