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 $C$.

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 from the mean of the series $(a), (a+d), (a+2d), \ldots, (a+2nd)$ is

$A$ batsman scores runs in $10$ innings: $38, 70, 48, 34, 42, 55, 63, 46, 54, 44$. The mean deviation about the median is:

If the mean of the data $p, 6, 6, 7, 8, 11, 15, 16$ is $3$ times $p$,then the mean deviation of the data from its mean is

If the mean deviation about the median of the numbers $a, 2a, 3a, \dots, 50a$ is $50$,then what is the value of $|a|$?

The mean deviation from the mean for the data $ 3, 10, 10, 4, 7, 10, 5 $ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module