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Question

Given: $\sigma_{x}^{2}=4$ and $\sigma_{y}^{2}=5$

Also given that $\frac{\Sigma x_{i}}{5}=2$ and $\frac{\Sigma y_{i}}{5}=4$

$\Rightarrow \Sigma x

Vedclass · Accepted Answer

$\frac{{11}}{2}$

Vedclass · Answer

Given: $\sigma_{x}^{2}=4$ and $\sigma_{y}^{2}=5$

Also given that $\frac{\Sigma x_{i}}{5}=2$ and $\frac{\Sigma y_{i}}{5}=4$

$\Rightarrow \Sigma x_{i}=\bar{x}=10$ and $\Sigma y_{i}=\bar{y}=20$

$\sigma_{x}^{2}=\left(\frac{1}{5} \Sigma x_{i}^{2}\right)-(\bar{x})^{2}$

$=\left(\frac{1}{5} \Sigma x_{i}^{2}\right)-(2)^{2}$      ......$(i)$

$\sigma_{y}^{2}=\frac{1}{5}\left(\Sigma y_{i}^{2}\right)-(\bar{y})^{2}$

$=\frac{1}{5}\left(\Sigma y_{i}^{2}\right)-16$         ..........$(ii)$

Substituting $\sigma_{x}^{2}=4$ in $(i)$ we get

$4=\left(\frac{1}{5} \Sigma x_{i}^{2}\right)-4$

$\Rightarrow 4+4=\frac{1}{5} \Sigma x_{i}^{2}$

$\Rightarrow \Sigma x_{i}^{2}=40$

Similarly by substituting $\sigma_{y}^{2}=5$ in $(ii)$ we have

$5=\frac{1}{5} \Sigma y_{i}^{2}-16$

$\Rightarrow 5+16=\frac{1}{5} \Sigma y_{i}^{2}$

$\Rightarrow 21=\frac{1}{5} \Sigma y_{i}^{2}$

$\Rightarrow \Sigma y_{i}^{2}=105$

Combined varience $=\sigma_{z}^{2}=\frac{1}{10}\left(\Sigma x_{i}^{2}+\Sigma y_{i}^{2}\right)-\left(\frac{\bar{x}+\bar{y}}{2}\right)^{2}$

$=\frac{1}{10}(40+105)-\left(\frac{2+4}{2}\right)^{2}$

$=\frac{145-90}{10}$

$=\frac{55}{10}=\frac{11}{2}$

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