Question

Find the mean deviation from the mean for the following data:
(i)

Classes	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Frequencies	4	8	9	10	7	5	4	3

(ii)

Classes	95-105	105-115	115-125	125-135	135-145	145-155
Frequencies	9	13	16	26	30	12

(iii)

Classes	0-10	10-20	20-30	30-40	40-50	50-60
Frequencies	6	8	14	16	4	2

Answer 1

(i) We will compute the mean deviation from the mean in the following way:

Classes	$f_i$	Midpoints $x_i$	$f_i{x}_i$	$\left|x_i-\overline{X}\right|$ = $\left|x_i-358\right|$	$f_i\left|x_i-\overline{X}\right|$
0−100	4	50	200	308	1232
100−200	8	150	1200	208	1664
200−300	9	250	2250	108	972
300−400	10	350	3500	8	80
400−500	7	450	3150	92	644
500−600	5	550	2750	192	960
600−700	4	650	2600	292	1168
700−800	3	750	2250	392	1176
	$\sum _{i=1}^8{f}_i=50$		$\sum _{i=1}^8{f}_i{x}_i=17900$		$\sum _{i=1}^8{f}_i\left|x_i-\overline{X}\right|=7896$

$N=\sum _{i=1}^6{f}_i=50$ and $\sum _{i=1}^6{f}_i{x}_i=17900$

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^8}{f}_i{x}_i}{{\displaystyle \underset{i=1}{\overset{8}{\sum f_i}}}}=\frac{17900}{50}=358$



$$\begin{gathered} \therefore \mathrm{Mean}\mathrm{deviation}=\frac{1}{N}\sum _{i=1}^8{f}_i\left|x_i-\overline{X}\right| \\ =\frac{7896}{50} \\ =157.92 \end{gathered}$$


(ii) We will compute the mean deviation from the mean in the following way:

Classes	Frequency $f_i$	Midpoints $x_i$	$f_i{x}_i$	$\left|x_i-\overline{X}\right|$ =$\left|x_i-128.58\right|$	$f_i\left|x_i-\overline{X}\right|$
95−105	9	100	900	28.58	257.22
105−115	13	110	1430	18.58	241.54
115−125	16	120	1920	8.58	137.28
125−135	26	130	3380	1.42	36.92
135−145	30	140	4200	11.42	342.6
145−155	12	150	1800	21.42	257.04
	$\sum _{i=1}^6{f}_i=106$		$\sum _{i=1}^6{f}_i{x}_i=13630$		$\sum _{i=1}^8{f}_i\left|x_i-\overline{X}\right|=1272.6$

$N=\sum _{i=1}^6{f}_i=106$ and $\sum _{i=1}^6{f}_i{x}_i=13630$

$$\begin{gathered} \therefore \overline{X}=\frac{{\displaystyle \sum _{i=1}^6}{f}_i{x}_i}{{\displaystyle \sum _{i=1}^6}{f}_i} \\ =\frac{13630}{106} \\ =128.58 \end{gathered}$$


$$\begin{gathered} \therefore \mathrm{Mean}\mathrm{deviation}=\frac{1}{N}\sum _{i=1}^8{f}_i\left|x_i-\overline{X}\right| \\ =\frac{1272.6}{106} \\ =12.005 \end{gathered}$$



(iii) We will compute the mean deviation from the mean in the following way:

Classes	Frequency $f_i$	Midpoints $x_i$	$f_i{x}_i$	$\left|x_i-\overline{X}\right|$ = $\left|x_i-358\right|$	$f_i\left|x_i-\overline{X}\right|$
0−10	6	5	30	22	132
10−20	8	15	120	12	96
20−30	14	25	350	2	28
30−40	16	35	560	8	128
40−50	4	45	180	18	72
50−60	2	55	110	28	56
	$\sum _{i=1}^6{f}_i=50$		$\sum _{i=1}^6{f}_i{x}_i=1350$		$\sum _{i=1}^8{f}_i\left|x_i-\overline{X}\right|=512$

$N=\sum _{i=1}^6{f}_i=50$ and $\sum _{i=1}^6{f}_i{x}_i=1350$ ​

$$\begin{gathered} \overline{X}=\frac{{\displaystyle \sum _{i=1}^6}{f}_i{x}_i}{{\displaystyle \sum _{i=1}^6}{f}_i} \\ =\frac{1350}{50} \\ =27 \end{gathered}$$


$$\begin{gathered} \therefore \mathrm{Mean}\mathrm{deviation}=\frac{1}{N}\sum _{i=1}^8{f}_i\left|x_i-\overline{X}\right| \\ =\frac{512}{50} \\ =10.24 \end{gathered}$$