Question

The time period (in $\text{s}$) of oscillations of a simple pendulum in an experiment is recorded as $2.36,2.65,2.24,2.17$ and $2.08$. The mean absolute error of the measurements is

• A. $0.3\text{s}$
• B. $0.16\text{s}$
• C. $0.2\text{s}$
• D. $0.164\text{s}$

Answer 1

The correct option is B $0.16\text{s}$

Arithmetic mean,
$a_{mean}=\frac{a_1+a_2+a_3+.....+a_n}{n}$
$\Rightarrow a_{mean}=\frac{2.36+2.65+2.24+2.17+2.08}{5}$
$\Rightarrow a_{mean}=2.30\text{s}$
Now, absolute error,
$|\Delta a_1|=|a_1-a_{mean}|=|2.36-2.30|=0.06\text{s}$
Similarly,
$|\Delta a_2|=0.35\text{s}$
$|\Delta a_3|=0.06\text{s}$
$|\Delta a_4|=0.13\text{s}$
$|\Delta a_5|=0.22\text{s}$
So, mean absolute error,
$\Delta a_{mean}=\frac{|\Delta a_1|+|\Delta a_2|+|\Delta a_3|+.....+|\Delta a_n|}{n}$
$\Rightarrow \Delta a_{mean}=\frac{0.06+0.35+0.06+0.13+0.22}{5}$
$\Rightarrow \Delta a_{mean}=0.164\text{s}$
On rounding off to correct decimal places, we get, $\Delta a_{mean}=0.16\text{s}$