Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Suatu “age-at-failure” variabel acak mempunyai distribusi sebagai berikut:
\({F_X}(x) = 1 – 0,1{(100 – x)^{\frac{1}{2}}}\) , \(0 \le x \le 100\)
Tentukan nilai dari \(E[X]\) dan median dari distribusi tersebut
- 100/3 ; 75
- 100/3 ; 100
- 200/3 ; 75
- 200/3 ; 100
- 200/3 ; 50
| Step 1 | \({S_X}(t) = 1 – {F_X}(t)\) \({S_X}(t) = 1 – \left( {1 – 0,1{{(100 – t)}^{\frac{1}{2}}}} \right)\) \({S_X}(t) = 0,1{(100 – t)^{\frac{1}{2}}}\) |
| Step 2 | \(E[X] = \int\limits_0^{100} {{S_X}(t)dt} \) \(E[X] = \int\limits_0^{100} {0,1{{(100 – t)}^{\frac{1}{2}}}dt} \) \(E[X] = 0,1\left( {\frac{1}{{0,5 + 1}}} \right)( – 1)\left( {{{(100 – 100)}^{\frac{3}{2}}} – {{(100 – 0)}^{\frac{3}{2}}}} \right)\) \(E[X] = 0,1\left( {\frac{2}{3}} \right)1.000\) \(E[X] = \frac{{200}}{3}\) |
| Step 3 | Median, \({F_X}({x_{med}}) = 0,5\) \(1 – 0,1{(100 – x)^{\frac{1}{2}}} = 0,5\) \({(100 – x)^{\frac{1}{2}}} = \frac{{0,5}}{{0,1}}\) \(x = 75\) |
| Jawaban | c. 200/3 ; 75 |