The standard deviations of x_i (i=1, 2, ldots | vedclass

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(B) Given that the standard deviations of $x_i$ and $y_i$ are $a$ and $b$ respectively,we have $a^2 = \frac{1}{10} \sum x_i^2 - \bar{x}^2$ and $b^2 =

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$\sqrt{a^2 + b^2 - \frac{c}{5}}$

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(B) Given that the standard deviations of $x_i$ and $y_i$ are $a$ and $b$ respectively,we have $a^2 = \frac{1}{10} \sum x_i^2 - \bar{x}^2$ and $b^2 = \frac{1}{10} \sum y_i^2 - \bar{y}^2$. We are given $\sum_{i=1}^{10} (x_i - \bar{x})(y_i - \bar{y}) = c$. Let $d_i = x_i - y_i$. The mean of $d_i$ is $\bar{d} = \bar{x} - \bar{y}$. The variance of $d_i$ is $\sigma_d^2 = \frac{1}{10} \sum (d_i - \bar{d})^2 = \frac{1}{10} \sum ((x_i - y_i) - (\bar{x} - \bar{y}))^2$. $\sigma_d^2 = \frac{1}{10} \sum ((x_i - \bar{x}) - (y_i - \bar{y}))^2 = \frac{1}{10} \sum (x_i - \bar{x})^2 + \frac{1}{10} \sum (y_i - \bar{y})^2 - \frac{2}{10} \sum (x_i - \bar{x})(y_i - \bar{y})$. Since $\frac{1}{10} \sum (x_i - \bar{x})^2 = a^2$ and $\frac{1}{10} \sum (y_i - \bar{y})^2 = b^2$,and $\sum (x_i - \bar{x})(y_i - \bar{y}) = c$,we get: $\sigma_d^2 = a^2 + b^2 - \frac{2c}{10} = a^2 + b^2 - \frac{c}{5}$. Therefore,the standard deviation is $\sqrt{a^2 + b^2 - \frac{c}{5}}$.

Class Interval	Frequency
$0 - 6$	$10$
$6 - 12$	$8$
$12 - 18$	$6$
$18 - 24$	$4$
$24 - 30$	$2$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$

Similar Questions

The variance of the following frequency distribution is
Class Interval Frequency
$0 - 6$ $10$
$6 - 12$ $8$
$12 - 18$ $6$
$18 - 24$ $4$
$24 - 30$ $2$

If the mean of the frequency distribution is $28$,then its variance is $........$.

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$

Frequency $2$ $3$ $x$ $5$ $4$

Suppose a population $A$ has $100$ observations $101, 102, . . ., 200$ and another population $B$ has $100$ observations $151, 152, . . ., 250$. If $V_A$ and $V_B$ represent the variances of the two populations,respectively,then $V_A / V_B$ is:

If the mean and variance of the data $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$ where $\alpha > \beta$ are $56$ and $66.2$ respectively,then $\alpha^2 + \beta^2$ is equal to

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