For two data sets, each of size of $5$, the variances are given to be $4$ and $5$ and the corresponding means are given to be $2$ and $4$, respectively. The variance of the combined data set is :

The correct option is A $\frac{11}{2}$
We know that variance is given by $\frac{\sum x_i^2}{n}-(\frac{\sum x_i}{n}{)}^2$

Let $V_1$ and $V_2$ be the variances and $m_1$ and $m_2$ be their corresponding means.

Given $V_1=4$ and $V_2=5$

Given $m_1=2$ and $m_2=4$

Given size of the data $n=5$

$V_1=\frac{\sum x_i^2}{n}-(m_1{)}^2$

$⟹4=\frac{\sum x_i^2}{5}-(2{)}^2$

$⟹8=\frac{\sum x_i^2}{5}$

$⟹\sum x_i^2=40$ -----(1)

Similarly $V_2=\frac{\sum y_i^2}{n}-(m_2{)}^2$

$⟹5=\frac{\sum y_i^2}{5}-(4{)}^2$

$⟹21=\frac{\sum y_i^2}{5}$

$⟹\sum y_i^2=40+105=105$ -----(2)

adding (1) and (2) we get

$\sum (x_i^2+y_i^2)=145$ -----(3)

$m_1=\frac{\sum x_i}{n}⟹\sum x_i=2\ast 5=10$

$m_2=\frac{\sum y_i}{n}⟹\sum y_i=4\ast 5=20$

Therefore, $\sum (x_i+y_i)=10+20=30$ -----(4)

Variance of the combined data is $\frac{\sum (x_i^2+y_i^2)}{10}-(\frac{\sum (x_i+y_i)}{10}{)}^2$

$=\frac{145}{10}-(\frac{30}{10}{)}^2=\frac{55}{10}=\frac{11}{2}$