GATE Overflow | Mock GATE | Test 1 | Question: 44

Question

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 is?

• A. $11/2$
• B. $6$
• C. $13/2$
• D. $5/2$

Answer 1

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 variance $= \sigma_z^2 = \large\frac{1}{10}$ $ ( \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}$

Answer 2

$\text{Combined Mean and Combined variance of two populations:}$

If we put all the respective values, we'll get $(A) \frac{11}{2}$ as the correct answer.