Notes on Coefficient of Variation , Numerical Problems on Measures of Dispersion ( Range , Quartile Deviation , M.D., S.d., Variance) | Grade 12 > Mathematics > Dispersion, correlation and regression | KULLABS.COM

Coefficient of Variation , Numerical Problems on Measures of Dispersion ( Range , Quartile Deviation , M.D., S.d., Variance)

  • Note
  • Things to remember

Coefficient of variation (C.V)

Standard deviation is the absolute measure of dispersion. The relative measure of dispersion based on the standard deviation is known as the coefficient of standard deviation. Symbolically it is defined as

$$Coefficient\;of\;S.d.=\frac{S.D}{Mean}=\frac{\sigma}{\overline{x}}$$

The coefficient of dispersion based on the standard deviation multiplied by 100 is known as the coefficient of variation (C.V.) If \(\overline{x}\) be the arithmetic mean and \(\sigma\), the standard deviation of the distribution, then C.V. is defined by

$$C.V.=\frac{\sigma}{\overline{x}}\times\;100$$

It is independent of the unit, so, two distributions can be compared with the help of C.V. for their variability. Less the C.V. more will be the uniformity, consistency etc. and more the C.v. less will be the uniformity, consistency etc.

Examples

Example 1 (Range):

Find the range and its coefficient for the following weight of patients coming to the hospital.

Weight on kg

10 – 20

20 – 30

30 – 40

40 – 50

50 – 100

Patients

2

3

5

4

2

Solution:

Range = L – S = 100 – 10 = 90 Kg and

$$Coefficient\;of\;range=\frac{L-S}{L+S}=\frac{90}{110}=\frac{9}{11}=0.818$$

Example 2 ( Quartile Deviation):

From the following table giving the heights of students, calculate the quartile deviation and the coefficient of Q.D.

Height (cm)

153

155

157

159

161

163

No of students

25

21

28

20

18

20

Solution:

Height

(cms.)

Frequency ( f )

C.f.

153

155

157

159

161

163

25

21

28

20

18

24

25

46

74

94

112

136

$$\frac{N+1}{4}=\frac{136+1}{4}=34.25$$

$$\frac{3(N+1)}{4}=\frac{3(136+1)}{4}=102.75$$

Q­1­ = Size of (34.25)th item = 155

Q­3­ = Size of (102.75)th item = 161

$$now,\;Quartile\;devation=\frac{Q_3-Q_1}{2}=\frac{161-155}{2}=3\;cms$$

$$And,\;Coeff.\;of\;Q.D.\;=\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{161-155}{161+155}=0.02$$

Example 3: (Quartile Deviation)

Below is a given data of income of workers in a factory

Income Rs.(‘000)

10 – 12

12 – 14

14 – 16

16 – 18

18 - 20

Workers

14

18

23

18

7

Calculate

  • First and third quartiles
  • Inter Quartile range and Quartile deviation
  • Coefficient of Q.D.

Solution:

First, we calculate Q­1­ and Q­3­ and then the required measures of dispersion:

Income Rs. (‘000)

Mid value Income

Workers

C.f.

10 – 12

12 – 14

14 – 16

16 – 18

18 – 20

11

13

15

17

19

14

18

23

18

7

14

32

55

73

80

$$Here,\;N=\Sigma\;f=80,So\;that\;Q_1\;and\;Q_3\;will\;lie\;in\;the\;interval\;12-14\;and\;16-18\;respectively\;and\;their\;values\;will\;be,$$

$$Q_1=L+\frac{\frac{N}{4}-c.f.}{f}\times\;h\;=12+\frac{60-55}{18}\times\;2=12.67$$

$$Q_3=L+\frac{\frac{3N}{4}-c.f.}{f}\times\;h\;=16+\frac{20-14}{18}\times\;2=16.56$$

$$Hence\;Interquartile\;range\;=Q_3-Q_1=16.56-12.67=3.89$$

$$now,\;Quartile\;devation=\frac{Q_3-Q_1}{2}=\frac{3.89}{2}=1.945$$

$$And,\;Coeff.\;of\;Q.D.\;=\frac{Q_3-Q_1}{Q_3+Q_1}=\frac{3.89}{29.23}=0.133$$

Example 4: (Mean Deviation from Mean )

Marks obtained by 25 students are given below. Find the mean deviation from the mean of the following data:

Marks

10

15

20

25

30

No of students

2

4

6

8

5

Solution:

Marks

(x)

f

fx

$$\vert\;x-\overline{x}(22)\vert$$

$$f\vert\;x-\overline{x}\vert$$

10

15

20

25

30

2

4

6

8

5

20

60

120

200

150

12

7

2

3

8

24

28

12

24

40

N=25

$$\Sigma\;fx=550$$

$$\Sigma\;f\vert\;x-\overline{x}\vert=128$$

$$Here,\;\Sigma\;fx=550,\;N=25,\;\overline{x}=?$$

$$Now,\;Mean\;=\;\overline{x}=\frac{\Sigma\;fx}{25}=22$$

$$Again,\;N=25,\;\Sigma\;f\vert\;x-\overline{x}\vert=128$$

Mean Deviation from mean = ?

$$M.D\;from\;mean\;=\frac{\Sigma\;f\vert\;x-\overline{x}\vert}{N}$$

$$=\frac{128}{25}=5.12$$

Example 5: (Mean Deviation From median )

The following table gives the marks obtained by 50 students in a certain examination. Calculate the median and mean deviation from median of the following data .

Marks

0-10

10-20

20-30

30-40

40-50

No. of students

5

8

15

16

8

Solution:

$$x=mid\;value,\;d=x-M_d$$

Computation of mean deviation:

Marks

x

f

C.f.

$$\vert\;d\vert$$

$$f\vert\;d\vert$$

0-10

10-20

20-30

30-40

40-50

5

15

25

30

35

5

8

15

16

6

5

13

28

44

50

23

13

3

7

17

115

104

45

112

102

N = 50

$$\Sigma\;f\vert\;d\vert=478$$

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

$$\therefore\;Median\;lies\;in\;the\;class\’20-30$$

$$Median=M_d=L+\frac{\frac{N}{2}-c.f.}{f}\times\;h$$

$$=20+\frac{25-13}{15}\times=28$$

$$M.D.\;from\;median=\frac{\Sigma\;f\vert\;d\vert}{N}=\frac{478}{50}$$

$$=9.56\;marks$$

Example 6 : ( Standard Deviation)

Find out the mean and Standard deviation from the following data:

X:

10

11

12

13

14

F:

3

12

18

12

2

Solution:

Computation of mean and S.D.

x

f

d = x -\(\overline{x}\)

fd

fd2

10

11

12

13

14

3

12

18

12

3

-2

-1

0

1

2

-6

-12

0

12

6

12

12

0

12

12

N=48

\(\Sigma\)fd=0

\(\Sigma\)fd2=48

$$Mean\;=\overline{x}=a+\frac{\Sigma\;fd}{N}=12+\frac{0}{48}=12$$

$$S.D.=\sigma=\sqrt{\frac{\Sigma\;d^2}{n}-\bigg(\frac{d}{n}\bigg)^2}$$

$$\;=\sqrt{\frac{48}{48}-0}=1$$

Example 7: (Coefficient of Varitation)

$$If\;n=10,\;\Sigma\;x=120,\;\Sigma\;x^2=1530,\;find\;the\;standard\;deviation\;and\;the\;coefficient\;of\;variation.$$

Solution:

$$Here,\;Mean=\overline{x}=\frac{\Sigma\;x}{n}=\frac{120}{10}=12$$

$$Standard\;deviation\;\sigma\;=\;\sqrt{\frac{\Sigma(x-\overline{x})^2}{n}}=\sqrt{\frac{\Sigma\;x^2}{n}-\bigg(\frac{x}{n}\bigg)^2}$$

$$\;=\sqrt{\frac{1530}{10}-\bigg(\frac{120}{10}\bigg)^2}=\sqrt{153—144}=3$$

$$Now,\;C.V.=\frac{\sigma}{\overline{x}}\times\;100$$

$$=\frac{3}{12}\times\;100=25%$$

Example 8:

Scores of two golfers for 6 rounds were as follows:

Golfer A

74

75

80

78

72

77

Golfer B

86

84

80

88

87

85

Find which golfer may be considered a more consistent player.

Golfer A

Golfer B

x

d=x-77(mean)

d2

x

d=x-86

d2

74

75

80

78

72

77

-3

-2

3

1

-5

0

9

4

9

1

25

0

86

84

80

88

87

85

0

-2

-6

2

1

-1

0

4

36

4

1

1

\(\Sigma\)d=-6

\(\Sigma\)d2=48

\(\Sigma\)d=-6

\(\Sigma\)d2=46

For Golfer A:

Here, a = 77, n = 6, \(\Sigma\)d=-6, \(\Sigma\)d2=48

$$Mean\;=\overline{x}=a+\frac{\Sigma\;fd}{N}=77+\frac{-6}{6}=76$$

$$S.D.=\sigma=\sqrt{\frac{\Sigma\;d^2}{n}-\bigg(\frac{d}{n}\bigg)^2}$$

$$\;=\sqrt{\frac{48}{6}-\bigg(\frac{-6}{6}\bigg)^2}=\sqrt{7}=2.64$$

$$Now,\;C.V.(A)=\frac{\sigma}{\overline{x}}\times\;100$$

$$=\frac{2.64}{76}\times\;100=3.47%$$

For Golfer B:

Here, a = 86, n = 6, \(\Sigma\)d=-6, \(\Sigma\)d2=46

$$Mean\;=\overline{x}=a+\frac{\Sigma\;fd}{N}=86+\frac{-6}{6}=85$$

$$S.D.=\sigma=\sqrt{\frac{\Sigma\;d^2}{n}-\bigg(\frac{d}{n}\bigg)^2}$$

$$\;=\sqrt{\frac{46}{6}-\bigg(\frac{-6}{6}\bigg)^2}=\sqrt{6.67}=2.58$$

$$Now,\;C.V.(B)=\frac{\sigma}{\overline{x}}\times\;100$$

$$=\frac{2.58}{85}\times\;100=3.03%$$

Since C.V.(B)<C.V.(A). So, B is a Consistent player.

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )



  • Standard deviation is the absolute measure of dispersion. The relative measure of dispersion based on the standard deviation is known as the coefficient of standard deviation. Symbolically it is defined as

    $$Coefficient\;of\;S.d.=\frac{S.D}{Mean}=\frac{\sigma}{\overline{x}}$$

.

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