The mean deviation from the mean the of observations -1-0-4 is Less than 3Less than 1Greater than 2-5Greater than 4-9

Question

Toppr Admin · Accepted Answer

We have -x-x2212-1-0-43-1-xA0-xi-xA0-xi-x2212-x-xA0-x2212-1-xA0-x2212-1-x2212-1-2-xA0-0-xA0-0-x2212-1-1-xA0-4-xA0-4-x2212-1-3Mean deviation from mean -x2211-xi-x2212-x-n-2-1-33-2So- option A is correct-

$x_{i}$	$\| x_{i} -_x \|$
$- 1$	$\| - 1 - 1 \| = 2$
$0$	$\| 0 - 1 \| = 1$
$4$	$\| 4 - 1 \| = 3$

The mean deviation from the mean for the set of observations $- 1, 0, 4$ is
Less than $3$
Less than $1$
Greater than $2.5$
Greater than $4.9$

We have $_x = \frac{- 1 + 0 + 4}{3} = 1$
$x_{i}$ $| x_{i} -_x |$
$- 1$ $| - 1 - 1 | = 2$
$0$ $| 0 - 1 | = 1$
$4$ $| 4 - 1 | = 3$
Mean deviation from mean $= \frac{\sum | x_{i} -_x |}{n} = \frac{2 + 1 + 3}{3} = 2$
So, option A is correct.

The mean deviation from the mean for the set of observations −1,0,4 is Less than 3Less than 1Greater than 2.5Greater than 4.9

We have _x=−1+0+43=1 xi |xi−_x| −1 |−1−1|=2 0 |0−1|=1 4 |4−1|=3Mean deviation from mean =∑|xi−_x|n=2+1+33=2So, option A is correct.

The mean deviation from the mean for the set of observations $- 1, 0, 4$ is
Less than $3$
Less than $1$
Greater than $2.5$
Greater than $4.9$

We have $_x = \frac{- 1 + 0 + 4}{3} = 1$
$x_{i}$ $| x_{i} -_x |$
$- 1$ $| - 1 - 1 | = 2$
$0$ $| 0 - 1 | = 1$
$4$ $| 4 - 1 | = 3$
Mean deviation from mean $= \frac{\sum | x_{i} -_x |}{n} = \frac{2 + 1 + 3}{3} = 2$
So, option A is correct.