TU Wien:Statistik und Wahrscheinlichkeitstheorie VO (Dutter)/Prüfung 2012-04-25 Lösung

Angabe

Datei:TU Wien-Statistik und Wahrscheinlichkeitstheorie VO (Dutter) - Prüfung 2012-04-25.pdf

After Test Thread

[1]

Beispiel-2-Lösung von Dboon

Angabe in bisschen anderer Form:

$f(x)={\begin{cases}0&:x\leq 1\\{\frac {1}{2}}&:1<x\leq 2\\0&:2<x\leq 3\\{\frac {1}{2}}&:3<x\leq 4\\0&:x>4\\\end{cases}}$

Verteilungsfunktion

$F(x)={\begin{cases}0&:x\leq 1\\{\frac {1}{2}}\int _{1}^{x}1\,\mathrm {d} t={\underline {{\frac {1}{2}}(x-1)}}&:1<x\leq 2\\{\color {Blue}{{\frac {1}{2}}(2-1)}}+\int _{2}^{x}0={\underline {\frac {1}{2}}}&:2<x\leq 3\\{\color {Blue}{\frac {1}{2}}}+{\frac {1}{2}}\int _{3}^{x}1={\frac {1}{2}}+{\frac {1}{2}}(x-3)={\underline {{\frac {1}{2}}(x-2)}}&:3<x\leq 4\\1&:x>4\\\end{cases}}$

Wobei der blaue Teil immer der Teil bis zu dem Anfang des jeweiligen Abschnitts ist, also einfach die untere Grenze eingesetzt.

Erwartungswert

${\begin{aligned}E[X]&=\int _{1}^{4}x\,f(x)\,\mathrm {d} x\\&={\frac {1}{2}}\int _{1}^{2}x\,\mathrm {d} x+{\frac {1}{2}}\int _{2}^{3}0\,\mathrm {d} x+{\frac {1}{2}}\int _{3}^{4}x\,\mathrm {d} x\\&={\frac {1}{2}}\left({\frac {x^{2}}{2}}{\Big |}_{1}^{2}+{\frac {x^{2}}{2}}{\Big |}_{3}^{4}\right)\\&={\frac {(4-1)+(16-9)}{4}}\\&={\tfrac {5}{2}}=2.5\end{aligned}}$

Median

${\begin{aligned}F(2)={\tfrac {1}{2}}(2-1)&=0.5\\F(3)={\tfrac {1}{2}}(3-2)&=0.5\\F(2.5)&=0.5\end{aligned}}$

Standardabweichung

$Var[X]=E\left[(X-E[X])^{2}\right]=E\left[X^{2}\right]-E[X]^{2}$

${\begin{aligned}E\left[X^{2}\right]&=\int _{1}^{4}x^{2}\,f(x)\,\mathrm {d} x\\&={\frac {1}{2}}\int _{1}^{2}x^{2}\,\mathrm {d} x+{\frac {1}{2}}\int _{3}^{4}x^{2}\,\mathrm {d} x\\&={\frac {1}{2}}({\frac {x^{3}}{3}}{\Big |}_{1}^{2}+{\frac {x^{3}}{3}}{\Big |}_{3}^{4})\\&={\frac {8-1+64-27}{6}}={\tfrac {22}{3}}\end{aligned}}$

$Var[X]={\tfrac {22}{3}}-\left({\tfrac {5}{2}}\right)^{2}={\tfrac {88-75}{12}}={\tfrac {13}{12}}$

$\sigma ={\sqrt {\tfrac {13}{12}}}={\tfrac {\sqrt {39}}{6}}\approx 1.0408$

IQR

${\begin{aligned}{\frac {1}{2}}(x-1)&={\tfrac {1}{4}}\\x-1&={\tfrac {1}{2}}\\x_{.25}&={\tfrac {3}{2}}\end{aligned}}$

${\begin{aligned}{\frac {1}{2}}(x-2)&={\tfrac {3}{4}}\\x-2&={\tfrac {3}{2}}\\x_{.75}&={\tfrac {7}{2}}\end{aligned}}$

$IQR={\tfrac {7}{2}}-{\tfrac {3}{2}}={\tfrac {4}{2}}=2$