If 10 is the mean deviation of n observations | vedclass

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Question

(C) Let the mean of the observations $x_1, x_2, \ldots, x_n$ be $\bar{x}$.Given that the mean deviation ($M$.$D$.) of $x_i$ is $\frac{1}{n} \sum_{i=1}

Vedclass · Accepted Answer

$\frac{20}{3}$

Vedclass · Answer

(C) Let the mean of the observations $x_1, x_2, \ldots, x_n$ be $\bar{x}$.Given that the mean deviation ($M$.$D$.) of $x_i$ is $\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| = 10$.Let the new observations be $y_i = \frac{2x_i + 5}{3}$.The mean of the new observations is $\bar{y} = \frac{2\bar{x} + 5}{3}$.The mean deviation of the new observations is given by:$	ext{New M.D.} = \frac{1}{n} \sum_{i=1}^{n} |y_i - \bar{y}|$$= \frac{1}{n} \sum_{i=1}^{n} |\frac{2x_i + 5}{3} - \frac{2\bar{x} + 5}{3}|$$= \frac{1}{n} \sum_{i=1}^{n} |\frac{2(x_i - \bar{x})}{3}|$$= \frac{2}{3} 	imes (\frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}|)$$= \frac{2}{3} 	imes 10 = \frac{20}{3}$.Thus,the correct option is $C$.

$x$	$6$	$12$	$18$	$24$	$30$	$36$	$42$
$f$	$4$	$7$	$9$	$18$	$15$	$10$	$5$

If $10$ is the mean deviation of $n$ observations $x_1, x_2, x_3, \ldots, x_n$,then the mean deviation of the observations $\frac{2x_1+5}{3}, \frac{2x_2+5}{3}, \frac{2x_3+5}{3}, \ldots, \frac{2x_n+5}{3}$ is

Similar Questions

The mean deviation about the median of the following distribution is:
$x$ $6$ $12$ $18$ $24$ $30$ $36$ $42$
$f$ $4$ $7$ $9$ $18$ $15$ $10$ $5$

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