If 10 is the mean deviation of n observations | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $C$.

$x_i$	$3$	$9$	$17$	$23$	$27$
$f_i$	$8$	$10$	$12$	$9$	$5$

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

If the mean deviation of the numbers $1, 1 + d, . . . , 1 + 100d$ from their mean is $255,$ then a value of $d$ is

The following data represents the frequency distribution of $20$ observations.
$x_i$ $3$ $4$ $5$ $8$ $10$ $11$
$f_i$ $\alpha+2$ $(\alpha-1)^2$ $4$ $\alpha-1$ $2$ $\alpha$
Then its mean deviation about the mean is

What is the mean deviation of a frequency distribution equal to?

The mean deviation from the median is

Find the mean deviation about the mean for the following data:

$x_i$ $3$ $9$ $17$ $23$ $27$

$f_i$ $8$ $10$ $12$ $9$ $5$

Vedclass Test Series

Exam Paper Generator

Online Exam Module