Question

Weight of nine students of a class is given below. Calculate mean deviation, using median and arithmetic mean of the series. Also calculate coefficient of mean deviation:
Weight (kg) : 47, 50, 58, 45, 53, 59, 47, 60, 49

Answer 1

Sr. No	Weight (X)	Deviation from Median $\left|\mathit{d}\mathit{m}\right|\mathbf{=}\left|\mathit{X}\mathbf{-}\mathbf{M}\right|$ (M = 50)
1 2 3 4 5 6 7 8 9	45 47 47 49 50 53 58 59 60	5 3 3 1 0 3 8 9 10
N=9		$\mathrm{\Sigma }\left|dm\right|=42$

Sr. No	Weight (X)	Deviation from Mean $$\begin{gathered} \left|{\mathit{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\left|\mathit{X}\mathbf{-}\overline{\mathbf{X}}\right| \\ \left(\overline{\mathit{X}}\mathbf{=}\mathbf{52}\right) \end{gathered}$$
1 2 3 4 5 6 7 8 9	45 47 47 49 50 53 58 59 60	7 5 5 3 2 1 6 7 8
N=9	$\sum _{}X$=468	$\mathrm{\Sigma }\left|d_{\overline{x}}\right|=44$

From Median
(a) M = Size of $\left(\frac{N+1}{2}\right)\mathrm{th}$ item
= Size of $\left(\frac{9+1}{2}\right)\mathrm{th}$ item
= Size of 5th item
= 50 kgs

(b) ${\mathrm{MD}}_m=\frac{\mathrm{\Sigma }\left|dm\right|}{N}=\frac{42}{9}=4.666=4.67kgsapprox.$

(c) $\mathrm{Coefficient}\mathrm{of}{\mathrm{MD}}_m=\frac{{\mathrm{MD}}_m}{\mathrm{M}}=\frac{4.67}{50}=0.0934$

From Mean

(a) $\overline{X}=\frac{\Sigma X}{N}=\frac{468}{9}=52kgs$

(b) ${\mathrm{MD}}_{\overline{\mathrm{X}}}=\frac{\mathrm{\Sigma }\left|d_{\overline{X}}\right|}{N}=\frac{44}{9}=4.89kgs$

(c) $\mathrm{Coefficient}\mathrm{of}{\mathrm{MD}}_{\overline{\mathrm{X}}}=\frac{{\mathrm{MD}}_{\overline{\mathrm{X}}}}{\overline{\mathit{X}}}=\frac{4.89}{52}=0.094$