Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
Probabilitas dan Statistika |
Periode Ujian |
: |
November 2018 |
Nomor Soal |
: |
21 |
SOAL
X memiliki distribusi diskret seragam pada bilangan bulat (integer) 0,1,2, … , n dan Y memiliki distribusi diskret seragam pada bilangan bulat (integer) 1,2,3, … , n.
Hitunglah Var[X] – Var[Y]
- \(\frac{{2n + 1}}{{12}}\)
- \(\frac{1}{{12}}\)
- 0
- \(– \frac{1}{{12}}\)
- \(– \frac{{2n + 1}}{{12}}\)
Rumus |
Rumus Distribusi Seragam (1,n)
\(Var[A] = \frac{{{{\left( n \right)}^2} – 1}}{{12}}\) |
Step 1 |
\(X = 0,1,2,…,n\)
\(X = 1,2,…,n + 1\)
\(Var[X] = \frac{{{{\left( {n + 1} \right)}^2} – 1}}{{12}}\) |
Step 2 |
\(Y = 1,2,3,…,n\)
\(Var[Y] = \frac{{{{\left( n \right)}^2} – 1}}{{12}}\) |
Maka |
\(Var[X] – Var[Y] = \frac{{{{\left( {n + 1} \right)}^2} – 1}}{{12}} – \frac{{{{\left( n \right)}^2} – 1}}{{12}}\)
\(Var[X] – Var[Y] = \frac{{{n^2} + 2n + 1 – 1}}{{12}} – \frac{{{n^2} – 1}}{{12}}\)
\(Var[X] – Var[Y] = \frac{{2n + 1}}{{12}}\) |
Jawaban |
a. \(\frac{{2n + 1}}{{12}}\) |