The refractive index (n) of glass is found to have the values 1.49,1.50,1.52,1.54 and 1.48. Calculate
(i) the mean value of refractive index,
(ii) absolute error in each measurement,
(iii) mean absolute error,
(iv) fractional error and
(v) percentage error

Explanation

(i) Mean value of refractive index,
`n_(m)=(1.49+1.50+1.52+1.54+1.48)/(5)`
1.505=1.51 (rounded off to two decimal places)
(ii) Taking `n_(m)` as the true value, the absolute erros in different observations are,
`Deltan_(1)=1.51-1.49=+0.02`
`Deltan_(2)=1.51-1.50=+0.01`
`Deltan_(3)=1.51-1.52=+0.01`
`Deltan_(4)=1.51-1.54=+0.03`
`Deltan_(5)=1.51-1.48=+0.03`
9ii) Mean absolute error, `Deltan_(mean)=(|Deltan_(1)|+|Deltan_(2)|+|Deltan_(3)|+|Deltan_(4)|+|Deltan_(5)|)/(5)`
`=(0.02+0.01+0.01+0.03+0.03)/(5)=0.02`
(iv) Fractional error= `(pmDeltan_(mean))/(n_(m))=(pm0.02)/(1.51)=pm0.0132`
(v) percentage error=`(pm0.0132xx100)=pm1.32%`