Question

In an experiment, refractive index of glass was observed to be 1.45, 1.56, 1.54, 1.44, 1.54 and 1.53. Calculate (i) Mean value of refractive index; (ii) Mean absolute error; (iii) Fractional error; (iv) Percentage error. Express the term as absolute error and percentage error.

Answer 1

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Answer 2

Given,

The refractive index of glass was observed to be 1.45, 1.56, 1.54, 1.44, 1.54, and 1.53.

To find,

(a) Mean value of the refractive index

(b) Mean absolute error

(c) Fractional error

(d) The percentage error

Solution,

We can simply solve the numerical problem by following the steps below.

We know that,

Glass refractive indices were found to be 1.45, 1.56, 1.54, 1.44, 1.54, and 1.53.

(a) Mean (X) = $\frac{1.45+1.56+1.54+1.44+1.54+1.53}{6}$

                    = $\frac{9.06}{6}$

                    = 1.51

Thus, the mean of the observations is 1.51.

(b) Mean absolute error (X') = √ ((1/6) ∑ (x₁-x)²)

                                              = 0.04

As a result, the mean absolute error is computed to be 0.04.

(c) Relative or fractional error = $\frac{X'}{X}$

                                                 = $\frac{0.04}{1.51}$

                                                 = 0.026

                                                 = 0.03

As a result, the relative or fractional inaccuracy is 0.03.

(d) Percentage error = relative error x 100 percent

                                  = 0.03 x 100 percent

                                  = 3 percent

Thus, the percentage error is calculated to be 3 percent.