The average speed of a train is measured by 5 students. The results of measurements are given below

Mean value,$v_m=\frac{10.2+10.4+9.8+10.6+10.8}{5}$

$=\frac{51.8}{5}=10.4m{s}^{-1}$

$\Delta v_1=v_m-v_1=10.4-10.2=0.2$

$\Delta v_2=v_m-v_2=10.4-10.4=0.0$

$\Delta v_3=v_m-v_3=10.4-9.8=0.6$

$\Delta v_4=v_m-v_4=10.4-10.6=-0.2$

$\Delta v_5=v_m-v_5=10.4-10.8=-0.4$

Mean absolute error,

$\overline{\Delta v}=\frac{\left|\Delta v_1\right|+\left|\Delta v_2\right|+\left|\Delta v_3\right|+\left|\Delta v_4\right|+\left|\Delta v_5\right|}{5}$

$=\frac{0.2+0.0+0.6+0.2+0.4}{5}=\frac{1.4}{5}=0.28m{s}^{-1}$

Relative error $=\pm \frac{\overline{\Delta v}}{v_m}=\pm \frac{0.28}{10.4}$

Percentage error $=\pm \frac{\overline{\Delta v}}{v_m}\times 100=\pm \frac{0.28}{10.4}\times 100=\pm 2.6\%$