The standard deviations of two sets of observ | vedclass

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(C) Given standard deviations $\sigma_x = 5$ and $\sigma_y = 6$ for $n = 100$ observations.$\sum_{i=1}^{100}(x_i-\bar{x})^2 = n \sigma_x^2 = 100 	ime

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Given standard deviations $\sigma_x = 5$ and $\sigma_y = 6$ for $n = 100$ observations.$\sum_{i=1}^{100}(x_i-\bar{x})^2 = n \sigma_x^2 = 100 	imes 25 = 2500$.$\sum_{i=1}^{100}(y_i-\bar{y})^2 = n \sigma_y^2 = 100 	imes 36 = 3600$.Given $\sum_{i=1}^{100}(x_i-\bar{x})(y_i-\bar{y}) = 600$.Let $z_i = x_i - y_i$. Then $\bar{z} = \bar{x} - \bar{y}$.The variance of $Z$ is $\sigma_z^2 = \frac{1}{n} \sum_{i=1}^{100}(z_i - \bar{z})^2$.$\sigma_z^2 = \frac{1}{100} \sum_{i=1}^{100}((x_i - y_i) - (\bar{x} - \bar{y}))^2 = \frac{1}{100} \sum_{i=1}^{100}((x_i - \bar{x}) - (y_i - \bar{y}))^2$.Expanding the square: $\sigma_z^2 = \frac{1}{100} [\sum(x_i - \bar{x})^2 + \sum(y_i - \bar{y})^2 - 2 \sum(x_i - \bar{x})(y_i - \bar{y})]$.$\sigma_z^2 = \frac{1}{100} [2500 + 3600 - 2(600)] = \frac{1}{100} [6100 - 1200] = \frac{4900}{100} = 49$.Therefore,$\sigma_z = \sqrt{49} = 7$.

$x_i$	$6$	$10$	$14$	$18$	$24$	$28$	$30$
$f_i$	$2$	$4$	$7$	$12$	$8$	$4$	$3$

The standard deviations of two sets of observations $X=\{x_i\}$ and $Y=\{y_i\}$ $(i=1, 2, \ldots, 100)$ are respectively $5$ and $6$. If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}(x_i-\bar{x})(y_i-\bar{y})=600$,then the standard deviation of $Z=\{z_i \mid z_i=x_i-y_i\}$ is

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