Question

The diameter of a wire as measured by screw gauge was found to be $2.620,2.625,2.628and2.626cm$. Calculate mean absolute error.

• A. $0.3$
• B. $0.03$
• C. $0.003$
• D. $0.025$

Answer 1

The correct option is C $0.003$

Here, $d_1=2.620,d_2=2.625,d_3=2.628,d_4=2.626$

Mean of the diameter, $d=\frac{2.620+2.625+2.628+2.626}{4}=2.625$

So, $\Delta d_1=d_1-d=2.620-2.625=-0.005$

$\Delta d_2=d_2-d=2.625-2.625=0$

$\Delta d_3=d_3-d=2.628-2.625=0.003$

$\Delta d_4=d_4-d=2.626-2.625=0.001$

Thus, mean absolute error $=\frac{|\Delta d_1|+|\Delta d_2|+|\Delta d_3|+|\Delta d_4|}{4}=\frac{0.005+0+0.003+0.001}{4}=0.0025\sim 0.003$