Mathematics - Statistics Question with Solution | TestHub

Given that,

For set $1:$

Mean $=12$ and Variance $=14$

For set $2:$

Mean $=14$ and Variance $={\sigma }^2$

Now, For set $1$ we have,

$\Rightarrow \frac{{\displaystyle \sum _{i=1}^{i=15}}{x}_i}{15}=12$

$\Rightarrow {\displaystyle \sum _{i=1}^{i=15}}{x}_i=15\times 12$

Also, $\frac{{\displaystyle \sum _{i=1}^{i=15}}{x_i}^2}{15}-{\left(\frac{{\displaystyle \sum _{i=1}^{i=15}}{x}_i}{15}\right)}^2=14$

 $\Rightarrow \frac{{\displaystyle \sum _{i=1}^{i=15}}{x_i}^2}{15}-{12}^2=14$

$\Rightarrow \sum _{i=1}^{i=15}{x_i}^2=\left(14+144\right)\times 15$

Now for set $2:$

$\Rightarrow \frac{{\displaystyle \sum _{i=1}^{i=15}{y}_i}}{15}=14$

and ${\displaystyle \Rightarrow }{\displaystyle \sum _{i=1}^{i=15}}{\displaystyle {{\displaystyle y}}_i}=15\times 14$

Also, $\frac{{\displaystyle \sum _{i=1}^{i=15}{y_i}^2}}{15}-{\left(\frac{{\displaystyle \sum _{i=1}^{i=15}{y}_i}}{15}\right)}^2={\sigma }^2$

$\Rightarrow \frac{{\displaystyle \sum _{i=1}^{i=15}}{y_i}^2}{15}-{14}^2={\sigma }^2$

$\Rightarrow \sum _{i=1}^{i=15}{y_i}^2=\left({\sigma }^2+196\right)\times 15$

Now, 

The combined variance will be

​​​​​​$\Rightarrow 13=\frac{\left({\displaystyle \sum _{i=1}^{i=15}}{x_i}^2\right)\times 15+\left({\displaystyle \sum _{i=1}^{i=15}}{y_i}^2\right)\times 15}{15+15}-{\left({\displaystyle \frac{{\displaystyle \sum _{i=1}^{i=15}}{x}_i+{\displaystyle \sum _{i=1}^{i=15}}{y}_i}{15+15}}\right)}^2$

$\Rightarrow 13=\frac{\left(14+144\right)\times 15+\left({\sigma }^2+196\right)\times 15}{30}-{13}^2$

$\therefore {\sigma }^2=10$

Hence, ${\sigma }^2=10$