The mean and SD of $$63$$ children on an arithmetic test are respectively $$27,\,6$$  and $$7.1.$$  To them are added a new group of $$26$$ who had less training and whose mean is $$19.2$$  and SD $$6.2.$$  The values of the combined group differ from the original as to $$\left( {\text{i}} \right)$$ the mean and $$\left( {\text{ii}} \right)$$ the SD is :

Solution :
$$\eqalign{ & {\text{Mean and SD }}\sigma {\text{ of the combined group are}} \cr & m = \frac{{63 \times 27.6 + 26 \times 19.2}}{{63 + 26}} = 25.1 \cr & {\text{Thus, AM is decreased by}}\,\,27.6 - 25.1 = 2.5 \cr & {\sigma ^2} = \frac{{63 \times {{\left( {7.1} \right)}^2} + 26 \times {{\left( {6.2} \right)}^2}}}{{89}} + \frac{{63{{\left( {25.1 - 27.6} \right)}^2} + 26{{\left( {25.1 - 19.2} \right)}^2}}}{{89}} \cr & \Rightarrow \sigma = 7.8\,\,\,\left( {{\text{approx}}{\text{.}}} \right) \cr} $$