According to the question all three readings have 3 significant nos each. so the average of the tree readings should also have 3 significant figures.
so we carry on the computation like this:
average reading = \(\frac{2.17~+~2.18~+~2.17}{3}~\)
= \(2.173333333....\)
= 2.17 {correct no of significant nos}
NOTE that 3 is a sure no and has infinite significant nos.
so we carry on the computation like this:
average reading = \(\frac{2.17~+~2.18~+~2.17}{3}~\)
= \(2.173333333....\)
= 2.17 {correct no of significant nos}
NOTE that 3 is a sure no and has infinite significant nos.
Thanks for pointing out about sure no 3 which was creating trouble.