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