Question

Match the following continuous series given in Column I with their corresponding mean deviations about mean or median in Column II and choose the correct option from the codes given below.

• A. A - (2), B - (3), C - (1)
• B. A - (3), B - (2), C - (1)
• C. A - (3), B - (1), C - (2)
• D. A - (1), B - (2), C - (3)

Answer 1

Answer: (A)

A .

Class	$f_i$	Mid value $(x_i)$	$d_i=\frac{x_i-A}{h}A=350,h=100$	$f_i{d}_i|$	|x_i - \bar{x}|$</th><thcolspan="1">$f_i |x_i - \bar{x}|$</th></tr></thead><tbody><tr><td>0-100</td><td>4</td><td>50</td><td>-3</td><td>-12</td><td>308</td><td>1232</td></tr><tr><td>100-200</td><td>8</td><td>150</td><td>-2</td><td>-16</td><td>208</td><td>1664</td></tr><tr><td>200-300</td><td>9</td><td>250</td><td>-1</td><td>-9</td><td>108</td><td>972</td></tr><tr><td>300-400</td><td>10</td><td>350</td><td>0</td><td>0</td><td>8</td><td>80</td></tr><tr><td>400-500</td><td>7</td><td>450</td><td>1</td><td>7</td><td>92</td><td>644</td></tr><tr><td>500-600</td><td>5</td><td>550</td><td>2</td><td>10</td><td>192</td><td>960</td></tr><tr><td>600-700</td><td>4</td><td>650</td><td>3</td><td>12</td><td>292</td><td>1168</td></tr><tr><td>700-800</td><td>3</td><td>750</td><td>4</td><td>12</td><td>392</td><td>1176</td></tr><tr><td>Total</td><td>$\Sigma f_i = 50$</td><td></td><td></td><td>4</td><td></td><td>7896</td></tr></tbody></table></div><br/><br/>Mean$ \bar{x} =A+\frac{\Sigma f d_i}{\Sigma f_i} \times h=350+\frac{4}{50} \times 100=350+8 $<br/>$\bar{x} =358$<br/>Meandeviationaboutthemean<br/>$=\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}=\frac{7896}{50}=157.92$<br/><divclass="table-responsive"><tableclass="tabletable-bordered"><thead><tr><thcolspan="1">Class</th><thcolspan="1">$f_i$</th><thcolspan="1">Midvalue$(x_i)$</th><thcolspan="1">$d_i = \frac{x_i-A}{h} A = 350, h = 100$</th><thcolspan="1">$f_i d_i |$</th><thcolspan="1">|x_i-\bar{x}|$	$f_i|x_i-\bar{x}|$
95- 105	9	100	-3	-27	25.3	227.7
105-115	13	110	-2	-26	15.3	198.9
115-125	26	120	-1	-26	5.3	137.8
125-135	30	130	0	0	4.7	141.0
135-145	12	140	1	12	14.7	176.4
145-155	10	150	2	20	24.7	247.0
Total	100			-47		1128.8

Mean $(\bar{x})=A+\frac{\Sigma f_i{d}_i}{\Sigma f_i}\times h$
$=130+\frac{(-47)}{100}\times 10=130-4.7=125.3$
Mean deviation about mean $=\frac{\Sigma f_i|x_i-\bar{x}|}{\Sigma f_i}=\frac{1128.8}{100}$
$=11.288$

Class	$f_i$	$cf$	Mid value $(x_i)$	$|x_i-M|$	$f_i|x_i-M|$
0 - 10	6	6	5	|5 - 27.86|= 22.86	137.16
10 - 20	8	(14)C	15	|15 - 27.86| = 12.86	102.88
(20-30)	(14)	28	25	|25-27.86| = 2.86	40.04
30 - 40	16	44	35	|35-27.86| = 7.14	114.24
40 - 50	4	48	45	|45 - 27.86| = 17.14	68.56
50 - 60	2	50	55	|55 - 27.86| = 27.14	54.28
Total	$\Sigma f_i=50$				517.16

Here, $\frac{N}{2}=\frac{50}{2}=25$
$\Rightarrow C=14,f=14,I=20,h=10$
Median $(M)=I+\frac{\frac{N}{2}-C}{f}\times h$
$=20+\frac{25-14}{14}\times 10$
$=20+\frac{11\times 10}{14}$
$=20+7.86=27.86$
$\therefore$ Mean deviation about median $=\frac{\Sigma f_i|x_i-M|}{\Sigma f_i}$
$=\frac{517.16}{50}=10.34$