Correct Answer - Option 2 : ± 0.1
Range of error:
= ± |Exterme reading - Average reading|
Average reading
\(\frac{{110.10{\rm{ }}V + 110.20{\rm{ }}V + {\rm{ }}110.15{\rm{ }}V + 110.30{\rm{ }}V + {\rm{ }}110.25{\rm{ }}V}}{5}\)
= 110.20
Range of error
= ± |110.30 - 110.20|
= ± 0.10
Alternate method:
\(RE = ±\frac{{Max - Min}}{2}\)
\(RE = ±\frac{{110.30 - 110.10}}{2}\)
= ± 0.1