for two data sets,each of size 5, the variances are given to be 4 and 5 andthe correspondings means are given to be 2 and 4 respectively. the variances of the combined data set is

We have the formulas

$\begin{aligned} \sigma^{2} & =\frac{1}{n} \sum u_{i}^{2}-\left(\frac{\sum u_{i}}{n}\right)^{2} \\ \bar{u} & =\frac{\sum u_{i}}{n} \end{aligned}$

For first data

$\begin{array}{l} \therefore 2=\frac{\sum u_{i}}{5} \Rightarrow \sum u_{i}=10 \\ \therefore 4=\sigma^{2}=\frac{1}{5} \sum u_{i}^{2}-\left(\frac{10}{2}\right)^{2}=\frac{1}{5} \sum u_{i}^{2}-2^{2} \\ \quad \Rightarrow \sum u_{i}^{2}=5(4+2)=40 \end{array}$

Again, For second data

$\begin{array}{l} 4=\frac{\Sigma y_{i}}{5} \Rightarrow \Sigma y_{i}=20 \\ 5=\sigma^{2}=\frac{1}{5} \sum y_{i}^{2}-\left(\frac{20}{5}\right)^{2}=\frac{1}{5} \sum y_{i}^{2}-4^{2} \\ \Rightarrow \sum y_{i}^{2}=5\left(5+4^{2}\right)=105 \\ \end{array}$

$\therefore$ variance of combined data

$\begin{aligned} \sigma^{2} & =\frac{1}{10} \Sigma\left(u_{i}^{2}+y_{i}^{2}\right)-\left(\frac{\Sigma u_{i}+y_{i}}{10}\right)^{2} \\ & =\frac{105+40}{10}-\left(\frac{10+20}{10}\right)^{2} \\ & =\frac{11}{2} \end{aligned}$