$a)$  

$\overline{x}=\frac{4+5+6+7+8}{5}=\frac{30}{5}=6$  

Obliczamy wariancję:

${\sigma }^2=\frac{{\left(4-6\right)}^2+{\left(5-6\right)}^2+{\left(6-6\right)}^2+{\left(7-6\right)}^2+{\left(8-6\right)}^2}{5}=\frac{{\left(-2\right)}^2+{\left(-1\right)}^2+0^2+1^2+2^2}{5}=\frac{4+1+0+1+4}{5}=\frac{10}{5}=2$  

Obliczamy odchylenie standardowe:

$\sigma =\sqrt{{\sigma }^2}=\sqrt{2}$  

$b)$  

$\overline{x}=\frac{3+6+6+6+9}{5}=\frac{30}{5}=6$  

${\sigma }^2=\frac{{\left(3-6\right)}^2+{\left(6-6\right)}^2+{\left(6-6\right)}^2+{\left(6-6\right)}^2+{\left(9-6\right)}^2}{5}=\frac{{\left(-3\right)}^2+0^2+0^2+0^2+3^2}{5}=\frac{9+0+0+0+9}{5}=\frac{18}{5}=3\frac{3}{5}=3,6$  

$\sigma =\sqrt{{\sigma }^2}=\sqrt{3,6}=\sqrt{36}\cdot \sqrt{0,1}=6\sqrt{0,1}$  

$c)$  

$\overline{x}=\frac{8+12+13+13+14}{5}=\frac{60}{5}=12$  

${\sigma }^2=\frac{{\left(8-12\right)}^2+{\left(12-12\right)}^2+{\left(13-12\right)}^2+{\left(13-12\right)}^2+{\left(14-12\right)}^2}{5}=\frac{{\left(-4\right)}^2+0^2+1^2+1^2+2^2}{5}=\frac{16+0+1+1+4}{5}=\frac{22}{5}=4\frac{2}{5}=4,4$  

$\sigma =\sqrt{{\sigma }^2}=\sqrt{4,4}=\sqrt{4}\cdot \sqrt{1,1}=2\sqrt{1,1}$