Sannsynlighetsfordeling til brettspill
Til et brettspill hører det med en spesiell terning med 6 sider. Det er en side med en ener, en side med en toer, en side med en treer og tre sider med seksere. Vi kaster terningen én gang. La \(X\) være antall øyne terningen viser.
| \(k\) | 1 | 2 | 3 | 6 |
|---|---|---|---|---|
| \(P(X=k)\) | \(\frac{1}{6}\) |
Oppgave
- Skriv av og fyll ut tabellen. Vis at \(E(X)=4\).
- Bestem \(\text{Var}(X)\).
Fasit
a) –
b) 13/3
Løsningsforslag
a
| \(k\) | \(\textcolor{orange}{1}\) | \(\textcolor{seagreen}{2}\) | \(\textcolor{steelblue}{3}\) | \(\textcolor{tomato}{6}\) | Sum |
|---|---|---|---|---|---|
| \(P(X=k)\) | \(\textcolor{orange}{\frac{1}{6}}\) | \(\textcolor{seagreen}{\frac{1}{6}}\) | \(\textcolor{steelblue}{\frac{1}{6}}\) | \(\textcolor{tomato}{\frac{3}{6}=\frac{1}{2}}\) | \(1\) |
| \(k \cdot P(X=k)\) | \(\textcolor{orange}{\frac{1}{6}}\) | \(\textcolor{seagreen}{\frac{2}{6}}\) | \(\textcolor{steelblue}{\frac{3}{6}}\) | \(\textcolor{tomato}{\frac{18}{6}}\) | \(\frac{24}{6}=4\) |
| \((k-\mu)^{2}\cdot P(X=k)\) | \(\textcolor{orange}{3^{2} \cdot \frac{1}{6}=\frac{9}{6}}\) | \(\textcolor{seagreen}{2^{2}\cdot \frac{1}{6}=\frac{4}{6}}\) | \(\textcolor{steelblue}{1^{2}\cdot \frac{1}{6}=\frac{1}{6}}\) | \(\textcolor{tomato}{2^{2}\cdot \frac{3}{6}=\frac{12}{6}}\) | \(\frac{26}{6}=\frac{13}{3}\) |
Vi finner forventningsverdien ved å finne summen av rad 3 siden \(E(X)=\sum k \cdot P(X=k)\)
\[E(X)=\textcolor{orange}{\frac{1}{6}}+ \textcolor{seagreen}{\frac{2}{6}}+ \textcolor{steelblue}{\frac{3}{6}}+ \textcolor{tomato}{\frac{18}{6}}=\frac{24}{6}=4 \]
Forventningsverdien \(\mathrm{E}(X)=\underline{\underline{4}}\)
b
Vi finner variansen ved å summere rad 4 i tabellen siden \(\text{Var}(X)=\sum (k-\mu)^{2}\cdot P(X=k)\)
\[\text{Var}(X)=\textcolor{orange}{\frac{9}{6}}+ \textcolor{seagreen}{\frac{4}{6}}+ \textcolor{steelblue}{\frac{1}{6}}+ \textcolor{tomato}{\frac{12}{6}}=\frac{26}{6}=\frac{13}{3} \]
Variansen er \(\mathrm{Var}(X)=\underline{\underline{\frac{13}{3}}}\)