Question

Compute the mean deviation from the median and from the mean for the following distribution of the scores of 50 college students:

Scores	140−150	150−160	160−170	170−180	180−190	190−200
Frequency	4	6	10	18	9	3

Answer 1

Mean Deviation from Median

Scores (X)	Frequency (f)	Cumulative Frequency (c.f.)	Mid -Values (m)	Deviation from Median $$\begin{gathered} \left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{M}\right| \\ \left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{172}\mathbf{.}\mathbf{78}\right| \end{gathered}$$	$\mathit{f}\mathbf{\times }\left|{\mathbf{d}}_{\mathbf{M}}\right|\mathbf{=}\mathit{f}\left|{\mathbf{d}}_{\mathbf{M}}\right|$
140−150	4	4	145	27.78	111.12
150−160	6	10	155	17.78	106.68
160−170	10	$\boxed{20}$	165	7.78	77.8
$\boxed{170}$−180	$\boxed{18}$	38	175	2.22	39.96
180−190	9	47	185	12.22	109.98
190−200	3	50	195	22.22	66.66
	N=Σf = 50				Σf|dM|=512.2

Median class is given by the size of ${\left(\frac{N}{2}\right)}^{th}$item, i.e. ${\left(\frac{50}{2}\right)}^{th}$ item, which is 25^th item.

This corresponds to the class interval of 170-180, so this is the median class.

$$\begin{gathered} \mathrm{Median}=l_1+\frac{{\displaystyle \frac{N}{2}}-c.f.}{f}\times i \\ \mathrm{or},\mathrm{Median}=170+\frac{25-20}{18}\times 10 \\ \mathrm{or},\mathrm{Median}=170+\frac{50}{18} \\ \Rightarrow \mathrm{Median}=172.78 \\ \mathrm{Mean}\mathrm{deviation}\mathrm{from}\mathrm{Median}\left(M.D_M\right)=\frac{\Sigma f{d}_M}{\Sigma f}=\frac{512.2}{50}=10.244marks \end{gathered}$$

Mean Deviation from Mean

Scores (X)	Frequency (f)	Mid -Values (m)	fm	Deviation from Mean $$\begin{gathered} \left|{\mathbf{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\overline{\mathbf{X}}\right| \\ \left|{\mathbf{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\left|\mathbf{m}\mathbf{-}\mathbf{171}\mathbf{.}\mathbf{2}\right| \end{gathered}$$	$\mathit{f}\mathbf{\times }\left|{\mathbf{d}}_{\overline{\mathbf{X}}}\right|\mathbf{=}\mathit{f}\left|{\mathbf{d}}_{\overline{\mathbf{X}}}\right|$
140−150	4	145	580	26.2	104.8
150−160	6	155	930	16.2	97.2
160−170	10	165	1650	6.2	62
170−180	18	175	3150	3.8	68.4
180−190	9	185	1665	13.8	124.2
190−200	3	195	585	23.8	71.4
	Σf = 50		Σfm=8560		Σ$\mathit{f}\left|{\mathbf{d}}_{\overline{\mathbf{X}}}\right|$=528

$$\begin{gathered} \mathrm{Mean}\left(\overline{X}\right)=\frac{\Sigma fm}{\Sigma f}=\frac{8560}{50}=171.2 \\ \mathrm{Mean}\mathrm{deviation}\left(M.D_{\overline{X}}\right)=\frac{\Sigma fd\overline{{}_X}}{\Sigma f}=\frac{528}{50}=10.56marks \end{gathered}$$