Question

Calculate mean deviation and its coefficient from the following figures:

Class-Interval	Frequency
Less than 10	5
Less than 20	12
Less than 30	20
Less than 40	35
Less than 50	54
Less than 60	60

Answer 1

Converting less than cumulative frequency distribution into simple frequency distribution:

Class Interval	Cumulative Frequency (c.f.)	Frequency (f)	Mid-Values (m)	$\left|{\mathit{d}}_{\mathit{M}}\right|\mathbf{=}\left|\mathit{m}\mathbf{-}\mathit{M}\right|\mathbf{=}\left|\mathit{m}\mathbf{-}\mathbf{36}\mathbf{.}\mathbf{67}\right|$	$\mathit{f}\mathbf{\times }\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$
0−10	5	5	5	31.67	158.35
10−20	12	7	15	21.67	151.69
20−30	20	8	25	11.67	93.36
30−40	35	15	35	1.67	25.65
40−50	54	19	45	8.33	158.27
50−60	60	6	55	18.33	109.98
		Σf=N=60			Σf|dM|=696.7

Median class is given by the size of ${\left[\frac{\mathrm{N}}{2}\right]}^{\mathrm{th}}\mathrm{item}$, i.e. ${\left(\frac{60}{2}\right)}^{th}item$, which is the 30^th item.

This corresponds to the class interval of 30-40, so this is the median class.

$$\begin{gathered} \mathrm{Median}\left(M\right)=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ or,M=30+\frac{30-20}{15}\times 10 \\ or,M=30+\frac{100}{15} \\ \Rightarrow M=36.67 \end{gathered}$$

$$\begin{gathered} \mathrm{Mean}\mathrm{deviation}\mathrm{from}\mathrm{Median}=\frac{\Sigma f\left|d_M\right|}{\Sigma f}=\frac{696.7}{60}=11.61 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{M}.\mathrm{D}.=\frac{M.D_M}{Median}=\frac{11.61}{36.67}=0.32 \end{gathered}$$