Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diketahui variabel acak X berdistribusi Normal dengan mean 1 dan variansi 4. Maka nilai dari \(\Pr ({X^2} – 8 \le 2X)\) sama dengan …
- 0,13
- 0,43
- 0,75
- 0,86
- 0,93
| Diketahui | variabel acak X berdistribusi Normal dengan \(\mu = 1,{\sigma ^2} = 4\) |
| Rumus yang digunakan | \(\Pr ({X^2} – 8 \le 2X) = \Pr ({X^2} – 2X – 8 \le 0) = \Pr ((X – 4)(X + 2) \le 0)\) |
| Proses pengerjaan | \(\Pr ({X^2} – 8 \le 2X) = \Pr ({X^2} – 2X – 8 \le 0) = \Pr ((X – 4)(X + 2) \le 0)\) \(\Pr ((X – 4)(X + 2) \le 0) = \Pr ( – 2 \le X \le 4)\) \(\Pr ( – 2 \le X \le 4) = \Pr \left( {\frac{{ – 2 – \mu }}{\sigma } \le \frac{{X – \mu }}{\sigma } \le \frac{{4 – \mu }}{\sigma }} \right)\) \(\Pr ( – 2 \le X \le 4) = \Pr \left( {\frac{{ – 2 – 1}}{2} \le Z \le \frac{{4 – 1}}{2}} \right)\) \(= \Pr ( – 1,5 \le Z \le 1,5) = 2(0,4332) = 0,8664\) |
| Jawaban | D. 0,86 |