In an experiment, the values of refractive in | vedclass

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Question

Mean $\,\bar n = \frac{{1.54 + 1.53 + 1.44 + 1.54 + 1.56 + 1.45}}{6}\,\,$

$\,	herefore \,\bar n = 1.51$

$\Delta n_1 = 1.51 - 1.54 = -0.03,$

Vedclass · Accepted Answer

$0.04$

Vedclass · Answer

Mean $\,\bar n = \frac{{1.54 + 1.53 + 1.44 + 1.54 + 1.56 + 1.45}}{6}\,\,$

$\,	herefore \,\bar n = 1.51$

$\Delta n_1 = 1.51 - 1.54 = -0.03,$

$\Delta n_2 = 1.51 - 1.53 = -0.02,$

$\Delta n_3 = 1.51 - 1.44 = +0.07$

$\Delta n_4 = 1.51 - 1.54 = -0.03,$

$\Delta n_5 = 1.51 - 1.56 = -0.05,$ 

$\Delta n_6 = 1.51 - 1.45 = +0.06$

$\Delta \bar n = \frac{{\left| {\,\Delta {{	ext{n}}_{	ext{1}}}\,} ight|{	ext{ }} + \left| {\,\Delta {{	ext{n}}_{{	ext{2}}\,}}} ight| + .... + \left| {\,\Delta {{	ext{n}}_{{	ext{6}}\,}}} ight|}}{6}$

$\, = \,\,\frac{{\left| {\, - 0.03\,} ight|{	ext{ }} + \left| {\, - 0.02\,} ight| + \left| {\,0.07\,} ight| + \left| {\, - 0.03\,} ight| + \left| {\, - 0.05\,} ight| + \left| {\,0.06\,} ight|}}{6}\,$

$ = \,\frac{{0.26}}{6}\, = \,0.043\,\, \approx \,0.04\,\,\,	herefore \,\,\Delta \bar n\, = \,0.04$

In an experiment, the values of refractive indices of glass were found to be $1.54, 1.53,$$1.44,1.54,1.56$ and $1.45$ in successive measurements, then Mean absolute error is

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