$\,\therefore \,\bar n = 1.51$

$\Delta n_1 = 1.51 - 1.54 = -0.03,$

$\Delta n_2 = 1.51 - 1.53 = -0.02,$

$\Delta n_3 = 1.51 - 1.44 = +0.07$

$\Delta n_4 = 1.51 - 1.54 = -0.03,$

$\Delta n_5 = 1.51 - 1.56 = -0.05,$ 

$\Delta n_6 = 1.51 - 1.45 = +0.06$

$\Delta \bar n = \frac{{\left| {\,\Delta {{\text{n}}_{\text{1}}}\,} \right|{\text{ }} + \left| {\,\Delta {{\text{n}}_{{\text{2}}\,}}} \right| + .... + \left| {\,\Delta {{\text{n}}_{{\text{6}}\,}}} \right|}}{6}$

$\, = \,\,\frac{{\left| {\, - 0.03\,} \right|{\text{ }} + \left| {\, - 0.02\,} \right| + \left| {\,0.07\,} \right| + \left| {\, - 0.03\,} \right| + \left| {\, - 0.05\,} \right| + \left| {\,0.06\,} \right|}}{6}\,$

$ = \,\frac{{0.26}}{6}\, = \,0.043\,\, \approx \,0.04\,\,\,\therefore \,\,\Delta \bar n\, = \,0.04$