a) Dla Jurka

${\overline{x}}_J=\frac{0+3+2+3+1+0}{6}=\frac{9}{6}=1\frac{1}{2}$  

Dla Wojtka

${\overline{x}}_W=\frac{2+1+2+3+0+2}{6}=\frac{10}{6}=1\frac{2}{3}$  

Stąd

${\overline{x}}_W>{\overline{x}}_J$  

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b) Dla Jurka wariancja

${\sigma }_J^2=\frac{2\cdot {\left(0-1,5\right)}^2+{\left(1-1,5\right)}^2+{\left(2-1,5\right)}^2+2\cdot {\left(3-1,5\right)}^2}{6}=\frac{2\cdot 2,25+0,25+0,25+2\cdot 2,25}{6}=\frac{9+0,5}{6}=\frac{19}{12}$  

Odchylenie standardowe

${\sigma }_J=\sqrt{\frac{19}{12}}\approx 1,258$  

Dla Wojtka wariancja

${\sigma }_W^2=\frac{{\left(0-1\frac{2}{3}\right)}^2+{\left(1-1\frac{2}{3}\right)}^2+3\cdot {\left(2-1\frac{2}{3}\right)}^2+{\left(3-1\frac{2}{3}\right)}^2}{6}=\frac{\frac{25}{9}+\frac{4}{9}+3\cdot \frac{1}{9}+\frac{16}{9}}{6}=\frac{48}{9}\cdot \frac{1}{6}=\frac{8}{9}$  

Odchylenie standardowe

${\sigma }_W=\sqrt{\frac{8}{9}}\approx 0,943$  

Stąd

${\sigma }_J>{\sigma }_W$