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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Statistics
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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

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

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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)
answered Jan 28, 2014 by thanvigandhi_1
edited Mar 26, 2014 by rvidyagovindarajan_1
 

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