Question

Following data relate to the age-difference of husbands and wives of a particular community. Find out mean deviation from mean.

Age-difference	0−5	5−10	10−15	15−20	20−25	25−30	30−35	35−40
Frequency	449	705	507	281	109	52	16	4

Answer 1

X	Mid Value m	f	fm	$\left|m-\overline{X}\right|=\left|{\mathit{d}}_{\overline{\mathit{X}}}\right|$	$\mathit{f}\left|{\mathit{d}}_{\overline{\mathit{X}}}\right|$
0 − 5 5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35 35 − 40	2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5	449 705 507 281 109 52 16 4	1122.5 5287.5 6337.5 4917.5 2452.5 1430 520 150	7.96 2.96 2.04 7.04 12.02 17.04 22.04 27.04	3574.04 2086.8 1034.28 1978.24 1312.36 886.08 352.64 108.16
		∑f = 2123	∑fm = 22217.5		∑$\mathit{f}\left|{\mathit{d}}_{\overline{\mathit{X}}}\right|$ = 11332.6

$$\begin{gathered} \mathrm{Mean}\left(\overline{\mathit{X}}\right)=\frac{{\sum }_{}fm}{{\sum }_{}f}=\frac{22217.5}{2123}=10.46 \\ \mathrm{Mean}\mathrm{Deviation}\mathrm{from}\mathrm{Mean}\left(\mathit{M}{\mathit{D}}_{\overline{\mathit{X}}}\right)=\frac{{\sum }_{}f\left|d_{\overline{X}}\right|}{\overline{X}}=\frac{11332.6}{2123}=5.34 \end{gathered}$$