Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diketahui \({l_x} = 2.500{(64 – 0,8x)^{\frac{1}{3}}},\,0 \le x \le 80\). Tentukanlah angka yang paling dekat untuk \(Var[X] – {(E[X])^2}\).
- 60
- 514
- 3.600
- 4.114
- 7.714
| Step 1 | \({l_0}\, = 2.500{(64 – 0,8(0))^{\frac{1}{3}}}\) \({l_0}\, = 10.000\) |
| Step 2 | \(E[X]\, = \int\limits_0^{80} {\frac{{{l_{0 + t}}}}{{{l_0}}}} dt\) \(E[X] = \,\frac{{\int\limits_0^{80} {2.500{{(64 – 0,8x)}^{\frac{1}{3}}}} }}{{10.000}}\,\) \(E[X] = \,\frac{{2.500\frac{3}{4}\left( {\frac{{10}}{8}} \right)\left[ {{{(64 – 0,8(0))}^{\frac{4}{3}}} – {{(64 – 0,8(80))}^{\frac{4}{3}}}} \right]}}{{10.000}}\) \(E[X] = \frac{{600.000}}{{10.000}}\) \(E[X] = 60\) |
| Step 3 | \(E[{X^2}]\,\, = 2\int\limits_0^{80} {x\frac{{{l_{0 + t}}}}{{{l_0}}}} dt\,\) \(E[{X^2}] = \,\frac{{2\int\limits_0^{80} {x\,\,\left[ {2.500{{(64 – 0,8x)}^{\frac{1}{3}}}} \right]} dt}}{{10.000}}\,\,\) \(E[{X^2}] = \,4.114,285711\) |
| Step 4 | \(Var[X] = E[{X^2}]\,\, – \,E{[X]^2}\) \(Var[X] = 4.114 – {60^2}\) \(Var[X] = 514\) |
| Maka | \(Var[X] – E{[X]^2} = 514 – {60^2}\) \(Var[X] – E{[X]^2} = – 3.086\) |
| Jawaban | Anulir |