Question

In an experiment, the refractive index of glass was observed to be $1.45,1.56,1.54,1.44,1.54,1.44,1.54$ and $1.53$ Calculate mean absolute error.

Answer 1

$\begin{array}{c}\text{Refractive index}=\mu \\ {\mu }_1=1\cdot 45{\mu }_5=1.54\\ {\mu }_2=1.46{\mu }_6=1.44\\ {\mu }_3=1.54{\mu }_7=1.54\\ {\mu }_4=1.44\\ {\mu }_8=1\cdot 53\end{array}$

$\begin{array}{c}{\mu }_{\text{average}}=\frac{1.45+1.46+1.54+1.44+1.54+1.44+1.54+1.53}{8}\\ \overline{u}=\frac{11\cdot 94}{8}=1.4925=1.50\end{array}$

$\begin{array}{c}\Delta \mu =\overline{u}\mu \\ \Delta {\mu }_1\Rightarrow 1.50-1.45=0.05\\ \Delta {\mu }_2=1.50-1.46=0.04\\ \Delta {\mu }_3=1.50-1.54=-0.04\\ \Delta {\mu }_4=1.50-1.44=0.06\end{array}$

$\begin{array}{c}\Delta {\mu }_5=1.50-1.54=-0.04\\ \Delta {\mu }_6=1.50-1.44=0.06\\ \Delta {\mu }_7=1.50-1.54=-0.04\\ △{\mu }_8=1.50-1.53=-0.03\end{array}$

$\begin{array}{c}\text{Error is always}\text{added}\end{array}$

$\Delta \overrightarrow{\mu }=\frac{0.05+0.04+0.04+0.06+0.04+0.03+0.06+0.04}{8}$

$\begin{array}{cc}=\frac{0.36}{8}& =0.045\\ & =0.04.\end{array}$