Using screw gauge the observation of the diameter of a wire are `1.324,1.326,1.334,1.336cm` respectively Find the average diameter the mean error the relative error and % error .

Explanation

`vecD=(sum(D))/(N)=(1.324+1.326+1.334+1.336)/(4)=1.330`
`DeltaD_(1) = 1.324 -1.330 = - 0.006`
`DeltaD_(2) = 1.326 -1 1.330 = - 0.004`
`DeltaD_(3) = 1.334 - 1.330 = 0.004`
`DeltaD_(4) = 1.336 - 1336 -1 1.330 = 0.006`
`DeltabarD=(|DeltaD_(1)|+|DeltaD_(2)|+|DeltaD_(3)|+|DeltaD_(4)|)/(4)`
`= (0.006 + 0.004 + 0.004 + 0.006)/(4) = (0.020)/(4)=0.005cm`
`(DeltabarD)/(D) = (0.005)/(1.330) = 0.004`
`%` error `= (DeltabarD)/(D) xx 100 = 0.4%` .