Let the mean and variance of $5$ observations $x_1,x_2,x_3,x_4,x_5$ be $2$ and $4$ respectively. If the mean and variance of $5$ other observations $y_1,y_2,y_3,y_4,y_5$ are $2$ and $4$ respectively, and the combined mean of all $10$ observations is $2$, then the variance of the combined $10$ observations is:

Solution

1. Given two sets of $$\begin{array}{rl}{n}_1& =n_2\\ & =5\end{array}$$ observations with means $$\begin{array}{rl}\bar{x}& =\bar{y}\\ & =2\end{array}$$ and variances $$\begin{array}{rl}{\sigma }_1^2& ={\sigma }_2^2\\ & =4\end{array}$$.
2. Since the combined mean $\bar{z}=2$ is equal to the individual means, the deviations $$\begin{array}{rl}{d}_1& =\bar{x}-\bar{z}\\ & =0\end{array}$$ and $$\begin{array}{rl}{d}_2& =\bar{y}-\bar{z}\\ & =0\end{array}$$.
3. The combined variance is calculated using the formula ${\sigma }^2=\frac{n_1({\sigma }_1^2+d_1^2)+n_2({\sigma }_2^2+d_2^2)}{n_1+n_2}$.
4. Substituting the values, we get $$\begin{array}{rl}{\sigma }^2& =\frac{5(4+0)+5(4+0)}{10}\\ & =\frac{20+20}{10}\\ & =4\end{array}$$.
5. Hence the answer is (B).