The correct option is A 0.01 m
Arithmetic mean,
amean=a1+a2+a3+.....+ann
⇒amean=1.22+1.23+1.23+1.24+1.255
⇒amean=1.234 m
On rounding off to correct decimal places, we get, amean=1.23 m
Now, absolute error,
|Δa1|=|a1−amean|=|1.22−1.23|=0.01 m
Similarly,
|Δa2|=0.00 m
|Δa3|=0.00 m
|Δa4|=0.01 m
|Δa5|=0.02 m
So, mean absolute error,
Δamean=|Δa1|+|Δa2|+|Δa3|+.....+|Δan|n
⇒Δamean=0.01+0.00+0.00+0.01+0.025
⇒Δamean=0.008 m
On rounding off to correct decimal places, we get, Δamean=0.01 m