Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Metoda Statistika |
| Periode Ujian |
: |
November 2014 |
| Nomor Soal |
: |
15 |
SOAL
Jika diketahui \({l_x} = 2.500{\left( {64 – 0,8x} \right)^{\frac{1}{3}}},0 \le x \le 80\) tentukan \(f\left( x \right)\)
- \(\frac{1}{{15}}{\left( {64 – 0,8x} \right)^{\frac{2}{3}}}\)
- \(\frac{1}{{15}}{\left( {64 – 0,8x} \right)^{ – \frac{1}{3}}}\)
- \(\frac{1}{{15}}{\left( {64 – 0,8x} \right)^{ – \frac{2}{3}}}\)
- \(\frac{1}{{15}}{\left( {64 – 0,8x} \right)^{\frac{1}{3}}}\)
- Tidak ada jawaban yang benar
| Diketahui |
\({l_x} = 2.500{\left( {64 – 0,8x} \right)^{\frac{1}{3}}},0 \le x \le 80\) |
| Rumus yang digunakan |
- \(f\left( x \right) = {}_x{p_0} \cdot {\mu _x}\)
- \({}_x{p_0} = S\left( x \right) = \frac{{{l_x}}}{{{l_0}}}\)
- \({\mu _x} = \frac{1}{{{l_x}}}\left( { – \frac{d}{{dx}}{l_x}} \right)\)
|
| Proses pengerjaan |
\({}_x{p_0} = S\left( x \right) = \frac{{{l_x}}}{{{l_0}}} = \frac{{2,500{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}}}{{2,500{{\left( {64 – 0.8\left( 0 \right)} \right)}^{\frac{1}{3}}}}} = \frac{{{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}}}{4}\)
\({\mu _x} = \frac{1}{{{l_x}}}\left( { – \frac{d}{{dx}}{l_x}} \right) = \frac{1}{{2,500{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}}}\left( { – \frac{d}{{dx}}\left( {2,500{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}} \right)} \right)\)
\({\mu _x} = \frac{1}{{2,500{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}}}\left( {\frac{{2,500\left( {0.8} \right)}}{{3{{\left( {64 – 0.8x} \right)}^{\frac{2}{3}}}}}} \right)\)
\({\mu _x} = \frac{{0.8}}{{3\left( {64 – 0.8x} \right)}} = \frac{4}{{15\left( {64 – 0.8x} \right)}}\)
\(f\left( x \right) = {}_x{p_0} \cdot {\mu _x} = \frac{{{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}}}{4} \cdot \frac{4}{{15\left( {64 – 0.8x} \right)}}\)
\(f\left( x \right) = \frac{1}{{15}}{\left( {64 – 0.8x} \right)^{ – \frac{2}{3}}}\) |
| Jawaban |
C. \(\frac{1}{{15}}{\left( {64 – 0,8x} \right)^{ – \frac{2}{3}}}\) |
