To measure the diameter of a thin wire using screw gauge, in successive measurements; the values of the diameter of the wire are 2.04 mm, 2.06mm, 2.08mm, 2.07mm and 2.05mm. Find i)average diameter of the wire ii)absolute errors, iii)mean absolute error, iv)relative error and v)percentage error.

Mean Absolure Error
nmean​ = 2.04 + 2.06 + 2.08 + 2.07 +2.05/5
         = 10.3/5
nmean = 2.06
Absolute Error
(delta)n1 = n1 - nmean
                = 2.04 - 2.06
                = 0.02
(delta)n2 = 2.06 - 2.06
              = 0.00
(delta)n3 = 2.08 - 2.06
               = 0.02
(delta)n4 = 2.07 - 2.06
                = 0.01
(delta)n5 = 2.05 - 2.06
              = 0.01
(delta)nmean = 0.02 + 0.00 + 0.02 + 0.01 + 0.01/5
                   = 0.06/5
                    = 0.012 
(delta)nmean= 0.01
Result = 2.06 + or - 0.01
Relative Error = (delta)nmean/nmean
                        = 0.01/2.06
                       = 0.004854
Percentage Error = Relative Error x 100
                            = 0.004854 x 100
                           = 0.4854 %
 Reducing it to 3 significant figures
Percentage error = 0.485%

 

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