Question

In an experiment, the value of refractive index of a plastic has been found 1.33,1.30, 1.34 and 1.29 in successive measurements. Find the mean absolute error for refractive index.

• A. $0.12$
• B. $0.02$
• C. $0.10$
• D. $0.01$

Answer 1

The correct option is B $0.02$

Here the mean value, $\overline{n}=\frac{1.33+1.30+1.34+1.29}{4}=1.31$
The absolute error in each measurements are,
$\Delta n_1=\overline{n}-n_1=1.31-1.33=-0.02$;
$\Delta n_2=\overline{n}-n_2=1.31-1.30=0.01$;
$\Delta n_3=\overline{n}-n_3=1.31-1.34=-0.03$;
$\Delta n_4=\overline{n}-n_4=1.31-1.29=0.02$;
The mean absolute error, $\Delta \overline{n}=\frac{|\Delta n_1|+|\Delta n_2|+|\Delta n_3|+|\Delta n_4|}{4}=\frac{0.02+0.01+0.03+0.02}{4}=0.02$