Find the mean deviation about the median for | vedclass

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(A) The number of observations $n = 11$,which is odd.Arranging the data in ascending order: $3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21$.Median $(M) = \left

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The number of observations $n = 11$,which is odd.Arranging the data in ascending order: $3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21$.Median $(M) = \left(\frac{n+1}{2}ight)^{	ext{th}}$ observation $= 6^{	ext{th}}$ observation $= 9$.The absolute deviations $|x_i - M|$ are:$|3-9|=6, |3-9|=6, |4-9|=5, |5-9|=4, |7-9|=2, |9-9|=0, |10-9|=1, |12-9|=3, |18-9|=9, |19-9|=10, |21-9|=12$.Sum of absolute deviations $\sum |x_i - M| = 6 + 6 + 5 + 4 + 2 + 0 + 1 + 3 + 9 + 10 + 12 = 58$.Mean Deviation about median $= \frac{1}{n} \sum |x_i - M| = \frac{58}{11} \approx 5.27$.

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$
Freq	$6$	$7$	$15$	$16$	$4$	$2$

$x_i$	$2$	$4$	$6$	$8$	$10$
$f_i$	$1$	$2$	$3$	$2$	$1$

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

Find the mean deviation about the median for the following data: $3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21$ (in $.27$)

Similar Questions

Calculate the mean deviation about the median for the following data:
Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$ $50-60$
Freq $6$ $7$ $15$ $16$ $4$ $2$

If the mean deviation of the numbers $1, 1+d, 1+2d, \dots, 1+100d$ from their mean is $255$,then $d$ is equal to:

Consider the following distribution:
$x_i$ $2$ $4$ $6$ $8$ $10$
$f_i$ $1$ $2$ $3$ $2$ $1$

The sum of the mean deviation from the mean and the mean deviation from the median of this distribution is:

Let the mean and the standard deviation of the observations $2, 3, 3, 4, 5, 7, a, b$ be $4$ and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is:

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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