Diagram częstości:

 

Obliczmy średni czas, z jakim Bartek pokonywał trasę szkoła-dom. 

$\overline{x}=\frac{14\cdot 12+15\cdot 18+16\cdot 25+17\cdot 16+18\cdot 13+19\cdot 10+20\cdot 3+21\cdot 2+22\cdot 1}{12+18+25+16+13+10+3+2+1}$  

$\overline{x}=\frac{168+270+400+272+234+19+60+42+22}{100}$  

$\overline{x}=\frac{1658}{100}$  

$\overline{x}=16,58$  

$\overline{x}\approx 17$  

 

Obliczmy wariancję.

${\delta }^2=\frac{12\cdot {\left(14-17\right)}^2+18\cdot {\left(15-17\right)}^2+25\cdot {\left(16-17\right)}^2+16\cdot {\left(17-17\right)}^2+13\cdot {\left(18-17\right)}^2+10\cdot {\left(19-17\right)}^2+3\cdot {\left(20-17\right)}^2+2\cdot {\left(21-17\right)}^2+{\left(22-17\right)}^2}{100}$  

${\delta }^2=\frac{12\cdot {\left(-3\right)}^2+18\cdot {\left(-2\right)}^2+25\cdot {\left(-1\right)}^2+13\cdot 1^2+10\cdot 2^2+3\cdot 3^2+2\cdot 4^2+5^2}{100}$  

${\delta }^2=\frac{12\cdot 9+18\cdot 4+25+13+10\cdot 4+3\cdot 9+2\cdot 16+25}{100}$  

${\delta }^2=\frac{108+72+25+13+40+27+32+25}{100}$  

${\delta }^2=\frac{342}{100}$  

${\delta }^2=3,42$  

Obliczmy odchylenie standardowe.

$\delta =\sqrt{3,42}$  

$\delta \approx 1,85$