Pembahasan Soal Ujian Profesi Aktuaris
Institusi : Persatuan Aktuaris Indonesia (PAI) Mata Ujian : Metoda Statistika Periode Ujian : November 2014 Nomor Soal : 4
SOAL
Sebuah Survival Distribution didefinisikan sebagai \(S\left( t \right) = 0,10{\left( {100 – t} \right)^{\frac{1}{2}}}\) , di dalam daerah domain \(0 \le t \le 100\) . Tentukan \(Var\left( T \right)\)
333,33 666,67 777,78 889,77 998,89 Kunci Jawaban & Pembahasan
Diketahui Sebuah Survival Distribution didefinisikan sebagai \(S\left( t \right) = 0,10{\left( {100 – t} \right)^{\frac{1}{2}}}\) , di dalam daerah domain \(0 \le t \le 100\) . Rumus yang digunakan \(Var\left( T \right) = E\left[ {{T^2}} \right] – E{\left[ T \right]^2}\) \(E\left[ T \right] = \int\limits_0^\infty {S\left( t \right)dt} \) \(E\left[ {{T^2}} \right] = \int\limits_0^\infty {{t^2} \cdot f\left( t \right)dt} \) \(f\left( t \right) = – \frac{d}{{dx}}S\left( t \right)\) Proses pengerjaan \(f\left( t \right) = – \frac{d}{{dt}}S\left( t \right) = – \frac{d}{{dt}}0.10{\left( {100 – t} \right)^{\frac{1}{2}}} = \frac{1}{{20\sqrt {100 – t} }}\) \(E\left[ T \right] = \int\limits_0^\infty {S\left( t \right)dt} = \int\limits_0^{100} {0.10{{\left( {100 – t} \right)}^{\frac{1}{2}}}dt} \)
\({E\left[ T \right] = 0.10\int\limits_{\sqrt {100} }^0 { – 2{u^2}dt} }\)
\({{\rm{misal }}{u^2} = 100 – t \Rightarrow 2udu = – dt}\)
\(E\left[ T \right] = – 0.20\left[ {0 – \frac{{{{\left( {\sqrt {100} } \right)}^3}}}{3}} \right] = \frac{{200}}{3}\) \(E\left[ {{T^2}} \right] = \int\limits_0^\infty {{t^2} \cdot f\left( t \right)dt} = \int\limits_0^{100} {\frac{{{t^2}}}{{20\sqrt {100 – t} }}dt} \)
\({E\left[ {{T^2}} \right] = \frac{1}{{20}}\int\limits_{\sqrt {100} }^0 {\frac{{ – 2u{{\left( {100 – {u^2}} \right)}^2}}}{u}du} }\)
\({{\rm{misal\_}}{u^2} = 100 – t \Rightarrow 2udu = – dt}\)
\(E\left[ {{T^2}} \right] = \frac{1}{{20}}\int\limits_{\sqrt {100} }^0 {\left[ { – 2{u^4} + 400{u^2} – 20,000} \right]du} \)
\(E\left[ {{T^2}} \right] = \frac{1}{{20}}\left[ { – \frac{{2{u^5}}}{5} + \frac{{400{u^3}}}{3} – 20,000u} \right]_{\sqrt {100} }^0\)
\(E\left[ {{T^2}} \right] = \frac{1}{{20}}\left[ {0 + 40,000 – \frac{{400,000}}{3} + 200,000} \right] = \frac{{16,000}}{3}\) \(Var\left( T \right) = E\left[ {{T^2}} \right] – E{\left[ T \right]^2} = \frac{{16,000}}{3} – {\left( {\frac{{200}}{3}} \right)^2} = 888.888889\) Jawaban D. 889,77
