If the standard deviation of 10 observations x1,x2,.....,x10 is 10 and that of another set of 10 observations y1,y2,.......y10 is 4 and also Xi=(xi−x¯)(yi−y¯) . Given that x¯ is mean of all xi ’s and y¯ is mean all yi’s. ∑i=110Xi=320 . The variance of observations x1+y1,x2+y2,......,x10+y10 is equal to ________

σx=10⇒σx2=100⇒∑xi−x¯210=100σy=4⇒σy2=16⇒∑yi−y¯210=16 Let Zi=xi+yi Then σz=∑zi−z¯2n=∑xi-x¯+yi−y210=∑xi−x¯210+∑yi−y¯210+2∑xi−x¯yi−y¯10=100+16+2×32=180σz2=180

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If the standard deviation of 10 observations $x_{1}, x_{2}, ....., x_{10}$ is 10 and that of another set of 10 observations $y_{1}, y_{2}, ....... y_{10}$ is 4 and also $X_{i} = (x_{i} - \bar{x}) (y_{i} - \bar{y})$ . Given that $\bar{x}$ is mean of all $x_{i}$ ’s and $\bar{y}$ is mean all $y_{i}$ ’s. $\sum_{i = 1}^{10} X_{i} = 320$ . The variance of observations $x_{1} + y_{1}, x_{2} + y_{2}, ......, x_{10} + y_{10}$ is equal to ________

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answer is 180.

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Detailed Solution

$σ_{x} = 10 \Rightarrow σ_{x}^{2} = 100 \Rightarrow \sum \frac{{(x_{i} - \bar{x})}^{2}}{10} = 100$

$σ_{y} = 4 \Rightarrow σ_{y}^{2} = 16 \Rightarrow \sum \frac{{(y_{i} - \bar{y})}^{2}}{10} = 16$

$Let Z i = (x_{i} + y_{i}) Then σ_{z} = \sqrt{\frac{\sum {(z_{i} - \bar{z})}^{2}}{n}}$

$\begin{aligned} = \sqrt{\frac{\sum {[(x_{i} - \bar{x}) + (y_{i} - y)]}^{2}}{10}} \\ = \sqrt{\frac{\sum {(x_{i} - \bar{x})}^{2}}{10} + \frac{\sum {(y_{i} - \bar{y})}^{2}}{10} + 2 \frac{\sum (x_{i} - \bar{x}) (y_{i} - \bar{y})}{10}} \\ = \sqrt{100 + 16 + 2 \times 32} = \sqrt{180} \end{aligned}$