Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
A20 – Probabilitas dan Statistika |
| Periode Ujian |
: |
Juni 2016 |
| Nomor Soal |
: |
7 |
SOAL
Misalkan dan ialah peubah acak saling bebas dengan
\({\mu _X} = 1,{\mu _Y} = – 1,\sigma _X^2 = \frac{1}{2},\sigma _Y^2 = 2\)
Hitung \(E[{(X + 1)^2}{(Y – 1)^2}].\)
- 1
- 9/2
- 16
- 17
- 27
PEMBAHASAN
| Kalkulasi |
\(E[{(X + 1)^2}{(Y – 1)^2}] = E[{(X + 1)^2}]E[{(Y – 1)^2}]\)
\(E[{(X + 1)^2}] = E[{X^2} + 2X + 1]\)
\(E[{(X + 1)^2}] = E[{X^2}] + 2E[X] + 1\)
\(E[{(X + 1)^2}] = (\sigma _X^2 + \mu _X^2) + 2{\mu _X} + 1\)
\(E[{(X + 1)^2}] = \left( {\frac{1}{2} + 1} \right) + 2(1) + 1\)
\(E[{(X + 1)^2}] = 4,5\)
\(E[{(Y – 1)^2}] = E[{Y^2} – 2Y + 1]\)
\(E[{(Y – 1)^2}] = E[{Y^2}] – 2E[Y] + 1\)
\(E[{(Y – 1)^2}] = (\sigma _Y^2 + \mu _Y^2) – 2{\mu _Y} + 1\)
\(E[{(Y – 1)^2}] = (2 + {( – 1)^2}) – 2( – 1) + 1\)
\(E[{(Y – 1)^2}] = 6\)
\(E[{(X + 1)^2}{(Y – 1)^2}] = \left( {4,5} \right)\left( 6 \right)\)
\(E[{(X + 1)^2}{(Y – 1)^2}] = 27\) |
| Jawaban |
e. 27 |
