Question Number 118165 by andilizhaa last updated on 15/Oct/20
Commented by Dwaipayan Shikari last updated on 15/Oct/20
$${Translate} \\ $$
Answered by Olaf last updated on 15/Oct/20
$$\mathrm{Mean}\:\overset{\_} {{x}}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{8}+\mathrm{5}+\mathrm{2}+\mathrm{6}+\mathrm{7}+\mathrm{10}+\mathrm{6}+\mathrm{4}\right) \\ $$$$\overset{\_} {{x}}\:=\:\frac{\mathrm{48}}{\mathrm{8}}\:=\:\mathrm{6} \\ $$$$\mathrm{Average}\:\mathrm{deviation}\:\mathrm{of}\:\mathrm{data}\:: \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}+\mathrm{1}+\mathrm{4}+\mathrm{0}+\mathrm{1}+\mathrm{4}+\mathrm{0}+\mathrm{2}\right)\:=\:\frac{\mathrm{14}}{\mathrm{8}}\:=\:\frac{\mathrm{7}}{\mathrm{4}} \\ $$
Answered by Dwaipayan Shikari last updated on 15/Oct/20
$$\overset{−} {{X}}=\frac{\mathrm{8}+\mathrm{5}+\mathrm{2}+\mathrm{6}+\mathrm{7}+\mathrm{10}+\mathrm{6}+\mathrm{4}}{\mathrm{8}}=\mathrm{6} \\ $$$${Mean}\:{Deviation}\:=\frac{\mathrm{1}}{{n}}\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mid{X}_{{r}} −\overset{−} {{X}}\mid\:\:\left({Generally}\right) \\ $$$${So}\:{Mean}\:{deviation} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}+\mathrm{1}+\mathrm{4}+\mathrm{1}+\mathrm{4}+\mathrm{2}\right)=\frac{\mathrm{7}}{\mathrm{4}} \\ $$