If the standard deviation of 10 observations x1,x2,......,x10 is 4 and that of another set of 10 observations y1,y2,......,y10 is 3 and also Xi=(xi−x¯)(yi−y¯) and x¯ is mean of all xi 's and y¯ is mean all yi's and ∑i=110Xi=80. Then standard deviation of observations x1−y1,x2−y2,.......,x10−y10 is

∑i=110(xi−x¯)(yi−y¯)=80 ∑i=110xiyi−y¯∑i=110xi−x¯∑i=110yi+∑i=110xy¯=80 which implies ∑i=110xiyi−10yx¯=80 σ2=∑i=110(xi−yi)210−(y¯−x¯)=9

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If the standard deviation of 10 observations $x_{1}, x_{2}, ......, x_{10}$ is 4 and that of another set of 10 observations $y_{1}, y_{2}, ......, y_{10}$ is 3 and also $X_{i} = (x_{i} - \bar{x}) (y_{i} - \bar{y})$ and $\bar{x}$ is mean of all $x_{i}' s a n d \bar{y}$ is mean all $y_{i}' s a n d \sum_{i = 1}^{10} X_{i} = 80$ . Then standard deviation of observations $x_{1} - y_{1}, x_{2} - y_{2}, ......., x_{10} - y_{10}$ is

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Detailed Solution

$\sum_{i = 1}^{10} (x_{i} - \bar{x}) (y_{i} - \bar{y}) = 80 \sum_{i = 1}^{10} x_{i} y_{i} - \bar{y} \sum_{i = 1}^{10} x_{i} - \bar{x} \sum_{i = 1}^{10} y_{i} + \sum_{i = 1}^{10} \bar{x y} = 80 w h i c h i m p l i e s \sum_{i = 1}^{10} x_{i} y_{i} - 10 \bar{y x} = 80 σ^{2} = \frac{\sum_{i = 1}^{10} {(x_{i} - y_{i})}^{2}}{10} - (\bar{y} - \bar{x}) = 9$