Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | November 2014 |
| Nomor Soal | : | 17 |
SOAL
Berdasarkan soal nomor 15. Tentukan \(Var\left( X \right)\)
- 518,2457
- 517,2854
- 515,2478
- 514,2857
- Tidak ada jawaban yang benar
| Diketahui |
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| Rumus yang digunakan |
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| Proses pengerjaan | \(E\left[ {{X^2}} \right] = 2\int\limits_0^\infty {x \cdot {}_x{p_0}dx} = 2\int\limits_0^{80} {\frac{{x{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}}}{4}dx} \)
\({E\left[ {{X^2}} \right] = \frac{1}{2}\int\limits_4^0 {\left[ {\left( {\frac{{64 – {u^3}}}{{0.8}}} \right)u} \right]\left( { – \frac{{3{u^2}}}{{0.8}}} \right)du} }\)
\({{\rm{misal\_}}{u^3} = 64 – 0.8x \Rightarrow 3{u^2}du = – 0.8dx}\)
\(E\left[ {{X^2}} \right] = \frac{1}{2}\int\limits_4^0 {\left[ {\left( {\frac{{64 – {u^3}}}{{0.8}}} \right)u} \right]\left( { – \frac{{3{u^2}}}{{0.8}}} \right)du} \)
\(E\left[ {{X^2}} \right] = \frac{1}{2}\int\limits_4^0 {\left( {4.6875{u^6} – 300{u^3}} \right)du} \)
\(E\left[ {{X^2}} \right] = \frac{1}{2}\left[ {0 – \frac{{4.6875{{\left( 4 \right)}^7}}}{7} + \frac{{300{{\left( 4 \right)}^4}}}{4}} \right]\)
\(E\left[ {{X^2}} \right] = 4,114.285714\)
\(Var\left( X \right) = E\left[ {{X^2}} \right] – {\left( {E\left[ X \right]} \right)^2} = 4,114.285714 – {60^2} = 514.285714\) |
| Jawaban | D. 514,2857 |
