Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | Mei 2018 |
| Nomor Soal | : | 14 |
SOAL
Dari fungsi-fungsi di bawah ini, manakah yang tidak dapat digunakan sebagai survival model untuk x > 0 ?
- \({S_X}\left( x \right) = {\left( {1 + x} \right)^{ – 3}}\)
- \({S_X}\left( x \right) = \exp \left[ {7,215 \cdot \left( {1 – {2^x}} \right)} \right]\)
- \({S_X}\left( x \right) = {e^{ – {x^2}}}\)
- \({S_X}\left( x \right) = \exp \left[ {x – 0,72\left( {{2^x} – 1} \right)} \right]\)
- \({S_X}\left( x \right) = \frac{1}{{1 + \sqrt x }}\)
| Diketahui | Yang dapat digunakan sebagai survival model untuk x > 0 adalah jika \({S_X}\left( 0 \right) = 1\) dan \({S_X}\left( \infty \right) = 0\) |
| Pembahasan | Dengan trial dan eror, - \({S_X}\left( 0 \right) = {\left( {1 + 0} \right)^{ – 3}} = 1\)
\({S_X}\left( \infty \right) = {\left( {1 + \infty } \right)^{ – 3}} = 0\)
- \({S_X}\left( 0 \right) = \exp \left[ {7,215 \cdot \left( {1 – {2^0}} \right)} \right] = {e^0} = 1\)
\({S_X}\left( \infty \right) = \exp \left[ {7,215 \cdot \left( {1 – {2^x}} \right)} \right] = {e^{\frac{{7,215}}{\infty }}} = 0\)
- \({S_X}\left( 0 \right) = {e^{ – {0^2}}} = 1\)
\({S_X}\left( \infty \right) = {e^{ – {\infty ^2}}} = {e^{ – \infty }} = \frac{1}{{{e^\infty }}} = 0\)
- \({S_X}\left( 0 \right) = \exp \left[ {0 – 0,72\left( {{2^0} – 1} \right)} \right] = 1\)
\({S_X}\left( \infty \right) = \exp \left[ {\infty – 0,72\left( {{2^\infty } – 1} \right)} \right] = {e^\infty } = \infty \) → tidak bisa
- \({S_X}\left( x \right) = \frac{1}{{1 + \sqrt 0 }} = 1\)
\({S_X}\left( x \right) = \frac{1}{{1 + \sqrt \infty }} = \frac{1}{\infty } = 0\)
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| Jawaban | d.\({S_X}\left( x \right) = \exp \left[ {x – 0,72\left( {{2^x} – 1} \right)} \right]\) |
