Consider a series of measurements of the length of a box in an...

Given that, number of observations, $n=5$$\begin{aligned}& a_1=2.4 \mathrm{~m},a_2=2.5 \mathrm{~m},a_3=2.6 \mathrm{~m},a_4=2.8 \mathrm{~m},\\ & a_5=3.0 \mathrm{~m} \end{aligned}$$$Mean value of observations,$$\begin{aligned} \bar{a}& =\frac{a_1+a_2+a_3+a_4+a_5}{5}\\ \bar{a} & =\frac{2.4+2.5+2.6+2.8+3.0}{5}\\ & =\frac{13.3}{5}=2.66 \end{aligned}$$Absolute errors in individual observed value$$\begin{aligned}& \left|\Deltaa_1\right|=\left|a_1-\bar{a}\right|=|2.4-2.66|=|-0.26|=0.26 \\ & \left|\Deltaa_2\right|=\left|a_2-\bar{a}\right|=|2.5-2.66|=|-0.16|=0.16 \\ & \left|\Deltaa_3\right|=\left|a_3-\bar{a}\right|=|2.6-2.66|=|-0.06|=0.06 \\ & \left|\Deltaa_4\right|=\left|a_4-\bar{a}\right|=|2.8-2.66|=0.14 \\ & \left|\Deltaa_5\right|=\left|a_5-\bar{a}\right|=|3.0-2.66|=0.34 \end{aligned}$$Mean absolute error,$$\begin{aligned}\Delta \bar{a}& =\frac{\left|\Delta a_1\right|+\left|\Delta a_2\right|+\left|\Delta a_3\right|+\left|\Delta a_4\right|+\left|\Delta a_5\right|}{n}\\ \Delta \bar{a}& =\frac{0.26+0.16+0.06+0.14+0.34}{5}=\frac{0.96}{5}\\ \Delta \bar{a}& =0.192 \end{aligned}$$\begin{aligned}\text { Relative error }& =\frac{\text { Mean absolute error }}{\text { Mean value }}\\ & =\frac{\Delta \bar{a}}{\bar{a}}=\frac{0.192}{2.66}=0.072 \end{aligned}