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Pembahasan Ujian PAI: A50 – No. 17 – November 2014

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A50 - Metoda StatistikaAktuariaEdukasiPembahasan Ujian Profesi Aktuaris (Persatuan Aktuaris Indonesia / PAI)

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 \(Var\left( X \right)\)

  1. 518,2457
  2. 517,2854
  3. 515,2478
  4. 514,2857
  5. Tidak ada jawaban yang benar
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Diketahui
  • \({l_x} = 2.500{\left( {64 – 0,8x} \right)^{\frac{1}{3}}},0 \le x \le 80\)
  • \({}_x{p_0} = \frac{{{{\left( {64 – 0.8x} \right)}^{\frac{1}{3}}}}}{4}\)  ,  \({\mu _x} = \frac{4}{{15\left( {64 – 0.8x} \right)}}\)  , dan \(f\left( x \right) = \frac{1}{{15}}{\left( {64 – 0.8x} \right)^{ – \frac{2}{3}}}\)
  • \(E\left[ X \right] = 60\)
Rumus yang digunakan
  • \(Var\left( X \right) = E\left[ {{X^2}} \right] – {\left( {E\left[ X \right]} \right)^2}\)
  • \(E\left[ {{X^2}} \right] = \int\limits_0^\infty {{x^2} \cdot f\left( x \right)dx} = \int\limits_0^\infty {{x^2} \cdot {}_x{p_0} \cdot {\mu _x}dx} = 2\int\limits_0^\infty {x \cdot {}_x{p_0}dx} \)
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
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