Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Permodelan dan Teori Risiko |
| Periode Ujian | : | November 2014 |
| Nomor Soal | : | 1 |
SOAL
X adalah continuous random variable dengan fungsi densitas.
Diketahui \(f(x) = 6x(1 – x);0 < x < 1.\) Hitung \({\rm P}\left[ {\left| {X – \frac{1}{2}} \right| > \frac{1}{4}} \right]\)
- 0,0521
- 0,1563
- 0,3125
- 0,5000
| Diketahui | X adalah continuous random variable dengan fungsi densitas.
Diketahui \(f(x) = 6x(1 – x);0 < x < 1.\) |
| Rumus yang digunakan | \({\rm P}\left[ {\left| {X – a} \right| > b} \right] = {\rm P}\left[ {X – a > b} \right] + {\rm P}\left[ {X – a < – b} \right]\) |
| Proses pengerjaan | \(F(x) = \int\limits_0^x {6t(1 – t)dt} \)
\(F(x) = 3{x^2} – 2{x^3}\) maka dapat mencari
\({\rm P}\left[ {\left| {X – \frac{1}{2}} \right| > \frac{1}{4}} \right] = {\rm P}\left[ {X – \frac{1}{2} > \frac{1}{4}} \right] + {\rm P}\left[ {X – \frac{1}{2} < – \frac{1}{4}} \right]\)
\({\rm P}\left[ {\left| {X – \frac{1}{2}} \right| > \frac{1}{4}} \right] = {\rm P}\left[ {X > \frac{3}{4}} \right] + {\rm P}\left[ {X < \frac{1}{4}} \right]\)
\({\rm P}\left[ {\left| {X – \frac{1}{2}} \right| > \frac{1}{4}} \right] = \left( {1 – F\left( {\frac{3}{4}} \right)} \right) + F\left( {\frac{1}{4}} \right)\)
\({\rm P}\left[ {\left| {X – \frac{1}{2}} \right| > \frac{1}{4}} \right] = 0,3125\) |
| Jawaban | c. 0,3125 |
