In an experiment, refractive index of glass wavs observed to be `1.54,1.53, 1.44, 1.54, 1.56 & 1.45,` So, its absolute error, relative error & percentage error are ..... ........ & ...... Respectively.

Explanation

Average absolute error
First of all, we have to find average refractive index `barn`
Now, `bar n=(1.54+1.53+1.44+1.54+1.56+1.45)/(6)`
`:. barn=1.51`
Now the aboslute error for each observation will be :
`Deltan_(1)=1.51-1.54=0.03`
`Deltan_(2)=1.51-1.44=-0.02`
`Deltan_(3)=1.51-1.54=-0.07`
`Deltan_(4)=1.51-1.54=-0.03`
`Deltan_(5)=1.51-1.56=0.05`
`Deltan_(_(6)=1.51-1.45=+0.06`
For averae absolute error, we ll take only magnitudes.
`:.Deltan barn=(|Deltan_(1)|+|Deltan_(2)|+....+|Deltan_(6)|)/(6)`
`(|-0.03|+|-0.02|+|+0.07|+|-0.03|+|-0.05|+|+0.06|)/(6)`
`:. Delta barn=(0.26)/(6)=0.043~~0.04`
So, refrative index of glass with absolute, error,
`n=1.51+-0.04`
Relative error :
Relative error `=(Deltabarn)/(n)=(0.04)/(1.51)=0.0264`
Percentage error :
Percenatage error `=0.03xx100=3%`