Question

Find the mean deviation from mean and its coefficient for the given data:

Marks (more than)	0	10	20	30	40	50	60
No. of students	200	180	150	100	40	15	5

Answer 1

Marks (X)	Frequency (f)	Mid-Values (m)	fm	$\left|{\mathit{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\left|\mathit{m}\mathbf{-}\overline{\mathit{X}}\right|\mathbf{=}\left|\mathit{m}\mathbf{-}\mathbf{29}\mathbf{.}\mathbf{5}\right|$	$\mathit{f}\mathbf{\times }\left|\mathit{d}\overline{{}_{\mathbf{X}}}\right|\mathbf{=}\mathit{f}\left|{\mathit{d}}_{\overline{\mathbf{X}}}\right|$
0−10	20	5	100	24.5	490
10−20	30	15	450	14.5	435
20−30	50	25	1250	4.5	225
30−40	60	35	2100	5.5	330
40−50	25	45	1125	15.5	387.5
50−60	10	55	550	25.5	255
60−70	5	65	325	35.5	177.5
	Σf=200		Σfm=5900		Σ$f\left|d_{\overline{X}}\right|$=2300

$$\begin{gathered} \mathrm{Mean},\overline{X}=\frac{\Sigma fm}{\Sigma f}=\frac{5900}{200}=29.5marks \\ \mathrm{Mean}\mathrm{deviation}\left(\mathrm{M}.\mathrm{D}{.}_{\overline{X}}\right)=\frac{\Sigma f\left|d\overline{{}_X}\right|}{\Sigma f}=\frac{2300}{200}=11.5marks \\ \mathrm{Coefficient}\mathrm{of}\mathrm{M}.\mathrm{D}.=\frac{\mathrm{M}.\mathrm{D}{.}_{\overline{X}}}{\overline{X}}=\frac{11.5}{29.5}=0.39 \end{gathered}$$