Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
A20 – Probabilitas dan Statistika |
| Periode Ujian |
: |
November 2017 |
| Nomor Soal |
: |
19 |
SOAL
Misalkan fungsi distribusi X untuk x>0 adalah \(F(x) = 1 – \sum {_{k = 0}^3} \frac{{{x^k}{e^{ – x}}}}{{k!}}\) . Apa fungsi kepadatan X untuk x > 0?
- \({e^{ – x}}\)
- \(\frac{{{x^2}{e^{ – x}}}}{2}\)
- \(\frac{{{x^3}{e^{ – x}}}}{6}\)
- \(\frac{{{x^3}{e^{ – x}}}}{6} – {e^{ – x}}\)
- \(\frac{{{x^3}{e^{ – x}}}}{6} + {e^{ – x}}\)
PEMBAHASAN
| Kalkulasi |
\(F(x) = 1 – \sum {_{k = 0}^3} \frac{{{x^k}{e^{ – x}}}}{{k!}}\)
\(F(x) = 1 – \left( {\frac{{{x^0}{e^{ – x}}}}{{0!}} + \frac{{{x^1}{e^{ – x}}}}{{1!}} + \frac{{{x^2}{e^{ – x}}}}{{2!}} + \frac{{{x^3}{e^{ – x}}}}{{3!}}} \right)\)
\(f(x) = \frac{d}{{dx}}\left[ {1 – \left( {{e^{ – x}} + x{e^{ – x}} + \frac{{{x^2}{e^{ – x}}}}{2} + \frac{{{x^3}{e^{ – x}}}}{6}} \right)} \right]\)
\(f(x) = – \left( { – {e^{ – x}} + {e^{ – x}} – x{e^{ – x}} + x{e^{ – x}} – \frac{{{x^2}{e^{ – x}}}}{2} + \frac{{{x^2}{e^{ – x}}}}{2} – \frac{{{x^3}{e^{ – x}}}}{6}} \right)\)
\(f(x) = – \left( { – \frac{{{x^3}{e^{ – x}}}}{6}} \right)\)
\(f(x) = \frac{{{x^3}{e^{ – x}}}}{6}\) |
| Jawaban |
C. \(\frac{{{x^3}{e^{ – x}}}}{6}\) |
