Pembahasan Ujian PAI: A20 – No. 6 – Juni 2014

Akan dicari nilai dari \(E(|X–m|)\) \(E(|X–\mu |) = \int\limits_0^1 {\left| {X – \mu } \right|} f(x)dx\) \(E(|X–\mu |) = \int\limits_0^\mu {(\mu – x)2x} dx + \int\limits_\mu ^1 {(x – \mu )2x} dx\) \(E(|X–\mu |) = \int\limits_0^{\frac{2}{3}} {\left( {\frac{2}{3} – x} \right)2x} dx + \int\limits_{\frac{2}{3}}^1 {\left( {x – \frac{2}{3}} \right)2x} dx\) \(E(|X–\mu |) = \frac{8}{{81}} + \frac{8}{{81}} = \frac{{16}}{{81}}\)

Karena sudah diketahui \(E(|X–\mu |) = \frac{{16}}{{81}}\) maka bisa didapatkan nilai
\(\frac{{E(|X–\mu |)}}{{Var(X)}} = \frac{{\frac{{16}}{{81}}}}{{\frac{1}{{18}}}} = \frac{{32}}{9}\)