Question

Find out the mean deviation from the median and its coefficient.

Marks	10	11	12	13	14
No. of students	3	12	18	12	3

Answer 1

Marks (X)	Frequency (f)	Cumulative frequency (c.f.)	Deviation from Median $$\begin{gathered} \left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{X}\mathbf{-}\mathbf{M}\right| \\ \mathit{M}\mathit{e}\mathit{d}\mathit{i}\mathit{a}\mathit{n}\mathbf{=}\mathbf{12} \end{gathered}$$	$\mathit{f}\mathbf{\times }\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$
10	3	3	2	6
11	12	15	1	12
12	18	33	0	0
13	12	45	1	12
14	3	48	2	6
	N=Σf = 48			Σf|dM| =36

$$\begin{gathered} \mathrm{Median}=\mathrm{Size}\mathrm{of}{\left[\frac{\mathrm{N}+1}{2}\right]}^{\mathrm{th}}\mathrm{item} \\ =\mathrm{Size}\mathrm{of}{\left[\frac{48+1}{2}\right]}^{\mathrm{th}}\mathrm{item} \\ =\mathrm{Size}\mathrm{of}{\left[\frac{49}{2}\right]}^{\mathrm{th}}\mathrm{item} \\ =\mathrm{Size}24.5^{\mathrm{th}}\mathrm{item} \end{gathered}$$

Now, we need to locate this item in the column of Cumulative Frequency. The item just exceeding 24.5 th item is 33 (in the c.f. column), which is corresponding to 12.

Hence, median is 12.
Thus we calculate the deviation of the values from 12.

$$\begin{gathered} \mathrm{Mean}\mathrm{deviation}\mathrm{from}\mathrm{Median}\left(\mathrm{M}.D_M\right)=\frac{\Sigma f{d}_M}{\Sigma f}=\frac{36}{48}=0.75marks \\ \mathrm{Coefficien}\mathrm{of}\mathrm{Mean}\mathrm{deviation}=\frac{M.D_M}{Median}=\frac{0.75}{12}=0.0625 \end{gathered}$$