If the standard deviation of 20 observations x1,x2,x3,.....,x20 is 4 and that of another 20 observations y1,y2,y3,.....,y20 is 3. Also Xi=(xi−x¯)(yi−y¯) where x¯ and y¯ are means of xi's and yi's respectively. If ∑i=120Xi=90 , then the standard deviation of the observations x1−y1,x2−y2,x3−y3,...........,x20−y20 is divisible by

∑i=120(xi−x¯)(yi−y¯)=90⇒∑xiyi−x¯∑yi−y¯∑xi+x¯ y¯∑1=90⇒∑xiyi−20 x¯ y¯=90σ2=∑i=120(xi−yi)220−(x¯−y¯)2=∑xi220+∑yi220−2∑xiyi20−x¯2−y¯2+2x¯ y¯=16+9−220(90)=16⇒σ=4

If the standard deviation of 20 observations x1,x2,x3,.....,x20 is 4 and that of another 20 observations y1,y2,y3,.....,y20 is 3. Also Xi=(xi−x¯)(yi−y¯) where x¯ and y¯ are means of xi's and yi's respectively. If ∑i=120Xi=90 , then the standard deviation of the observations x1−y1,x2−y2,x3−y3,...........,x20−y20 is divisible by

∑i=120(xi−x¯)(yi−y¯)=90⇒∑xiyi−x¯∑yi−y¯∑xi+x¯ y¯∑1=90⇒∑xiyi−20 x¯ y¯=90σ2=∑i=120(xi−yi)220−(x¯−y¯)2=∑xi220+∑yi220−2∑xiyi20−x¯2−y¯2+2x¯ y¯=16+9−220(90)=16⇒σ=4

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If the standard deviation of 20 observations $x_{1}, x_{2}, x_{3}, ....., x_{20}$ is 4 and that of another 20 observations $y_{1}, y_{2}, y_{3}, ....., y_{20}$ is 3. Also $X_{i} = (x_{i} - \bar{x}) (y_{i} - \bar{y})$ where $\bar{x} a n d \bar{y}$ are means of ${x_{i}}^{'} s a n d {y_{i}}^{'} s$ respectively. If $\sum_{i = 1}^{20} X_{i} = 90$ , then the standard deviation of the observations $x_{1} - y_{1}, x_{2} - y_{2}, x_{3} - y_{3}, ..........., x_{20} - y_{20}$ is divisible by

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Detailed Solution

$\begin{array}{l} \sum_{i = 1}^{20} (x_{i} - \bar{x}) (y_{i} - \bar{y}) = 90 \\ \Rightarrow \sum x_{i} y_{i} - \bar{x} \sum y_{i} - \bar{y} \sum x_{i} + \bar{x} \bar{y} \sum 1 = 90 \\ \Rightarrow \sum x_{i} y_{i} - 20 \bar{x} \bar{y} = 90 \\ σ^{2} = \frac{\sum_{i = 1}^{20} {(x_{i} - y_{i})}^{2}}{20} - {(\bar{x} - \bar{y})}^{2} \\ = \frac{\sum {x_{i}}^{2}}{20} + \frac{\sum {y_{i}}^{2}}{20} - \frac{2 \sum x_{i} y_{i}}{20} - {\bar{x}}^{2} - {\bar{y}}^{2} + 2 \bar{x} \bar{y} \\ = 16 + 9 - \frac{2}{20} (90) = 16 \\ \Rightarrow σ = 4 \end{array}$