Well-structured S Chand ISC Maths Class 11 Solutions Chapter 21 Measures of Dispersion Ex 21(a) facilitate a deeper understanding of mathematical principles.
S Chand Class 11 ICSE Maths Solutions Chapter 21 Measures of Dispersion Ex 21(a)
Question 1.
15, 17, 19, 25, 30, 35, 48
Solution:
The table of values is given as under :
| xi | di = xi – \(\bar{x}\) | | di | |
| 15 | – 12 | 12 |
| 17 | – 10 | 10 |
| 19 | -8 | 8 |
| 25 | -2 | 2 |
| 30 | 3 | 3 |
| 35 | 8 | 8 |
| 48 | 21 | 21 |
| Σxi = 189 | Σ | di | = 64 |
Question 2.
21, 23, 25, 28,30, 32, 38, 39, 46, 48
Solution:
Here Mean \(\vec{x}\) = \(\frac{21+23+25+28+30+32+38+39+46+48}{10}\) = \(\frac{330}{10}\) = 3
The table of values is given as under:
| xi | xi = xi – \(\bar{x}\) | | di | |
| 21 | – 12 | 12 |
| 23 | – 10 | 10 |
| 25 | -8 | 8 |
| 28 | -5 | 5 |
| 30 | -3 | 3 |
| 32 | – 1 | 1 |
| 38 | 5 | 5 |
| 39 | 6 | 6 |
| 46 | 13 | 13 |
| 48 | 15 | 15 |
| Σ | di | = 78 |
∴ M.D. about Mean = \(\frac{\Sigma\left|d_i\right|}{n}\) = \(\frac{78}{10}\) = 7.8 and coeff. of M.D. = \(\frac{\text { M.D }}{\bar{x}}\) = \(\frac{7.8}{33}\) = 0.236
Question 3.
10, 70, 50, 53, 20, 95, 55, 42, 60, 48, 80
Calculate the mean deviation from the mean for the following frequency distributions.
Solution:
Mean \(\bar{x}\) = \(\frac{10+70+50+53+20+95+55+42+60+48+80}{11}\) = \(\frac { 583 }{ 11 }\) = 53
| xi | 10 | 70 | 50 | 53 | 20 | 95 | 55 | 42 | 60 | 48 | 80 | |
| xi – \(\bar{x}\) = di | -43 | 17 | -3 | 0 | -33 | 42 | 2 | – 11 | 7 | -5 | 27 | |
| | di | | 43 | 17 | 3 | 0 | 33 | 42 | 2 | 11 | 7 | 5 | 27 | Σ | di | = 190 |
∴ required M.D about Mean = \(\frac{\Sigma\left|d_i\right|}{n}\) = \(\frac{190}{11}\) = 17.27 and coeff. of M.D = \(\frac{\text { M.D }}{\text { Mean }}\) = \(\frac{190}{11 \times 53}\) = 0.326
Question 4.
| xi | 3 | 9 | 17 | 23 | 27 |
| fi | 8 | 10 | 12 | 9 | 5 |
Solution:
| xi | fi | fixi | | di | = | xi – \(\bar{x}\) | | fi | di | |
| 3 | 8 | 24 | 12 | 96 |
| 9 | 10 | 90 | 6 | 60 |
| 17 | 12 | 204 | 2 | 24 |
| 23 | 9 | 207 | 8 | 72 |
| 27 | 5 | 135 | 16 | 60 |
| Σfi = 44 | Σ fixi = 660 | Σ| di | = 40 | Σfi | di | = 312 |
∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{660}{44}\) = 15
Thus M.D about Mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{312}{44}\) = 7.09
Question 5.
| xi | 10 | 11 | 12 | 13 | 14 |
| fi | 3 | 12 | 18 | 12 | 3 |
Solution:
The table of values is given as under :
| xi | fi | di xi | | di | = | xi – \(\bar{x}\) | | fi | di | |
| 10 | 3 | 30 | 2 | 6 |
| 11 | 12 | 132 | 1 | 12 |
| 12 | 18 | 216 | 0 | 0 |
| 13 | 12 | 156 | 1 | 12 |
| 14 | 3 | 42 | 2 | 6 |
| Σfi = 48 | Σfixi = 576 | Σ fi | di | = 36 |
∴ \(\bar{x}\) = \(\frac { 576 }{ 48 }\) = 12
∴ M.D about Mean = \(\frac{\Sigma f_i\left|d_i\right|}{48}\) = \(\frac{36}{48}\) = \(\frac{3}{4}\) = 0.75
Question 6.
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| No. of students | 5 | 8 | 15 | 16 | 6 |
Solution:
We construct the table of values is given as under :
| Marks | Class Mark | fi | fixi | | xi – \(\bar{x}\) | | fi | xi – \(\bar{x}\) | |
| 0-10 | 5 | 5 | 25 | 22 | 110 |
| 10-20 | 15 | 8 | 120 | 12 | 96 |
| 20-30 | 25 | 15 | 375 | 2 | 30 |
| 30-40 | 35 | 16 | 560 | 8 | 128 |
| 40-50 | 45 | 6 | 270 | 18 | 108 |
| Σfi = 5o | Σfixi = 1350 | Σ fi | xi – \(\bar{x}\) | = 472 |
∴ by direct method, mean \(\bar{x}\) = \(\frac{\sum f_i x_i}{\sum f_i}\) = \(\frac{1350}{50}\) = 27
Thus, M.D from mean = \(\frac{\Sigma f_i\left|x_i-\bar{x}\right|}{\Sigma f_i}\) = \(\frac{472}{50}\) = 9.44
Question 7.
| Scores | 140-150 | 150-160 | 160-170 | 170-180 | 180-190 | 190-200 |
| NO. of students | 4 | 6 | 10 | 18 | 9 | 3 |
Solution:
The table of values is given as under:
| Scores | No. of students fi | xi | fixi | | di | = | xi – \(\bar{x}\) | | fi | di | |
| 140-150 | 4 | 145 | 580 | 26.2 | 104.8 |
| 150-160 | 6 | 155 | 930 | 16.2 | 97.2 |
| 160-170 | 10 | 165 | 1650 | 6.2 | 62 |
| 170-180 | 18 | 175 | 3150 | 3.8 | 68.4 |
| 180-190 | 9 | 185 | 1665 | 13.8 | 124.2 |
| 190-200 | 3 | 195 | 585 | 23.8 | 71.4 |
| Σfi = 50 | Σfixi = 8560 | Σfi | di | = 528 |
∴ Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{8560}{50}\) = 171.2
Thus M.D about mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{528}{50}\) = 10.56
Question 8.
| Class Interval | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
| Frequency | 3 | 50 | 84 | 32 | 10 | 3 |
Solution:
| Class | Frequency | Mid-Marks | fixi | | di | = | xi – \(\bar{x}\) | | fi | di | |
| 0-20 | 3 | 10 | 30 | 40.55 | 121.65 |
| 20-40 | 50 | 30 | 1500 | 20.55 | 1027.50 |
| 40-60 | 84 | 50 | 4200 | 0.55 | 46.2 |
| 60-80 | 32 | 70 | 2240 | 19.45 | 622.40 |
| 80-100 | 10 | 90 | 900 | 39.45 | 394.5 |
| 100- 120 | 3 | 110 | 330 | 59.45 | 178.35 |
| Σfi = 182 | Σfixi = 9200 | Σfi | di | = 2390.6 |
∴ Mean \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\) = \(\frac{9200}{182}\) = 50.55
∴ M.D about mean = \(\frac{\Sigma f_i\left|d_i\right|}{\Sigma f_i}\) = \(\frac{2390.6}{182}\) = 13.135
Question 9.
3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21
Solution:
Arranging the data in ascending order, we get 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
Here no. of observations = n = 11 (odd)
∴ \(\mathrm{M}_d=\left(\frac{n+1}{2}\right) \mathrm{th}\) observation = 6th observation = 9
| xi | 3 | 3 | 4 | 5 | 7 | 9 | 10 | 12 | 18 | 19 | 21 | |
| | xi – Md | | + 6 | 6 | 5 | 4 | 2 | 0 | 1 | 3 | 9 | 10 | 12 | Σ | xi – Md | = 58 |
∴ M.D about median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{58}{11}\) = 5.273
and Coefficient of M.D = \(\frac{\text { M.D }}{\text { Median }}\) = \(\frac{58}{11 \times 9}\) = 0.586
Question 10.
100, 150, 200, 250,360, 490, 500, 600, 671
Solution:
Data given is already in ascending order we have, no. of observations = n = 9 (odd)
∴ \(\mathrm{M}_d=\left(\frac{n+1}{2}\right) \mathrm{th}\) observation = 5th observation = 360
| xi | 100 | 150 | 200 | 250 | 360 | 490 | 500 | 600 | 671 | |
| | xi – Md | | 260 | 210 | 160 | 110 | 0 | 130 | 140 | 240 | 311 | Σ | xi – Md | = 58 |
∴ M.D about median = \(\frac{\Sigma\left|x_i-\mathrm{M}_d\right|}{n}\) = \(\frac{1561}{9}\) = 173.44
and Coeff. of M.D = \(\frac{\text { M.D }}{\text { median }}\) = \(\frac{1561}{9 \times 360}\) = 0.48
Question 11.
| x | 10 | 11 | 12 | 13 | 14 |
| f | 3 | 12 | 18 | 12 | 3 |
Solution:
The table of values is given as under :
| xi | fi | Cumulative freq. | | xi – Md | | fi | xi – Md | |
| 10 | 3 | 3 | 2 | 6 |
| 11 | 12 | 15 | 1 | 12 |
| 12 | 18 | 33 | 0 | 0 |
| 13 | 12 | 45 | 1 | 12 |
| 14 | 3 | 48 | 2 | 6 |
| Σ fi = n = 48 | Σ fi | xi – Md | = 36 |
Question 12.
| x | 3 | 6 | 9 | 12 | 13 | 15 | 21 | 22 |
| f | 3 | 4 | 5 | 2 | 4 | 5 | 4 | 3 |
Solution:
The table of values is given as under :
| x | f | c.f | | x – Md | | f | x – Md | |
| 3 | 3 | 3 | 10 | 30 |
| 6 | 4 | 7 | 7 | 28 |
| 9 | 5 | 12 | 4 | 20 |
| 12 | 2 | 14 | 1 | 2 |
| 13 | 4 | 18 | 0 | 0 |
| 15 | 5 | 23 | 2 | 10 |
| 21 | 4 | 27 | 8 | 32 |
| 22 | 3 | 30 | 9 | 27 |
| Σf = 30 | Σf| x – Md |= 149 |
