Question

Find out mean deviation of the following series from mean and median:

Size	4	6	8	10	12	14	16
Frequency	2	4	5	31	2	1	4

Answer 1

x	f	fx	${\mathit{d}}_{\overline{\mathit{X}}}\mathit{=}\left|\mathit{X}\mathit{-}\overline{\mathit{X}}\right|$	$\mathit{f}{\mathit{d}}_{\overline{\mathit{X}}}$
4 6 8 10 12 14 16	2 4 5 31 2 1 4	8 24 40 310 24 14 64	5.87 3.87 1.87 0.13 2.13 4.13 6.13	11.74 15.48 9.35 4.03 4.26 4.13 24.52
	∑f = 49	∑fx = 484		$\mathbf{\sum }_{}\mathit{f}{\mathit{d}}_{\overline{\mathit{X}}}$ = 73.51

$$\begin{gathered} \mathrm{Mean}\left(\overline{X}\right)=\frac{{\sum }_{}fx}{{\sum }_{}f}=\frac{484}{49}=9.87 \\ \mathrm{Thus}\mathrm{calculate}\mathrm{the}\mathrm{deviation}\mathrm{of}\mathrm{the}\mathrm{values}\mathrm{from}9.87. \\ \mathrm{Mean}\mathrm{deviation}\mathrm{from}\mathrm{mean}\left(M{D}_{\overline{X}}\right)=\frac{{\sum }_{}f{d}_{\overline{X}}}{{\sum }_{}f}=\frac{73.51}{49}=1.50 \end{gathered}$$

x	f	c.f.	dM = |x − M|	f|dM|
4 6 8 10 12 14 16	2 4 5 31 2 1 4	2 6 11 42 44 45 49	6 4 2 0 2 4 6	12 16 10 0 4 4 24
	∑f = 49			∑f|dM| = 70

N = 49
$$\begin{gathered} \mathrm{Median}=\mathrm{size}\mathrm{of}{\left(\frac{N+1}{2}\right)}^{th} \\ \mathrm{or},\mathrm{Median}=\mathrm{size}\mathrm{of}{\left(\frac{49+1}{2}\right)}^{th}\mathrm{item}=\mathrm{size}\mathrm{of}{25}^{th}\mathrm{item} \end{gathered}$$
Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 25^th is 42 (in the c.f. column), which is corresponding to 10.
Hence, median is 10.
Thus, we calculate the deviation of the values from 10.
$\mathrm{Mean}\mathrm{Deviation}\mathrm{from}\mathrm{median}=\left(M{D}_M\right)=\frac{{\sum }_{}f\left|d_M\right|}{{\sum }_{}f}=\frac{70}{49}=1.428(\mathrm{or}1.43\mathrm{approximately})$