# 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

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

= 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 | xi – 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.

5.

 Xi 5 10 15 20 25 fi 7 4 6 3 5

Sol.

 xi fi fixi $$|x_i-\bar x|$$ $$f_i|x_i-\bar x|$$ 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 Σfi=25 350 158

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

[N = Σfi = 25]

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

6.

 xi 10 30 50 70 90 fi 4 24 28 16 8

Sol.

 xi fi fixi $$|x_i-\bar x|$$ $$f_i|x_i-\bar x|$$ 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 Σfi=80 Σfixi=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.

7.

 xi 5 7 9 10 12 15 fi 8 6 2 2 2 6

Sol.

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

Here, N = Σfi = 26 (even)

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

 |xi-M| fi fi|xi- 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 Σfi|x-M|=84

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

8.

Sol.

 xi 15 21 27 30 35 fi 3 5 6 7 8
 xi fi CF 15 3 3 21 5 8 27 6 14 30 7 21 35 8 29 Total Σfi=29

Here, N = Σfi = 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

 |xi-M| fi fi|xi-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 Σfi|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 fi Mid value (xi) $$u_i=\frac{x_i-A}{h}\\\text{A = 350, h = 100}$$ fiui $$|x_i-\bar x|$$ $$f_i|x_i-\bar{x}|$$ 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 Σfi=50 Σfidi=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 fi Mid value (xi) $$u_i=\frac{x_i-A}{h}\\\text{A = 130, h = 10}$$ fiui $$|x_i-\bar x|$$ $$f_i|x_i-\bar x|$$ 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 fi CF Mid value xi $$|x_i-M|$$ $$f_i|x_i-M|$$ 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 Σfi = 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 fi CF Mid value xi $$|x_i-M|$$ $$f_i|x_i-M|$$ 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 Σfi|xi-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.

 xi $$(x_i-\bar x)$$ $$(x_i-\bar x)^2$$ 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.

 xi xi2 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.

 xi $$(x_i-\bar x)$$ $$(x_i-\bar x)^2$$ 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.

4.

 xi 6 10 14 18 24 28 30 fi 2 4 7 12 8 4 3

Sol.

 xi fi fixi xi2 fixi2 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 Σfi=40 Σfixi=760 Σfixi2=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

5.

 xi 92 93 97 98 102 104 109 fi 3 2 3 2 6 3 3

Sol.

 xi fi fixi xi2 fixi2 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 Σfi=22 Σfixi=2200 Σfixi2=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:

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

Sol.

 Mid-value xi Frequency fi Deviation from mean di = xi – A, A = 64 fidi di2 fidi2 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 Σfi=100 Σdi=0 Σfidi=0 Σfidi2=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.

7.

 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 fi Mid value xi Deviation from mean$$d_i=\frac{x_i-A}{h},\space\text{A=105}$$ fidi di2 fidi2 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 Σfidi=2 Σfidi2=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

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

8.

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

Sol.

 Class Frequency fi Mid value xi $$\text{Mid-value}\\d_i=\frac{x_i=25}{10}$$ fidi di2 fidi2 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 Σfi = 50 Σfidi=10 Σfidi2=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 fi Mid value xi $$\text{Mid-value}\\d_i=\frac{x_i-92.5}{5}$$ fidi di2 fidi2 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 Σfidi = 6 Σfidi2=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 fi Mid value xi $$\text{Deviation from mean}\\d_i=\frac{x_i-92.5}{5},\space\text{A=42.5, h=4}$$ fidi di2 fidi2 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 fi xi $$d_i=\frac{x_i-45}{10}$$ fidi di2 fidi2 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 Σfi = 150 Σfidi=-6 Σfidi2=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 fi xi $$d_i=\frac{x_i-45}{10}$$ fidi di2 fidi2 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 Σfidi=-6 Σfidi2=-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 di=X-A di2 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 Σdi=-10 Σdi2=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 Σdi=0 Σdi2 = 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 fixi xi2 fixi2 xi fi 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 Σfi=25 Σfixi=50 Σfixi2=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

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

⇒ Σxi = 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, … axn = $$\frac{ax_1+ax_2+...+ax_n}{n}\\=a\bigg(\frac{x_1+x_2+x_3+...+x_n}{n}\bigg)\\=a\bar x\space\bigg(\because\space\bar x=\frac{x_1+x_2+...+x_n}{n}\bigg)\\\text{Variance of ax}_1, ax_2, … ax_n\\=\frac{\Sigma(ax_i-a\bar x)^2}{n}\\=\frac{(ax_1-a\bar x)^2 + (ax_2-a\bar x)^2+...+ (ax_n-a\bar x)^2}{n}\\=\frac{[a^2(x_1-\bar x)^2+(x_2-\bar x)^2+...+(x_n-\bar x)^2]}{n}\\=a^2\frac{\Sigma(x_i-\bar x)^2}{n}=a^2\sigma^2$$

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

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.

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

⇒ Σxi = 20 × 10

⇒ Σxi = 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

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

⇒  Σxi2 = (4 + 100) × 20 = 104 × 20 = 2080

If observation 8 is omitted, then

Σxi2 = 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 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.

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

⇒ Σxi = 100 × 20

⇒ Σx = 2000

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

Σxi = 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

= 40900 – 441 – 441 – 324

= 40900 – 1206 = 39694

Now, correct SD for remaining 97 observations is

$$\sigma=\sqrt{\frac{39694}{97}-\bigg(\frac{1940}{97}\bigg)^2}\\\sqrt{409.2-(20)^2}\\=\sqrt{409.2-400}\\\sqrt{9.2}=3.03$$