Question

in an experment,refractive index og glass was iobserved to be 1.45,1.56,1.54,1.44,1.54and1.53 calculate (a)mean value of refractive index.(b)mean absolute.(C)fractional error(D)percentage error

Answer 1

(a) Mean value:
                    (1.45+1.56+1.54+1.44+1.54+1.53)/6=1.51
(b)  Mean absolute error:
                     |1.51-1.45|=0.06
                     |1.51-1.56|=0.05
                     |1.51-1.54|=0.03
                     |1.51-1.44|=0.07
                     |1.51-1.54|=0.03
                     |1.51-1.53|=0.02
                     
therefore mean absolute error:
                    (0.06+0.05+0.03+0.07+0.03+0.02)/6=0.0433
(c) Fractional error:
                      0.0433/1.51=0.0286
(d) Percentage error:
                       0.286*100=2.86%