Question

Calculate mean deviation from mean of the following series:

Size of Items	3−4	4−5	5−6	6−7	7−8	8−9	9−10
Frequency	3	7	22	60	85	32	9

Answer 1

Class Interval	Mid Value (m)	f	fm	$\left|m-\overline{X}\right|=\left|{\mathit{d}}_{\overline{\mathit{X}}}\right|$	$\mathit{f}\left|{\mathit{d}}_{\overline{\mathit{X}}}\right|$
3 − 4 4 − 5 5 − 6 6 − 7 7 − 8 8 − 9 9 − 10	3.5 4.5 5.5 6.5 7.5 8.5 9.5	3 7 22 60 85 32 9	10.5 31.5 121 390 637.5 272 85.5	3.6 2.6 1.6 0.6 0.4 1.4 2.4	10.8 18.2 35.2 36 34 44.8 21.6
		∑f = 218	∑fm = 1548		${\mathbf{\sum }}_{}\mathit{f}\left|{\mathit{d}}_{\overline{\mathit{X}}}\right|$ = 200.6

$$\begin{gathered} \mathrm{Mean}\left(\overline{X}\right)=\frac{{\sum }_{}fm}{{\sum }_{}f}=\frac{1548}{218}=7.10 \\ \mathrm{Thus},\mathrm{we}\mathrm{calcualte}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{deviations}\mathrm{from}7.10 \\ \mathrm{Mean}\mathrm{Deviation}\mathrm{from}\mathrm{Mean}\left(M{D}_{\overline{X}}\right)=\frac{{\sum }_{}f\left|d_{\overline{X}}\right|}{{\sum }_{}f}=\frac{200.6}{218}=0.920 \end{gathered}$$