Question

The standard deviations of $x_i(i=1.2,.....10)$ and $y_i(i,.....10)$ are respectively 'a' and 'b' $\bar{x},\bar{y}$ are the means of these two sets of observations respectively. If $z_i=(x_i-\bar{x})(y_i-\bar{y})$ and ${\sum }_{i=1}^{10}{z}_i=c$ , then the standard deviation of the observations $(x_i-y_i),(i=1,2,.....10)$ is

• A. $\sqrt{a^2+b^2+\frac{c}{5}}$
• B. $\sqrt{a^2+b^2-\frac{c}{5}}$
• C. $\sqrt{a^2+b^2-\frac{c^2}{5}}$
• D. $\sqrt{a^2+b^2+\frac{c^2}{5}}$

Answer 1

Answer: (B)

We have,
$\sum _{i=1}^{10}{z}_i=c\Rightarrow \sum _{i=1}^{10}\left[\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)\right]=c$
$\Rightarrow \sum _{i=1}^{10}\left(x_i{y}_i-x_i\bar{y}-\bar{x}{y}_i+\bar{x}\cdot \bar{y}\right)=c$
$\Rightarrow \sum _{i=1}^{10}\left(x_i{y}_i\right)-\bar{y}\sum _{i=1}^{10}{x}_i-\bar{x}\sum _{i=1}^{10}{y}_i+\bar{x}\bar{y}\sum _{i=1}^{10}l=c$
$\Rightarrow \sum _{i=1}^{10}\left(x_i{y}_i\right)-10\bar{x}\bar{y}-10\bar{x}\bar{y}+10\bar{x}\bar{y}=c$
$\Rightarrow \sum _{i=1}^{10}{x}_i{y}_i-10\bar{x}\bar{y}=c\dots$(i)
Again, ${\sigma }_{\left(x_i-y_i\right)}=\sqrt{\frac{1}{10}\Sigma {\left(x_i-y_i\right)}^2-(\bar{x}-\bar{y}{)}^2}$
$=\sqrt{\left(\frac{1}{10}\Sigma x_i^2+\frac{1}{10}\Sigma y_i^2-\frac{1}{5}\Sigma x_i{y}_i-{\bar{x}}^2-{\bar{y}}^2+2\bar{x}\bar{y}\right)}$
$=\sqrt{\left[(\frac{1}{10}\Sigma x_i^2-(\bar{x}{)}^2)+(\frac{1}{10}\Sigma y_i^2-(\bar{y}{)}^2)-\frac{1}{5}(\Sigma x_i{y}_i-10\bar{x}\bar{y})\right]}$
$=\sqrt{a^2+b^2-\frac{c}{5}}[$ By Eq. (i)]