1. Tardigrade
  2. Question
  3. Mathematics
  4. The standard deviations of xi(i = 1. 2 , ..... 10 ) and yi(i ,.....10) are respectively 'a' and 'b' barx , bary are the means of these two sets of observations respectively. If zi = (xi - barx)(yi - bary) and ∑10i=1 zi = c , then the standard deviation of the observations (xi -yi) , (i = 1, 2, .....10) is

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Q. 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 $\sum^{10}_{i=1} z_i = c$ , then the standard deviation of the observations $(x_i -y_i) , (i = 1, 2, .....10)$ is

AP EAMCETAP EAMCET 2019

Solution:

We have,
$\displaystyle\sum_{i=1}^{10} z_{i}=c \Rightarrow \displaystyle\sum_{i=1}^{10}\left[\left(x_{i}-\bar{x}\right)\left(y_{i}-\bar{y}\right)\right]=c $
$ \Rightarrow \displaystyle\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 \displaystyle\sum_{i=1}^{10}\left(x_{i} y_{i}\right)-\bar{y} \displaystyle\sum_{i=1}^{10} x_{i}-\bar{x} \displaystyle\sum_{i=1}^{10} y_{i}+\bar{x} \bar{y} \displaystyle\sum_{i=1}^{10} l =c$
$\Rightarrow \displaystyle\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 $
$\Rightarrow \displaystyle\sum_{i=1}^{10} x_{i} y_{i}-10 \bar{x} \bar{y}=c \dots$(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)]