In an experiment the refractive index of glass was observed to be `1.45 , 1.56 , 1.54 , 1.44 , 1.54 , and 1.53`. Calculate
(a). Mean value of refractive index
(b). Mean absolute error
( c ) Fractional error
(d) Percentage error
(e) Express the result in terms of absolute error and percentage error

Answer

(a). `1.51` ; (b) . `~ 0.04 `; (c ) . `0.03`; (d) `3%` ; (e) `mu = 1.51 +- 3%`

Explanation

(a). `mu_(m) = ( 1.45 + 1.56 + 1.54 + 1.44 + 1.54 + 1.53 )/(6) = 1.51`
(b). Absolute error in each measurement
`Delta mu_(1) = [ 1.45 - 1.51] = 0.06 , Delta mu_(2) = [ 1.56 - 1.51] = 0.05` ,
`Delta mu_(3) = [ 1.54 - 1.51] = 0.03 , Delta mu_(4) = [ 1.44 - 1.51] = 0.07` ,
`Delta mu_(5) = [ 1.54 - 1.51] = 0.03 , Delta mu_(6) = [ 1.53 - 1.51] = 0.02`,
`Delta mu_(m) = ( 0.06 + 0.05 +0.03 + 0.07 + 0.03 +0.02)/(6)`
`= (0.26)/(6) = 0.0433 ~ 0.04`
( c ). Fractional error , `( Delta mu _(m))/( mu) = ( 0.04)/(1.51) = 0.02649 = 0.03`
(e). `mu = 1.51 +- 0.04 or 1.51 +- 3%`