# For two data sets each of size 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

$\begin {array} {1 1} (A)\;\large\frac{5}{2} & \quad (B)\;\large\frac{11}{2} \\ (C)\;6 & \quad (D)\;\large\frac{13}{2} \end {array}$

Given: $\sigma_x^2=4$ and $\sigma_y^2=5$
Also given that $\large\frac{\Sigma x_i}{5}$$= 2 and \large\frac{ \Sigma y_i}{5}$$ =4$
$\Rightarrow \Sigma x_i=\overline x=10$ and $\Sigma y_i =\overline y= 20$
$\sigma_x^2= \bigg( \large\frac{1}{5}$$\Sigma x_i^2 \bigg)- (\overline x)^2 = \bigg( \large\frac{1}{5}$$ \Sigma x_i^2 \bigg) - (2)^2$..............(i)
$\sigma^2_y = \large\frac{1}{5}$$( \Sigma y_i^2)-(\overline y)^2 = \large\frac{1}{5}$$ ( \Sigma y_i^2) - 16$.............(ii)
Substituting $\sigma_x^2=4$ in (i) we get
$4 = \bigg( \large\frac{1}{5}$$\Sigma x_i^2 \bigg)-4 \Rightarrow\: 4+4 = \large\frac{1}{5}$$ \Sigma x_i^2$
$\Rightarrow\: \Sigma x_i^2 = 40$
Similarly by substituting $\sigma_y^2 = 5$ in (ii) we have
$5 = \large\frac{1}{5}$ $\Sigma y_i^2-16$
$\Rightarrow\: 5+16 = \large\frac{1}{5}$ $\Sigma y_i^2$
$\Rightarrow\:21 = \large\frac{1}{5}$ $\Sigma y_i^2$
$\Rightarrow\: \Sigma y_i^2=105$
Combined varience $= \sigma_z^2 = \large\frac{1}{10}$ $( \Sigma x_i^2 + \Sigma y_i^2 ) - \bigg( \large\frac{\overline x + \overline y}{2} \bigg)^2$
$= \large\frac{1}{10}$$(40+105)$ $- \bigg( \large\frac{2+4}{2} \bigg)^2$
$= \large\frac{145-90}{10}$
$= \large\frac{55}{10}$ $= \large\frac{11}{2}$
Ans : (B)
edited Mar 26, 2014