Question

The lengths (in cm) of 10 rods in a shop are given below:
40.0, 52.3, 55.2, 72.9, 52.8, 79.0, 32.5, 15.2, 27.9, 30.2
(i) Find mean deviation from median
(ii) Find mean deviation from the mean also.

Answer 1

i) Formula for the mean deviation from the median is as follows:
$MD=\frac{1}{n}\sum _{i=1}^n\left|d_i\right|,where\left|d_i\right|=\left|x_i-M\right|$

Arranging the data in ascending order for finding the median:
15.2, 27.9, 30.2, 32.5, 40, 52.3, 52.8, 55.2, 72.9, 79

Here, $n=10$.
Therefore, median is the average of the fifth and the sixth observations.

$M=\frac{40+52.3}{2}=46.15$

$x_i$	$\left|d_i\right|=\left|x_i-46.15\right|$
40	6.15
52.3	6.15
55.2	9.05
72.9	26.75
52.8	6.65
79	32.85
32.5	13.65
15.2	30.95
27.9	18.25
32	14.15
Total	164.6

$MD=\frac{1}{10}\times 164.6=16.46$

Mean deviation from median in 16.4 cm.

ii)
Let $\overline{x}$ be the mean of the given data set.

$\overline{x}=\frac{40+52.3+55.2+72.9+52.8+79+32.5+15.2+27.9+30.2}{10}=45.98$

$x_i$	$\left|d_i\right|=\left|x_i-45.98\right|$
40	5.98
52.3	6.32
55.2	9.22
72.9	26.92
52.8	6.82
79	33.02
32.5	13.48
15.2	30.78
27.9	18.08
32	13.98
Total	164.6

$MD=\frac{1}{10}\times 164.6=16.46$

Mean deviation from the mean is 16.4 cm.