Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Probabilitas dan Statistika |
| Periode Ujian |
: |
November 2018 |
| Nomor Soal |
: |
15 |
SOAL
Suatu fungsi distribusi \(X\) untuk \(x > 0\) adalah \(F(x) = 1 – \sum\limits_{k = 0}^3 {\frac{{{x^k}{e^{^{ – x}}}}}{{k!}}} \)
Tentukan fungsi densitas dari \(X\) untuk \(x > 0\)?
- \({e^{ – x}}\)
- \(\frac{{{x^2}{e^{ – x}}}}{2}\)
- \(\frac{{{x^3}{e^{ – x}}}}{6}\)
- \(\frac{{{x^3}{e^{ – x}}}}{6} – {e^{ – x}}\)
- \(\frac{{{x^3}{e^{ – x}}}}{6} + {e^{ – x}}\)
| Rumus |
\(f(x) = \frac{d}{{dx}}F(x)\) |
| Maka |
\(f(x) = \frac{d}{{dx}}\left( {1 – \sum\limits_{k = 0}^3 {\frac{{{x^k}{e^{^{ – x}}}}}{{k!}}} } \right)\)
\(f(x) = \frac{d}{{dx}}\left( {1 – \left( {{e^{ – x}} + x{e^{ – x}} + \frac{{{x^2}{e^{ – x}}}}{2} + \frac{{{x^3}{e^{ – x}}}}{6}} \right)} \right)\)
\(f(x) = \frac{d}{{dx}}\left( {1 – \left( {{e^{ – x}}\left( {1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right)} \right)} \right)\)
\(f(x) = 0 – \left( { – {e^{ – x}} \times \left( {1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}} \right) + {e^{ – x}} \times \left( {1 + x + \frac{{{x^2}}}{2}} \right)} \right)\)
\(f(x) = – \left( {{e^{ – x}} \times \left( { – \frac{{{x^3}}}{6}} \right)} \right)\)
\(f(x) = \frac{{{x^3}{e^{ – x}}}}{6}\) |
| Jawaban |
c. \(\frac{{{x^3}{e^{ – x}}}}{6}\) |
