આપેલ મુજબ $ \bar x\,\, = \,\,{\text{4,}}\,\,{\text{n}}\,\, = \,\,{\text{5}}\,\,$ અને ${\sigma ^{\text{2}}}\, = \,\,{\text{5}}{\text{.2}}\,{\text{.}}$ જો બાકીના અવલોકનો ${{\text{x}}_{\text{1}}}{\text{,}}\,\,{{\text{x}}_{\text{2}}}{\text{ }}$ હોય તો 

${\sigma ^{\text{2}}}\, = \,\,\frac{{\Sigma {{({x_i}\, – \,\,\bar x)}^2}}}{n}\,\,\, = \,\,5.2$

$ \Rightarrow \,\,\frac{{{{({x_1}\, – \,\,4)}^2}\, + \,\,{{({x_2}\, – \,\,4)}^2}\, + \,\,{{(1\,\, – \,\,4)}^2}\, + \,\,{{(2\,\, – \,\,4)}^2}\, + \,\,{{(6\,\, – \,\,4)}^2}}}{5}\,\,\, = \,\,5.2\,\,\,\,\,\,\,\,\,\,\,\,$

$ \Rightarrow \,\,{({x_1}\, – \,\,4)^2}\, + \,\,{({x_2}\, – \,\,4)^2}\, = \,\,9\,\,…….\,(1)$

ઉપરાંત $ \bar x $     $ = \,\,{\text{4}}\,\,\,\,\, \Rightarrow \,\,\frac{{{{\text{x}}_{\text{1}}}\, + \,\,{x_2}\, + \,\,1\,\, + \,\,2\,\, + \,\,6}}{5}\,\,\, = \,\,4\,\,$

$\, \Rightarrow \,\,{x_1}\, + \,\,{x_2}\,\, = \,\,11\,\,……….\,\,(2)$

સમીકરણ ${\text{ }}{\text{. (1), (2) }}$ પરથી ${{\text{x}}_{\text{1}}}\,,\,\,{x_2}\,\, = \,\,4,\,\,7$