NCERT Solutions for Class 11 Maths Chapter 13 - Statistics

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Exercise 15.1

Direction (Q. Nos. 1 and 2): Find the mean deviation about the mean for the following series.

1. 4, 7, 8, 9, 10, 12, 13, 17.

Sol. Mean of the given series

$$\bar{x}=\frac{\text{Sum of terms}}{\text{Number of terms}}=\frac{\Sigma x_i}{n}\\=\frac{4+7+8+9+10+12+13+17}{8}=10$$

The absolute values of the respective deviation from the mean are 6, 3, 2, 1, 2, 2, 3, 7.

$$\text{Mean deviation about mean =}\\\frac{1}{n}\displaystyle\sum_{i=1}^n|(x_i-\bar x)|\\=\frac{6+3+2+1+0+2+3+7}{8}\\=\frac{24}{8}=3.$$

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.

Sol. Mean of the given series

$$\bar x=\frac{\text{Sum of terms}}{\text{Number of terms}}=\frac{\Sigma x_i}{n}\\=\\\frac{38+70+48+40+42+55+63+46+54+44}{10}\\=50$$

The absolute values of the respective deviation from the mean are

12, 20, 02, 10, 08, 05, 13, 04, 04, 06

Mean deviation about mean =

$$\frac{1}{n}\space\displaystyle\sum_{i=1}^n|x_i-\bar x_i|\\=\frac{84}{10}$$

= 8·4.

Direction (Q. Nos. 3 and 4): Find the mean deviation about the median for the following data.

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

Sol. The given data is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Arranging in ascending order

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Number of observation = 12 (even)

$$\text{Median M}=\\\frac{\frac{\text{Nth}}{2}\text{observation} + \bigg(\frac{N}{2}+1\bigg)\text{th observation}}{2}\\=\frac{\bigg(\frac{12}{2}\bigg)\text{th observation} + \bigg(\frac{12}{2}+1\bigg)\text{th observation}}{2}\\=\frac{\text{6th observation + 7th observation}}{2}\\=\frac{13+14}{2}=\frac{27}{2}$$

⇒ M = 13.5

The abslute values of the respective deviation from the median | x_i – M | are

3·5, 2·5, 1·5, 0·5, 0·5, 0·5, 2·5, 2·5, 3·5, 3·5, 4·5

$$\text{Therefore}\space\displaystyle\sum_{i=1}^{12}|x_i- M|=28\\\therefore\space\text{Mean deviation about median =}\\\space\frac{\displaystyle\sum_{i=1}^n|x_i-M|}{n}=\frac{28}{12}=2.32$$

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Sol. The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Arranging the data in ascending order,

36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Number of observations = 10 (even)

$$\text{Median M}=\\\frac{\frac{N}{2}\text{th observation} + \bigg(\frac{N}{2}+1\bigg)\text{th observation}}{2}\\=\frac{\bigg(\frac{10}{2}\bigg)\text{th observation} + \bigg(\frac{10}{2}+1\bigg)\text{th observation}}{2}\\=\frac{\text{5th observation + 6th observation}}{2}\\=\frac{46+49}{2}=47.5$$

The absolute values of the respective deviation from the median | xi – M | are

11·5, 5·5, 2·5, 1·5, 1·5, 1·5, 3·5, 5·5, 12·5, 24·5

$$\text{Therefore}\space\displaystyle\sum_{i=1}^n|(x_i-M)|$$

= 11·5 + 5·5 + 2·5 + … + 12·5 + 24·5

= 70

$$\text{M.D. (about median) =}\space\frac{1}{n}\displaystyle\sum_{i=1}^n|x_i-M|\\=\frac{70}{10}$$

= 7.

Direction (Q. Nos. 5 and 6): Find the mean deviation about the mean for the data in following tables.

X_i	5	10	15	20	25
f_i	7	4	6	3	5

Sol.

x_i	f_i	f_ix_i	$$\|x_i-\bar x\|$$	$$f_i\|x_i-\bar x\|$$
5	7	35	\| 5-14 \| = 9	63
10	4	40	\| 10-14 \| = 4	16
15	6	90	\| 15-14 \| = 1	06
20	3	60	\| 20-14 \| = 6	18
25	5	125	\| 25-14 \| = 11	55
Total	Σf_i=25	350		158

$$\text{Mean}=\bar x = \frac{\Sigma f_ix_i}{\text{N}}\\=\frac{350}{25}=14$$

[N = Σf_i = 25]

$$\therefore\space\text{Mean deivation about mean = }\\\space\frac{\Sigma f_i|x_i-\bar x|}{\Sigma f_i}=\frac{158}{25}=6.32$$

x_i	10	30	50	70	90
f_i	4	24	28	16	8

Sol.

x_i	f_i	f_ix_i	$$\|x_i-\bar x\|$$	$$f_i\|x_i-\bar x\|$$
10	4	40	\|10-50\|=40	160
30	24	720	\|30-50\|=20	480
50	28	1400	\|50-50\|=00	000
70	16	1120	\|70-50\|=20	320
90	8	720	\|90-50\|=40	320
Total	Σf_i=80	Σf_ix_i=4000		1280

$$\text{Mean}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{40000}{80}=50\\\lbrack\text{N}=\Sigma f_i=80\rbrack\\\therefore\space\text{Mean deviation about mean =}\\\frac{\Sigma f_i|x_i-\bar x|}{\Sigma f_i}=\frac{1280}{80}=16.$$

Direction (Q. Nos. 7 and 8): Find the mean deviation about the median for the data in following tables.

x_i	5	7	9	10	12	15
f_i	8	6	2	2	2	6

Sol.

x_i	f_i	c.f.
5	8	8
7	6	14
9	2	16
10	2	18
12	2	20
15	6	26

Here, N = Σf_i = 26 (even)

$$\text{Median M}=\\\frac{\frac{N}{2}\text{th observation + }\bigg(\frac{N}{2}+1\bigg)\text{th observation}}{2}\\=\frac{\frac{26}{2}\text{th observation} + \bigg(\frac{26}{2}+1\bigg)\text{th observation}}{2}\\=\frac{13\text{th observation + } 14\text{th observation}}{2}\\=\frac{7+7}{2}=\frac{14}{2}=7$$

\|x_i-M\|	f_i	f_i\|x_i- M\|
\|5-7\|=2	8	16
\|7-7\|=0	6	00
\|9-7\|=2	2	04
\|10-7\|=3	2	06
\|12-7\|=5	2	10
\|15-7\|=8	6	48
		Σf_i\|x-M\|=84

$$\therefore\space\text{Median deviation about median =}\\\frac{\Sigma f_i|x_i-M|}{\Sigma f_i}=\frac{84}{26}=3.23$$

Sol.

x_i	15	21	27	30	35
f_i	3	5	6	7	8

x_i	f_i	CF
15	3	3
21	5	8
27	6	14
30	7	21
35	8	29
Total	Σf_i=29

Here, N = Σf_i = 29 (odd)

$$\therefore\space\text{Median M}=\bigg(\frac{\text{N+1}}{2}\bigg)\space \text{th observation}\\=\bigg(\frac{29+1}{2}\bigg)\space\text{th value}=15 \text{th}$$

⇒ M = 30

\|x_i-M\|	f_i	f_i\|x_i-M\|
\|15-30\|=15	3	45
\|21-30\|=9	5	45
\|27-30\|=3	6	18
\|30-30\|=0	7	0
\|35-30\|=5	8	40
		Σf_i\|x - M\|= 148

$$\therefore\space\text{Mean deviation about median =}\\\frac{\Sigma f_i|x_i-M|}{\Sigma f_i}=\frac{148}{29}=5.1$$

9. Find the mean deviation about the mean for the data in Exercises 9 to 10.

Income per day	Number of persons
0 – 100	4
100 – 200	8
200 – 300	9
300 – 400	10
400 – 500	7
500 – 600	5
600 – 700	4
700 – 800	3

Sol.

Class	f_i	Mid value (x_i)	$$u_i=\frac{x_i-A}{h}\\\text{A = 350, h = 100}$$	f_iu_i	$$\|x_i-\bar x\|$$	$$f_i\|x_i-\bar{x}\|$$
0 – 100	4	50	-3	-12	308	1232
100 – 200	8	150	-2	-16	208	1664
200-300	9	250	-1	-9	108	972
300 – 400	10	350	0	0	8	80
400 – 500	7	450	1	7	92	644
500 – 600	7	550	2	10	192	960
600 – 700	4	650	3	12	292	1168
700 – 800	3	750	4	12	392	1176
Total	Σf_i=50			Σf_id_i=4		$$\Sigma f_i\|x_i-\bar x\|=7896$$

$$\text{Mean}\space\bar x=A+\frac{\Sigma f_iu_i}{\Sigma f_i}×h\\=350+\frac{4}{50}×100\\=350+8\\\bar x= 358\\\text{Mean deviation about the mean =}\\\space\frac{\Sigma f_i|x_i-\bar x|}{\Sigma f_i}\\=\frac{7896}{50}=157.92$$

10.

Height of em	Number of boys
95 – 105	9
105 – 115	13
115 – 125	26
125 – 135	30
135 – 145	12
145 – 155	10

Sol.

Class	f_i	Mid value (x_i)	$$u_i=\frac{x_i-A}{h}\\\text{A = 130, h = 10}$$	f_iu_i	$$\|x_i-\bar x\|$$	$$f_i\|x_i-\bar x\|$$
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

$$\text{Mean}\space \bar x= A+\frac{\Sigma f_iu_i}{\Sigma f_i}×h\\=130+\frac{(-47)}{100}×10$$

= 130 – 4.7 = 125.3

$$\text{Mean deviation about mean =}\space\frac{\Sigma f_i|x_i-\bar x|}{\Sigma f_i}\\=\frac{1128.8}{100}=11.29$$

11. Find the mean deviation about median for the following data:

Marks	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Number of girls	6	8	14	16	4	2

Sol.

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	Σf_i = 50				517.16

$$\frac{N}{2}=\frac{50}{2}=25$$

⇒ C(cumulative frequnecy of pre-median class)

= 14, f = 14, l = 20, h = 10

$$\text{Median}\space M = l+\frac{\frac{N}{2}-C}{f} ×h\\=20+\frac{25-14}{14}×10\\=20+\frac{11×10}{14}$$

= 20 + 7.86 = 27.86

$$\therefore\space\text{Mean deivation about mean =}\space\frac{\Sigma f_i|x_i-M|}{\Sigma f_i}\\=\frac{517.16}{50}=10.34$$

12. Calculate the mean deviation about median for the age distribution of 100 persons given below:

Age	16 – 20	21 – 25	26 – 30	31 – 35	36 – 40	41 – 45	46 – 50	51 - 55
Number	5	6	12	14	26	12	16	9

Sol.

Class	f_i	CF	Mid value x_i	$$\|x_i-M\|$$	$$f_i\|x_i-M\|$$
15.5 – 20.5	5	5	18	\| 18 – 38 \| = 20	100
20.5 – 25.5	6	11	23	\| 23 – 38 \| = 15	90
25.5 – 30.5	12	23	28	\| 28 – 38 \| = 10	120
30.5 – 35.5	14	37	33	\| 33 – 38 \| = 5	70
35.5 – 40.5	26	63	38	\| 38 – 38 \| = 0	0
40.5 – 45.5	12	75	43	\| 43 – 38 \| = 5	60
45.5 – 50.5	16	91	48	\| 48 – 38 \| = 10	160
50.5 – 55.5	9	100	53	\| 53 – 38 \| = 15	135
Total	100				Σf_i\|x_i-M\|=735

Here, N = 100

$$\therefore\space\frac{N}{2}=50$$

⇒ l = 35.5, C = 37, f = 26, h = 5

$$\because\space M=l+\frac{\frac{N}{2}-C}{f}×h\\\text{M =}\space 35.5+\frac{50-37}{26}×5\\=35.5+\frac{13}{26}×5$$

= 35.5 + 2.5 = 38

$$\therefore\space\text{Mean deviation about median =}\space\frac{\Sigma f_i|x_i-M|}{\Sigma f_i}\\=\frac{735}{100}=7.35$$

Exercise 15.2

Find the mean and variance for each of the following data in Q. No. 1 to 5:

1. 6, 7, 10, 12, 13, 4, 8, 12.

Sol.

x_i	$$(x_i-\bar x)$$	$$(x_i-\bar x)^2$$
6	-3	9
7	-2	4
10	-1	1
12	3	9
13	4	16
4	-5	25
8	-1	1
12	3	9
Total 72	722	74

$$\text{Mean}=\frac{\Sigma x_i}{n}=\frac{72}{8}=9\\\text{and}\space\text{Variance}\space\sigma^2=\frac{1}{n}\space\displaystyle\sum_{i=1}^{n}(x_i-\bar x)^2\\=\frac{1}{8}×74$$

2. First n natural numbers.

Sol.

x_i	x_i²
1	1²
2	2²
3	3²
4	4²
⫶	⫶
⫶	⫶
⫶	⫶
n	n²
$$\text{Total}=\frac{n(n+1)}{2}$$	$$\frac{n(n+1)(2n+1)}{6}$$

$$\text{Mean}=\frac{\Sigma x_i}{n}\\\bar x=\frac{n(n+1)}{2n}\\=\frac{n+1}{2}\\\text{Variance}=\frac{\Sigma x_i^2}{n}-\bigg(\frac{\Sigma x_i}{n}\bigg)^2\\=\frac{n(n+1)(2n+1)}{6n}-\bigg[\frac{n(n+1)}{2n}\bigg]^2\\=\frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}\\=\frac{(n+1)}{2}\bigg[\frac{2n+1}{3}-\frac{n+1}{2}\bigg]$$

$$=\frac{(n+1)}{2}\space\bigg[\frac{4n+2-3n-3}{6}\bigg]\\=\bigg(\frac{n+1}{2}\bigg)\bigg[\frac{n-1}{6}\bigg]=\frac{n^2-1}{12}$$

3. First 10 multiples of 3.

Sol. First 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

x_i	$$(x_i-\bar x)$$	$$(x_i-\bar x)^2$$
3	– 13·5	182.25
6	-10.5	110.25
9	-7.5	56.25
12	-4.5	20.25
15	-1.5	2.25
18	1.5	2.25
21	4.5	20.25
24	7.5	56.25
27	10.5	110.25
30	13.5	182.25
Total = 165		$$\Sigma(x_i-\bar x)^2=742.5$$

$$\text{Mean}\space\bar x=\frac{\Sigma x}{n}\\=\frac{165}{10}=16.5\\\text{Variance}\space\sigma^2=\frac{1}{n}\displaystyle\sum_{i=1}^{n}(x_i-\bar x)^2\\=\frac{742.5}{10}$$

= 74·25.

x_i	6	10	14	18	24	28	30
f_i	2	4	7	12	8	4	3

Sol.

x_i	f_i	f_ix_i	x_i²	f_ix_i²
6	2	12	36	72
10	4	40	100	400
14	7	98	196	1372
18	12	216	324	3888
24	8	192	576	4608
28	4	112	784	3136
30	3	90	900	7200
Total	Σf_i=40	Σf_ix_i=760		Σf_ix_i²=16176

$$\text{Mean}=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{760}{40}=19\\\text{Variance}=\frac{\Sigma f_ix_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_ix_i}{\Sigma f_i}\bigg)\\=\frac{16176}{40}-\bigg(\frac{760}{40}\bigg)^2$$

σ² = 404.4 – (19)²

σ² = 404.4 – 361

= 43.4

x_i	92	93	97	98	102	104	109
f_i	3	2	3	2	6	3	3

Sol.

x_i	f_i	f_ix_i	x_i²	f_ix_i²
92	3	276	8464	25392
93	2	186	8649	17298
97	3	291	9409	28227
98	2	196	9604	19208
102	6	612	10404	62424
104	3	312	10816	32448
109	3	327	11881	35643
Total	Σf_i=22	Σf_ix_i=2200		Σf_ix_i²=220640

$$\text{Mean}\space\bar x=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{2200}{22}=100\\\text{Variance}=\frac{\Sigma f_ix_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_ix_i}{\Sigma f_i}\bigg)^2\\\sigma^2=\frac{220640}{22}-(100)^2$$

σ² = 10029.09 – 10000

= 29.09

6. Find the mean and standard deviation of following table using short-cut method:

x_i	60	61	62	63	64	65	66	67	68
f_i	2	1	12	29	25	12	10	4	5

Sol.

Mid-value x_i	Frequency f_i	Deviation from mean d_i = x_i – A, A = 64	f_id_i	d_i²	f_id_i²
60	2	-4	-8	16	32
61	1	-3	-3	9	9
62	12	-3	-24	4	48
63	29	-1	-29	1	29
64	25	0	00	0	00
65	12	1	12	1	12
66	10	2	20	4	40
67	4	3	12	9	36
68	5	4	20	16	80
Total	Σf_i=100	Σd_i=0	Σf_id_i=0		Σf_id_i²=286

$$\text{Mean}=A+\frac{\Sigma f_id_i}{\Sigma f_i}\\=\bar x= 64+\frac{0}{100}=64\\\text{Standard deviation}\space\sigma=\sqrt{\frac{\Sigma f_id_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_id_i}{\Sigma f_i}\bigg)^2}\\=\sqrt{\frac{286}{100}-\bigg(\frac{0}{100}\bigg)^2}\\=\sqrt{2.86}=1.69$$

Direction (Q. Nos. 7 and 8): Find the mean and variance for the following frequency distribution in following tables.

Class	0 – 30	30 – 60	60 – 90	90 – 120	120 – 150	150 – 180	180 – 210
Frequency	2	3	5	10	3	5	2

Sol.

Class	Frequency f_i	Mid value x_i	Deviation from mean $$d_i=\frac{x_i-A}{h},\space\text{A=105}$$	f_id_i	d_i²	f_id_i²
0 – 30	2	15	-3	-6	9	18
30-60	3	45	-2	-6	4	12
60 – 90	5	75	-1	-5	1	5
90 – 120	10	105	0	0	0	0
120 – 150	3	135	1	3	1	3
150 – 180	5	165	2	10	4	20
180 – 210	2	195	3	6	9	18
Total	30			Σf_id_i=2		Σf_id_i²=76

$$\text{Mean}= A+\frac{\Sigma f_i d_i}{\Sigma f_i}×h\\=105+\frac{2}{30}×30\space(\because h=30)$$

= 105 + 2 = 107

$$\text{Variance}=\bigg[\frac{\Sigma f_id_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_id_i}{\Sigma f_i}\bigg)^2\bigg]×h^2\\=\bigg[\frac{76}{30}-\bigg(\frac{2}{30}\bigg)^2\bigg]×(30)^2\\=\bigg[\frac{76}{30}-\frac{4}{30×30}\bigg]×900\\=\bigg[\frac{76×30-4}{900}\bigg]×900\\=\bigg(\frac{2280-4}{900}\bigg)\\=\frac{2276}{900}×900=2276$$

Class	0 – 10	10 – 20	20 –30	30 –40	40 – 50
Frequency	5	8	15	16	6

Sol.

Class	Frequency f_i	Mid value x_i	$$\text{Mid-value}\\d_i=\frac{x_i=25}{10}$$	f_id_i	d_i²	f_id_i²
0 – 10	5	5	-2	-10	4	20
10-20	8	15	-1	-8	1	8
20-30	15	25	0	0	0	0
30 – 40	16	35	1	16	1	16
40 – 50	6	45	2	12	4	24
Total	Σf_i = 50			Σf_id_i=10		Σf_id_i²=68

$$\text{Mean}\space(\bar x)= A + \frac{\Sigma f_id_i}{\Sigma f_i}×h\\=25+\frac{10}{50}×10\\\text{25 + }\frac{100}{50}=25+2=27\\\text{Variance =}\space\bigg[\frac{\Sigma f_id_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_id_i}{\Sigma f_i}\bigg)^2\bigg]×h^2\\=\bigg[\frac{68}{50}-\bigg(\frac{10}{50}\bigg)^2\bigg]×(10)^2\\=\frac{[68×50-100]}{50×50}×100\\=\frac{(3400-100)}{50}×2$$

$$=\frac{3300×2}{50}=\frac{6600}{50}=132.$$

9. Find the mean, variance and standard deviation using short-cut method.

Height in cm	Number of children
70 – 75	3
75 – 80	4
80 – 85	7
85 – 90	7
90 – 95	15
95 – 100	9
100 – 105	6
105 – 110	6
110 – 115	3

Sol.

Class	Frequency f_i	Mid value x_i	$$\text{Mid-value}\\d_i=\frac{x_i-92.5}{5}$$	f_id_i	d_i²	f_id_i²
70 – 75	3	72.5	-4	-12	16	48
75 – 80	4	77.5	– 3	– 12	9	36
80 – 85	7	82.5	– 2	– 14	4	28
85 – 90	7	87.5	– 1	– 7	1	7
90 – 95	15	92.5	0	0	0	0
95 – 100	9	97.5	1	9	1	9
100 – 105	6	102.5	2	12	4	24
105 – 110	6	107.5	3	18	9	54
110 – 115	3	112.5	4	12	16	48
Total	60			Σf_id_i = 6		Σf_id_i²=254

Let assumed mean, A = 92.5, h = 5

$$\text{Mean }(\bar x)= \text{A} +\frac{\Sigma f_id_i}{\Sigma f_i}×h\\=92.5+\frac{6}{60}×5\\=92.5+\frac{5}{10}\\=92.5 + 0.5=93\\\text{Variance}=\bigg[\frac{\Sigma f_id_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_id_i}{\Sigma f_i}\bigg)^2\bigg]×h^2\\=\bigg[\frac{254}{60}-\bigg(\frac{6}{60}\bigg)^2\bigg]×(5)^2\\=\bigg[\frac{254×60-36}{60×60}\bigg]×25$$

$$=\frac{(15240-36)}{60×12}×5\\=\frac{15204×5}{720}=105.58\\\text{Standard deviation}\space\sigma=\sqrt{\text{Variance}}\\=\sqrt{105.58}=10.27$$

10. The diameter of circles (in mm) drawn in design are given below:

Diameter (in cm)	33 – 36	37 – 40	41 – 44	45 – 48	49 – 52
Number of circles	15	17	21	22	25

Calculate the standard deviation and mean diameter of the circles.

Sol.

Class	f_i	Mid value x_i	$$\text{Deviation from mean}\\d_i=\frac{x_i-92.5}{5},\space\text{A=42.5, h=4}$$	f_id_i	d_i²	f_id_i²
32.5 – 36.5	15	34.5	-2	-30	4	60
36.5 – 40.5	17	38.5	-1	-17	1	17
40.5 – 44.5	21	42.5	0	0	0	0
44.5 – 48.5	22	46.5	1	22	1	22
48.5 – 52.5	25	50.5	2	50	4	100
Total	100			25		199

$$\text{Mean}\space\bar x = A+\frac{\Sigma f_id_i}{\Sigma f_i}×h\\=42.5+\frac{25}{100}×4\\\text{= 42.5 + 1 = 43.5}\\\text{Standard deviation}\space \sigma =\\ \sqrt{\frac{\Sigma f_id_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_id_i}{\Sigma f_i}\bigg)^2}×h\\=\sqrt{\frac{199}{100}-\bigg(\frac{25}{100}\bigg)^2}×4\\=\sqrt{1.99-0.0625}×4\\=\sqrt{1.9275}×4$$

= 1·388 × 4

= 5·552

= 5·55.

Exercise 15.3

1. From the data given below state which group is more variable A or B ?

Marks	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80
Group A	9	17	32	33	40	10	9
Group B	10	20	30	25	43	15	7

Sol. For group A,

Class	f_i	x_i	$$d_i=\frac{x_i-45}{10}$$	f_id_i	d_i²	f_id_i²
10 – 20	9	15	-3	-27	9	81
20 – 30	17	25	-2	-34	4	68
30 – 40	32	35	-1	-32	1	32
40 – 50	33	45	0	0	0	0
50 – 60	40	55	1	40	1	40
60 – 70	10	65	2	20	4	40
70 – 80	9	75	3	27	9	81
Total	Σf_i = 150			Σf_id_i=-6		Σf_id_i²=342

A = 45, n = 10

$$\text{Mean}\space\bar x = A + \frac{\Sigma f_id_i}{\Sigma f_i}×h\\=45 + \frac{(-6)}{150}×10\\=45-\frac{60}{150}$$

= 45 – 0.4 = 44.6

$$\text{Standard deviation}\space\sigma=\\\sqrt{\frac{\Sigma f_id_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_id_i}{\Sigma f_i}\bigg)^2}×h\\\sqrt{\frac{342}{150}-\bigg(-\frac{6}{150}\bigg)^2}×10\\=\sqrt{\frac{342×150-36}{150×150}}×10\\=\frac{10}{150}\sqrt{51300-36}\\=\frac{1}{15}×\sqrt{51264}\\=\frac{226.42}{15}=15.09$$

∵ (Coefficient of variance) (CV for group A) =

$$\frac{\sigma}{x}×100=\frac{15.09}{44.6}×100\\=\frac{1509}{44.6}=33.83$$

For group B.

Class	f_i	x_i	$$d_i=\frac{x_i-45}{10}$$	f_id_i	d_i²	f_id_i²
10 – 20	10	15	-3	-30	9	90
20 – 30	20	25	– 2	– 40	4	80
30 – 40	30	35	– 1	– 30	1	30
40 – 50	25	45	0	00	0	00
50 – 60	43	55	1	43	1	43
60 – 70	15	65	2	30	4	60
70 – 80	7	75	3	21	9	63
Total	150			Σf_id_i=-6		Σf_id_i²=-366

$$\text{Mean}\space \bar x= A + \frac{\Sigma f_id_i}{\Sigma f_i}×h\\=45+\frac{(-6)}{150}×10\\=45-\frac{60}{150}$$

= 45 – 0.4 = 44.6

$$\text{Standard deviation}\space\sigma=\\\sqrt{\frac{\Sigma f_id_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_id_i}{\Sigma f_i}\bigg)^2}×h\\=\sqrt{\frac{366}{150}-\bigg(\frac{-6}{150}\bigg)^2}×10\\=\sqrt{\frac{366×150-36}{150×150}}×10\\=\frac{10}{150}\space\sqrt{54900-36}\\=\frac{1}{15}×\sqrt{54864}\\=\frac{234.23}{15}=15.61$$

$$\text{CV for group B}=\frac{\sigma}{x}×100\\=\frac{15.61}{44.6}×100\\=\frac{1561}{44.6}=35$$

∵ Coefficient of variation of group B is greater than the coefficient of variation from group A. Hence, group B is more variable than group A.

2. From the prices of shares X and Y below, find out which is more stable in value.

X	35	54	52	53	56	58	52	50	51	49
Y	108	107	105	105	106	107	104	103	104	101

For shares X
X	d_i=X-A	d_i²
35	-17	289
54	2	4
52	0	0
53	1	1
56	4	16
58	6	36
52	0	0
50	-2	4
51	-1	1
49	-3	9
Total	Σd_i=-10	Σd_i²=360

where, A = (assumed mean) = 52

$$\text{For share X,}\space\text{Mean}\space\bar X= A + \frac{\Sigma d_1}{n}\\=52-\frac{10}{10}$$

= 52 – 1 = 51

$$\text{Standard deviation}\space\sigma\\=\sqrt{\frac{\Sigma d_1^2}{n}-\bigg(\frac{\Sigma d_1}{n}\bigg)^2}\\=\sqrt{\frac{360}{10}-\bigg(\frac{-10}{10}\bigg)^2}\\=\sqrt{36-1}\\=\sqrt{35}$$

= 5.92

$$\text{Coefficient of variance for share X =}\frac{\sigma}{X}×100\\\frac{5.92}{51}×100\\=\frac{592}{51}=11.60$$

For shares Y
Y	d₂ = Y – A	d₂²
108	3	9
107	2	4
105	0	0
106	1	1
107	2	4
104	-1	1
103	-2	4
104	-1	1
101	-4	16
	Σd_i=0	Σd_i² = 40

$$\text{For share Y ,}\\\space\text{Mean}\space \bar Y = A + \frac{\Sigma d_2}{n}\\=105+\frac{0}{10}=105\space\text{where A = 105}\\\text{Standard deviation}\space\sigma=\\\sqrt{\frac{\Sigma d_2^2}{n}-\bigg(\frac{\Sigma d_2}{n}\bigg)^2}\\=\sqrt{\frac{40}{10}-0}\\\sqrt{4}=2\\\Rarr\space\text{CV for share = }\frac{\sigma}{Y}×100\\=\frac{2}{105}×100=\frac{200}{105}=1.90$$

∵ CV of share X > CV of share Y

∴ Share Y is more stable than share X.

3. An anlaysis of monthly wages paid to workers in two Firms A and B belonging to the same industry.

Gives the following results:

	Firm A	Firm B
Number of wages earners	586	648
Mean of monthly wages	₹ 5253	₹ 5253
Variance of distribution of wages	100	121

(i) Which firm A or B pays out larger amount as monthly wages ?

(ii) Which firm A or B shows greater variability in individual wages ?

Sol. For firm A,

Number of wages earners = 586

$$\text{Mean of monthly wages =}\bar x=₹5253$$

Variance of distribution of wages = 100

$$\text{Standard deviation}\space\sigma=\sqrt{100}=10$$

Amount paid by firm A = 586 × 5253 = ₹3078258 …(i)

$$\text{Coefficient of variation}=\frac{\sigma}{x}×100\\=\frac{10}{5253}×100\\=\frac{1000}{5253}=0.19\space\text{...(ii)}$$

For firm B,

Number of wages earners = 648

Mean of monthly wages = ₹5253

Variance of distribution of wages = 121

$$\text{Standard deviation}\sigma=\sqrt{\text{Variance}}\\=\sqrt{121}=11$$

Amount paid by firm B = 648 × ₹3403944 …(iii)

$$\text{Coefficient of vairation = }\frac{\sigma}{x}×100\\=\frac{11}{5253}×100\\=\frac{1100}{5253}=0.21\qquad\text{...(iv)}$$

From equation (i) and (ii),

Monthly wages paid by firm A = ₹3078258

and from firm B = ₹3403944

Thus, firm B pays out larger amount as monthly wages

From equation (ii) and (iv),

∵ CV of firm B > CV of firm A as 0.21 > 0.19

∴ Firm B shows greater variability in individual wages.

4. The following is the record of goals scored by team A in a football session:

Number of goal scored	0	1	2	3	4
Number of matches	1	9	7	5	3

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Sol.

Number of goals scored	Number of matches	f_ix_i	x_i²	f_ix_i²
x_i	f_i	f_ix_i	x_i²	f_ix_i²
0	1	0	0	0
1	9	9	1	9
2	7	14	4	28
3	5	15	9	45
4	3	12	16	48
Total	Σf_i=25	Σf_ix_i=50		Σf_ix_i²=130

$$\text{Mean}\space\bar x=\frac{\Sigma f_ix_i}{\Sigma f_i}=\frac{50}{25}=2\\\text{Standard deviation}\space\sigma=\sqrt{\frac{\Sigma f_ix_i^2}{\Sigma f_i}-\bigg(\frac{\Sigma f_ix_i}{\Sigma f_i}\bigg)^2}\\=\sqrt{\frac{130}{25}-\bigg(\frac{50}{25}\bigg)^2}\\=\sqrt{\frac{130}{25}-4}\\=\sqrt{\frac{130-100}{25}}=\sqrt{\frac{30}{25}}\\=\frac{1}{5}\sqrt{30}=\frac{5.48}{5}=1.096$$

$$\text{CV for team A = }\frac{\sigma}{x}×100\\=\frac{1.096}{2}×100\\=\frac{109.6}{2}=54.8$$

For team B,

Given, x = 2 and s = 1.25

$$\text{Given\space}\bar x=2\space\text{and}\space\sigma=1.25\\\therefore\space\text{CV for team B =}\frac{\sigma}{x}×100\\=\frac{1.25}{2}×100\\=\frac{125}{2}=62.5$$

∵ CV of team A < CV of team B as 54.8 < 62.5

Hence, team A is more consistent.

5. The sum and sum of squares corresponding to length x (in cm) and weight y (in g) of 50 plant products are given below:

$$\displaystyle\sum_{\textbf{i=1}}^{\textbf{50}}\space \textbf{x}_\textbf{i}\textbf{=212}\space\displaystyle\sum_{\textbf{i=1}}^{\textbf{50}}\space \textbf{x}_\textbf{i}^\textbf{2}\textbf{=902.8}\\\textbf{and}\space\displaystyle\sum_{\textbf{i=1}}^{\textbf{50}}\space \textbf{y}_\textbf{1}\textbf{=261}\space\displaystyle\sum_{\textbf{i=1}}^{\textbf{50}}\space \textbf{y}_\textbf{1}^\textbf{2}\textbf{=1457.6}$$

Which is more varying, the length or weight?

Sol. For length,

$$\text{Mean}\space \bar x=\frac{\Sigma x_i}{n}\space\\=\frac{212}{50}=4.24\\\text{Standard deviation} (\sigma) =\\\sqrt{\frac{\Sigma x_i^2}{n}-\bigg(\frac{\Sigma x_i}{n}\bigg)^2}\\=\sqrt{\frac{902.8}{50}-\bigg(\frac{212}{50}\bigg)^2}\\=\sqrt{\frac{902.8×50-(212)^2}{50×50}}\\=\frac{1}{50}\sqrt{45140-44944}$$

$$=\frac{1}{50}\sqrt{196}=\frac{14}{50}=\frac{14}{50}=0.28\\\therefore\space\text{Coefficient of variation for length }\\=\frac{\sigma}{x}×100=\frac{0.28}{4.24}=6.604\\\text{For weight,}\\\text{Mean}\space\bar y=\frac{\Sigma y_i}{n}=\frac{261}{50}=5.22\\\text{Standard deviation}(\sigma)=\sqrt{\frac{\Sigma y_i^2}{n}-\bigg(\frac{\Sigma y_i}{n}\bigg)^2}\\=\sqrt{\frac{1457.6}{50}-\bigg(\frac{261}{50}\bigg)^2}$$

$$=\frac{1457.6×50-(261)^2}{50×50}\\=\frac{1}{40}\space\sqrt{72880-68121}\\=\frac{1}{50}\sqrt{4759}\\=\frac{1}{50}×68.98=1.37\\\therefore\space\text{CV for weight =}\frac{\sigma}{y}×100\\=\frac{1.37}{5.22}×100\\=\frac{1.37}{5.22}×100=26.24$$

∵ CV of weight (26.24) > CV of length (6.6)

∴ Weight is more varying than length.

Miscellaneous Exercise

1. The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Sol. Let the remaining two observations are x and y.

Given,

$$\text{Given,\space} \bar x=9\space\text{and}\space\sigma^2=9.25\\\because\space\bar x = 9\\\Rarr\space\frac{\text{Sum of all observations}}{\text{Number of observation}}=9$$

⇒ Sum of all observations = 9 × 8

⇒ 6 + 7 + 10 + 12 + 12 + 13 + x + y = 72

⇒ 60 + x + y = 72

⇒ x + y = 72 – 60 …(i)

⇒ x + y = 12

$$\text{Again,\space}\sigma^2=\frac{\Sigma x_i^2}{n}-\bigg(\frac{\Sigma x_i}{n}\bigg)^2\\\text{or}\space\sigma^2=\frac{\Sigma x_i^2}{n}-(\bar x)^2\\\Rarr \space 9.25=\\\frac{36+49+100+144+144+169+x^2+y^2}{8}\\-(9)^2$$

$$\Rarr\space9.25=\frac{642+x^2+y^2}{8}-81\\\Rarr\space 9.25+81=\frac{642+x^2+y^2}{8} $$

⇒ 74 + 648 = 642 + x² + y²

⇒ 722 = 642 + x² + y²

⇒ 722 – 642 = x² + y²

⇒ x² + y² = 80 …(ii)

Now, put equation (i) put y = 12 – x in equation (ii),

x² + (12 – x)² = 80

⇒ x² + 144 + x² – 24x = 80

⇒ 2x² – 24x + 144 – 80 = 0

2x² – 24x + 64 = 0

⇒ x² – 12x + 31 = 0

⇒ x² – 8x – 4x + 32 = 0

⇒ x(x – 8) – 4(x – 8) = 0

⇒ (x – 4) (x – 8) = 0

⇒ x – 4 = 0 or x = 4

x – 8 = 0, x = 4

Put x = 4 and 8 in equation (i)

Then y = 8 and 4

Hence, observations are 4 and 8.

2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observation are 2, 4, 10, 12, 14. Find the remaining two observations.

Sol. Let the remaining two observations are x and y.

$$\text{Given\space} \bar x=8\space\text{and}\space\sigma^2=8\\\bar x=8$$

$$\frac{\text{Sum of all observations}}{\text{Number of observations}}=8\\\Rarr\space\frac{2+4+10+12+14+x+y}{7}=8$$

⇒ 42 + x + y = 56

⇒ x + y = 14 …(i)

$$\text{Again}\space \sigma^2=\frac{x_i^2}{n}-\bigg(\frac{\Sigma x_i}{n}\bigg)^2\\\Rarr\space \sigma^2=\frac{\Sigma x_i^2}{n}-(\bar{x})^2\\\Rarr\\\space 16=\frac{4+16+100+144+196+x^2+y^2}{n}-(8)^2\\\Rarr\space 16+64=\frac{460+x^2+y^2}{7}$$

⇒ 7 × (80) = 460 + x² + y²

⇒ x² + y² = 560 – 460

⇒ x² + y² = 100 …(ii)

Now, from equation (i), putting y = 14 – x in equation (ii),

x² + (14 – x²) = 100

⇒ x² + 196 + x² – 28x = 100

⇒ 2x² – 28x + 196 – 100 = 0

⇒ 2x² – 28x + 96 = 0

⇒ x² – 14x + 48 = 0

⇒ x² – 8x – 6x + 48 = 0

⇒ a(x – 8) – 6(x – 8) = 0

⇒ (x – 8) (x – 6) = 0

x – 8 = 0, x = 8

x – 6 = 0, x = 6

Put the value x cm in equation (i)

y = 6 and y = 8

Hence, the observations are 8 and 6.

3. The mean and standard deviation of six obseravtions are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Sol. Given,

$$\bar x=8\space\text{and}\space \sigma^2=4\space\text{and n=6}\\\text{Let the observation are}\\\space\bar x_1,\bar x_2,\bar x_3,\bar x_4,\bar x_5,\bar x_6\\\text{Then,}\\\text{Mean}\space\bar x=8\space\text{(Given)}\\\text{i.e.,}\space\text{Mean}=\frac{\bar x_1+\bar x_2+\bar x_3+\bar x_4+\bar x_5+\bar x_6}{6}=8$$

$$\Rarr\space \bar x_1,\bar x_2,\bar x_3,\bar x_4,\bar x_5,\bar x_6=48$$

Now, if and each observation is multiplied by 3, then mean is

$$3\bar x_1+3\bar x_2+3\bar x_3+3\bar x_4+3\bar x_5+\bar x_6\\=48×3$$

⇒ Σx_i = 144

$$\text{Now,}\space\text{new mean}\space\bar x=\frac{144}{6}=24$$

Since, Variance = 4² (Given)

$$\text{i.e}\space\frac{\Sigma x_i^2}{n}-\bigg(\frac{\Sigma x_i}{n}\bigg)^2=4^2\\\Rarr\space\frac{\Sigma x_i^2}{6}-(8)^2=4^2\\\Rarr\space\frac{\Sigma x^2}{6}=16+64$$

⇒ Σx² = 80 × 6

⇒ Σx² = 480 …(i)

$$\text{Now,}\qquad\text{New}\\\space\Sigma x^2=(3\bar x_1)^2+(3 \bar x_2)^2+(3 \bar x_3)^2+(3\bar x_4)^2+\\(3\bar x_5)^2+(3\bar x_6)$$

$$= 9(\bar x_1^2+\bar x_2^2+\bar x_3^2+\bar x_4^2+\bar x_5^2+\bar x_6^2)$$

= 9 × 480 [From equation (i)]

New Σx² = 4320

$$\therefore\space\text{New variance =}\space\frac{\text{(New)}\Sigma x^2}{n}-(\text{New mean})^2\\=\frac{4320}{6}-(24)^2$$

= 720 – 576 = 144

$$\therefore\space\text{New standard deviation }\\= \sqrt{\text{New variance}}\\\sqrt{144}$$

= 12.

$$\textbf{4. Given that}\space\bar x\space\textbf{is the mean and}\space\sigma^2\space\\ \textbf{is variance of n observations x}_1, \textbf{x}_2, …, \textbf{x}_\textbf{n.}\\\textbf{Prove that the mean and }\textbf{variance of the}\\\textbf{observations}\\\textbf{ax}_\textbf{1},\textbf{ax}_\textbf{2},\textbf{ax}_\textbf{3},...,\textbf{ax}_\textbf{n}\space\textbf{are}\\\textbf{a}\bar{x}\space\textbf{and a}^\textbf{2}\sigma^2,\textbf{respectively (a} ≠\textbf{ 0).}$$ Sol. Mean of ax₁, ax₂, … ax_n = $$\frac{ax_1+ax_2+...+ax_n}{n}\\=a\bigg(\frac{x_1+x_2+x_3+...+x_n}{n}\bigg)\\=a\bar x\space\bigg(\because\space\bar x=\frac{x_1+x_2+...+x_n}{n}\bigg)\\\text{Variance of ax}_1, ax_2, … ax_n\\=\frac{\Sigma(ax_i-a\bar x)^2}{n}\\=\frac{(ax_1-a\bar x)^2 + (ax_2-a\bar x)^2+...+ (ax_n-a\bar x)^2}{n}\\=\frac{[a^2(x_1-\bar x)^2+(x_2-\bar x)^2+...+(x_n-\bar x)^2]}{n}\\=a^2\frac{\Sigma(x_i-\bar x)^2}{n}=a^2\sigma^2$$

$$\bigg[\because\space \sigma^2=\frac{\Sigma(x_i-\bar x)^2}{n}\bigg]$$

5. The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted.

(ii) If it is replaced by 12.

Sol. (i) If wrong item is omitted.

$$\text{Given,}\space \bar x=10\space\text{and}\space\sigma=2,n=20\\\Rarr\space\frac{\Sigma x_i}{20}=10$$

⇒ Σx_i = 20 × 10

⇒ Σx_i = 200

If observation 8 is omitted, then

$$\Sigma x_i=200-8-192$$

Now, remaining number of observation = 19

$$\Rarr\space\text{Correct mean =}\frac{\Sigma x_i}{n}=\frac{192}{19}=10.10$$

Again, σ = 2

⇒ σ² = 4

$$\Rarr\space\frac{\Sigma x_i^2}{n}-(\bar x)^2=4\\\Rarr\space \frac{\Sigma x_i^2}{20}-(10)^2=4$$

⇒ Σx_i² = (4 + 100) × 20 = 104 × 20 = 2080

If observation 8 is omitted, then

Σx_i² = 2080 – 64 = 2016

$$\text{Now, correct standard deviation}\space\sigma=\\\sqrt{\frac{2016}{19}-\bigg(\frac{192}{19}\bigg)^2}\\=\sqrt{\frac{2016×19-(192)^2}{19×19}}\\=\frac{1}{19}\sqrt{38304-36864}\\=\frac{1}{19}\sqrt{1440}=\frac{3795}{19}=1.99$$

(ii) If it is placed by 12.

$$\bar x=10,\sigma=2, n=20\\\Rarr\space\frac{\Sigma x}{20}=10\\\Rarr\space\Sigma x=200 $$

If observation 8 is replaced by 12, then

Σx = 200 – 8 + 12 = 192 + 12 = 204

$$\text{Correct mean =}\frac{204}{20}\\=\frac{102}{10}=10.2$$

Again, σ = 2

⇒ σ² = 4

$$\Rarr\space\frac{\Sigma x^2}{n}-(\bar x)^2=4\\\Rarr\space \frac{\Sigma x^2}{20}-(10)^2=4$$

Σx² = 2080

If observation 8 is replaced by 12, then

Σx² = 2080 – (8)² + (12)²

= 2080 – 64 + 144

= 2224 –64

= 2160

$$\text{Now, correct standard deviation}\space\sigma=\\\sqrt{\frac{\Sigma x^2}{n}-(\bar x)^2}\\=\sqrt{\frac{2160}{20}-\bigg(\frac{204}{20}\bigg)^2}\\=\sqrt{\frac{2160×20-(204)^2}{20×20}}\\=\frac{1}{20}\sqrt{43200-41616}\\=\frac{1}{20}\sqrt{1584}=\frac{39.79}{20}=1.98$$

6. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

Subject	Mathematics	Physics	Chemistry
Mean Standard Deviation	42	32	40.9
Mean Standard Deviation	12	15	20

Which of these three subjects shows the highest variability in marks and which shows the lowest?

Sol. Here, n = 50

For mathematics,

$$\text{Coefficient of variation (CV) =}\frac{\sigma}{x}×100\\=\frac{12}{42}×100=\frac{1200}{42}=\frac{200}{7}\\\text{=28.57}\\\text{For Physics,}\space\text{CV}=\frac{\sigma}{x}×100\\=\frac{15}{32}×100\\=\frac{1500}{32}=46.87\space\text{...(ii)}$$

For Chemistry, CV =

$$\text{For Chemistry, CV = }\frac{\sigma}{x}×100\\=\frac{20}{40.9}×100=\frac{2000}{40.9}$$

= 48.89
…(iii)
From equations (i), (ii) and (iii), w e have

∵ CV of Chemistry > CV of Physcis > CV of Mathematics

∴ Chemistry shows the highest variability and Mathematics shows the lowest variability.

7. The mean and standard deviatoin of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation, if the incorrect observation and omitted.

$$\textbf{Sol.}\space\text{Given}\space n=100,\bar x=20,\sigma=3\\\because\space\bar x=3\\\Rarr\space\frac{\Sigma x}{100}=20$$

⇒ Σx_i = 100 × 20

⇒ Σx = 2000

Now, incorrect observations 21, 21 and 18 are omitted, then correct sum is

Σx_i = 2000 – 21 – 21 – 18 = 2000 – 60

= 1940

Now, correct mean of remaining 97 observations is

$$\bar x=\frac{1940}{97}=20$$

Again , σ = 3

$$\Rarr\space\sqrt{\frac{\Sigma x_i^2}{n}-(\bar x)^2}=3\\\Rarr\space \frac{\Sigma x_i^2}{100}-(20)^2=9\\\Rarr\space\frac{\Sigma x^2}{100}=9+400$$

= 409 × 100 = 40900

Now, correct Σx² is Σx² = 40900 – (21)² – (21)² – (18)²