Pembahasan Soal Ujian Profesi Aktuaris
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. \(\frac{{2n + 1}}{{12}}\) |