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

Subject: Mathematics

#### Overview

This note gives ou a slight idea on Coefficient of Variation and Numerical problems related to measures of Dispersion.
##### Coefficient of Variation , Numerical Problems on Measures of Dispersion ( Range , Quartile Deviation , M.D., S.d., Variance)

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 )

##### Things to remember
• 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

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

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.