Pembahasan Ujian PAI: A60 - No. 3 - Mei 2018

Pembahasan Soal Ujian Profesi Aktuaris

Institusi	:	Persatuan Aktuaris Indonesia (PAI)
Mata Ujian	:	Matematika Aktuaria
Periode Ujian	:	Mei 2018
Nomor Soal	:	3

SOAL

Diberikan \({l_x} = 2500{\left( {64 – 0,8x} \right)^{\frac{1}{3}}}\), dengan \(0 \le x \le 80\). Tentukan Var(X) !

16,2857
0,2857
4.114,2857
514,2857
3,2857

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Diketahui	\({l_x} = 2500{\left( {64 – 0,8x} \right)^{\frac{1}{3}}}\) dengan \(0 \le x \le 80\)
Step 1	\(E\left[ X \right] = \int\limits_0^{80} {{S_X}\left( x \right)dx} = \int\limits_0^{80} {\frac{{lx}}{{{l_0}}}dx} \) \(= \frac{1}{{{l_0}}}\int\limits_0^{80} {lx\,dx} \) → \({l_0} = 2500{\left( {64} \right)^{\frac{1}{3}}} = 10.000\) \(= \frac{1}{{10.000}}\int\limits_0^{80} {2500{{\left( {64 – 0,8x} \right)}^{\frac{1}{3}}}dx} \) \(= \frac{1}{{10.000}} \cdot 2500\left[ {\left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right){{\left( {64 – 0,8x} \right)}^{\frac{4}{3}}}\mathop {\left. {} \right\|}\nolimits_0^{80} \,} \right]\) \(= 0,25\left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right)\left[ {{{\left( {64 – 0,8\left( {80} \right)} \right)}^{\frac{4}{3}}} – {{\left( {64 – 0,8\left( 0 \right)} \right)}^{\frac{4}{3}}}} \right]\) \(= 0,25\left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right)\left[ { – 256} \right]\) \(= 60\)
Step 2	\(E\left[ {{X^2}} \right] = 2\int\limits_0^{80} {x \cdot \frac{{lx}}{{{l_0}}}\,dx = } \,\,2\int\limits_0^{80} {x \cdot \frac{{2500{{\left( {64 – 0,8x} \right)}^{\frac{1}{3}}}}}{{10.000}}\,dx} \) \(= \frac{2}{{10.000}}\left( {2500} \right)\int\limits_0^{80} {\underbrace x_u \cdot \underbrace {{{\left( {64 – 0,8x} \right)}^{\frac{1}{3}}}}_{dv}\,dx} \) \(u = x\) \(du = 1\) \(dv = {\left( {64 – 0,8x} \right)^{\frac{1}{3}}}\) \(v = \frac{3}{4}\left( {\frac{1}{{0,8}}} \right){\left( {64 – 0,8x} \right)^{\frac{4}{3}}}\) \(\int {u\,dv = uv – \int {v\,du} } \) \(= \frac{1}{2}\left[ {x\left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right){{\left( {64 – 0,8x} \right)}^{\frac{4}{3}}}\mathop {\left. {} \right\|}\nolimits_{80}^0 – \int\limits_0^{80} {\left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right){{\left( {64 – 0,8x} \right)}^{\frac{4}{3}}}\,dx\, \cdot 1} } \right]\) \(= \frac{1}{2}\left[ {0 – \left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right)\left( {\frac{3}{7} \cdot \frac{1}{{ – 0,8}} \cdot {{\left( {64 – 0,8x} \right)}^{\frac{7}{3}}}\,\,\mathop {\left. {} \right\|}\nolimits_{80}^0 } \right)} \right]\) \(= \frac{1}{2}\left[ { – \left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right)\left( {\frac{3}{7}} \right)\left( {\frac{1}{{ – 0,8}}} \right)\left( {{{\left( {64 – 0,8 \cdot 80} \right)}^{\frac{7}{3}}} – {{\left( {64 – 0} \right)}^{\frac{7}{3}}}} \right)} \right]\) \(= \frac{1}{2}\left[ { – \left( {\frac{3}{4}} \right)\left( {\frac{1}{{ – 0,8}}} \right)\left( {\frac{3}{7}} \right)\left( {\frac{1}{{ – 0,8}}} \right)\left( { – 16.384} \right)} \right]\) \(= 4114,285714\)
Step 3	\(Var\left[ X \right] = E\left[ {{X^2}} \right] – {\left( {E\left[ X \right]} \right)^2}\) \(= 4114,285714 – {\left( {60} \right)^2}\) \(= 514,285714\)
Jawaban	d. 514,2857