Measure of Dispersion

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Why to measure dispersion?

• An average can represent a data-set only as best as a single value can, but

it certainly cannot reveal the entire story of the data-set. For example,

consider the following data sets:

i) 5, 5, 5, 5, 5; (mean = 5, median 5)

ii) 3, 4, 5, 6, 7: (mean = 5, median =5)

iii) 1, 3,5, 7, 9: (mean = 5. median =5).

• In all three data-sets above, the average values (mean and median) are the

same.

• What we observe here is, an average fails to give any idea about the spread

or scatteredness of the observations in the data-set.

• This scatteredness is also called dispersion.

• Some important measures of dispersion are range, mean deviation, and

standard deviation

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Introduction

• ‘Measures of dispersion’ measures the degree or extent of scatterness (or

spread or variation) of items about the central value data.

• It measures the way into which the data in distributed and thus facilitates to

compare two or more distributions to get a precise information about them.

• Dispersion can also be called as variability, scatter or spread.

• It helps to interpret the variation of the data from one another that is to know

how much homogenous or heterogeneous the data is, and gives a clear

picture about the distribution of the data.

• The main goal of measure of dispersion is to get to know how the data is

scattered.

• It illustrates the variation in data from their average value.

• In simple wards, it shows how squeezed or scattered the variable is. Thus,

dispersion is the extent to which a distribution is stretched or squeezed.

• Dispersion is different than the location or central tendency. Dispersion with

central tendency is the most used property of distributions.

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Measures of dispersion are needed basically

(i) to determine the reliability of data.

(ii) as it serves as a basis to control the variability.

(iii) to compare two or more series with respect to variability.

(iv) many statistical tools are based on it.

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Types of Measures of dispersion

Absolute measure of dispersion Relative measure of dispersion

• obtained as ratios or percentages of the

• contains the same unit as the original average.

data set. • also known as coefficient of dispersion.

• expressed in the unit of variable itself • used to evaluate the distribution of two or

like kilograms, rupees, centimeters, more data sets.

marks etc. • data values are compared without their units.

• states the variations as average of These are pure numbers or percentages totally

deviations of observations like standard independent of the units of measurements.

or means deviations. • e.g. coefficients of range, variation, SD.

• E.g. Range, SD, quartile deviation, etc. quartile deviation and mean deviation,

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respectively.

Various Measures of Dispersion

• The various measures of dispersion are:

1. The range

2. Quartile and semi-quartile range or interquartile range

3. Mean Deviation

4. Variance and Standard deviation (S.D.) and

5. Coefficient of variation (C.V.)

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1. The Range

Range is defined as the difference between two extremes observations

of the data set such as lowest and highest value.

It is the simplest measure of dispersion.

Symbolically,

Range = L-S

Where L = Largest item

S = Smallest item

The relative measure of range is called the coefficient of range and is

given by

𝐋−𝐒

Coefficient of range = www.DuloMix.com 7

𝐋+𝐒

• In case of continuous series the range can be found out by

(i) finding the difference between the upper limit of the highest class

and the lower limit of the lowest class.

(ii) finding out the difference between the mid-point of the highest

class interval and mid-point of the lowest class internal.

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Merits of Range:

1. It is the simplest form of the measure of dispersion

2. It is easy to calculate and to understand.

3. It is independent of change of origin.

Demerits of Range:

1. It is based on two extreme observations, hence get affected by

fluctuations.

2. Although range is simplest measure of dispersion it is not reliable.

3. It is dependent on change of scale.

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Exercise: The percentage of marks obtained by six students in English

are 20, 25, 80, 30, 90 and 45. Calculate the range and coefficient of

range.

Solution: Here L = 90 and S 20

Range = L- S = 90 – 20 = 70% mark

𝐋−𝐒 𝟗𝟎−𝟐𝟎 𝟕𝟎

Coefficient of Range = = =

𝐋+𝐒 𝟗𝟎+𝟐𝟎 𝟏𝟏𝟎

= 0.64

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Exercise: Find the range and coefficient of range from the data given

below:

Weights (lbs) 85-95 95-105 105-115 115-125 125-135

No. of persons 4 8 12 14 7

Solution :

Here, L= 135 and S= 85

Range = L –S = 135-85

= 50 lbs

𝐋−𝐒

Coefficient of range=

𝐋+𝐒

𝟏𝟑𝟓−𝟖𝟓 𝟓𝟎

= =

𝟏𝟑𝟓+𝟖𝟓 𝟐𝟐𝟎

= 0.227

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Example: Age of sample of 10 patients from a population of 169

patients is given in Table. Determine range and coefficient of range.

Patient No. 1 2 3 4 5 6 7 8 9 10

Age (Years) 42 28 28 61 31 23 50 34 32 37

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Standard Deviation

• Standard deviation is most widely used measure of dispersion of a

series and is commonly denoted by the symbol ‘𝜎′(pronounced as

sigma).

• Standard deviation is defined as the square-root of the average of

squares of deviations, when such deviations for the values of

individual items in a series are obtained from the arithmetic average.

• The square od SD is called variance.

• So SD = 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒

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• SD is a measure of the amount of variation or how spread out

numbers in a set of values.

• The low value of SD indicates that the values tend to be close to the

mean, means high degree of uniformity also called the ‘expected

value’ of the data set.

• The high value of SD indicates that the values are spread out over a

wider range.

• The SDs of a random variable, statistical population, data set and

probability distribution is the square root of its variance.

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For the ungrouped data x1, x2, x3 ,…… xn, the standard deviation is

1

s = Ʃ 𝑥 − 𝑥ҧ 2

𝑛

This formula can be simplified as

Ʃ𝑥2 Ʃ𝑥 2

s = −

𝑛 𝑛

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Example: The reactions (in mm) of tuberculin test of 10 boys are as follows:

8,3,7,5,8,11,10,9,7,12 Find the standard deviation of the data.

Ʃ𝑥 80

Solution : Mean 𝑥ҧ = = = 8 mm

𝑛 10

𝒙 𝒙 − 𝒙ഥ (𝒙𝒊 − 𝒙ഥ)2 1

s = Ʃ 𝑥 − 𝑥ҧ 2

𝑛 𝑖

8 0 0

3 -5 25

66

7 -1 1 =

10

5 -3 9

8 0 0

= 6.6 = 2.57mm

11 3 9

10 2 4

9 1 1

7 -1 1

12 4 16

total 66

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Another method

Total

𝒙 8 3 7 5 8 11 10 9 7 12 80

𝒙2 64 9 49 25 64 121 100 81 49 144 706

Ʃ𝑥2 Ʃ𝑥 2

s = −

𝑛 𝑛

706 80 2

= −

10 10

= 70.6 − 64

= 6.6

= 2.57

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Example: Calculate variance and S.D. of the following series.

18,20,22,27,21,29,27,29,28,29

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• When deviations are taken from assumed mean the fooling formula is

applied

Ʃ𝑥2 Ʃ𝑥 2 Ʃ𝑑2 Ʃ𝑑 2

s = − = s = −

𝑛 𝑛 𝑛 𝑛

Where , X-A = d And A = Assumed mean

Ʃ𝑓𝑥2 Ʃ𝑓𝑥 2 Ʃ𝑓𝑑2 Ʃ𝑓𝑑

s = − 2

= − × 𝑐

𝑛 𝑛 𝑛 𝑛

𝑥−𝐴

Where, d = , c= width of the class

𝑐

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Example : Compute the arithmetic mean and the standard deviations

for the following data.

Scores 4-5 6-7 8-9 10-11 12-13 14-15

Frequency 4 10 20 15 8 3

Solution: we construct the following table taking A= 8.5 and c =2

Class Frequency f Mid-value x 𝒙 −𝑨

d = fd fd2 Ʃ𝑓𝑑2 Ʃ𝑓𝑑 2

𝒄 S = − × c

𝑛 𝑛

4-5 4 4.5 -2 -8 16

6-7 10 6.5 -1 -10 10 100 22 2

= − × 2

8-9 20 8.5 0 0 0 60 60

10-11 15 10.5 1 15 15 = 1.667 − 0.1344 × 2

12-13 8 12.5 2 16 32

14-15 3 14.5 3 9 27 = 1.5323 × 2

Total 60 22 100 = 1.24 × 2

= 2.48

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Example: Find the standard deviation of the following distribution.

Age 20-25 25-30 30-35 35-40 40-45 45-50

No.of persons 170 110 80 45 40 35

𝒙 −𝑨

Solution : A =32.5 , c = 5 and d =

𝒄

Class Frequency f Mid-value x 𝒙 −𝑨

d = fd fd2 Ʃ𝑓𝑑2 Ʃ𝑓𝑑 2

𝒄 S = − × c

𝑛 𝑛

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Example: Calculate the mean and standard deviation of the following

table giving the age distribution of 542 members.

Age in years 20-30 30-40 40-50 50-60 60-70 70-80 80-90

No.of 3 61 132 153 140 51 2

members

𝒙 −𝑨

Solution : A = 55 , c = 10 and d =

𝒄

Class Frequency f Mid-value x 𝒙 −𝑨

d = fd fd2 Ʃ𝑓𝑑2 Ʃ𝑓𝑑 2

𝒄 S = − × c

𝑛 𝑛

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Example : A sample of 20 persons drawn from the same universe showed the

number of attacks of cold per person as follows:

7,2,5,0,1,5,5,6,7,6,2,6,9,4,8,8,6,7,9,3

Calculate mean , Standard deviation and standard error of the mean for above

data and comment on your results.

Solution: Let x denote the number of attacks of cold per person.

For the given data Ʃx = 106 and Ʃx2 = 690

Ʃ𝑥 106

Mean 𝑥ҧ = = = 5.3

𝑛 20

Ʃ𝑥2 Ʃ𝑥 𝑠

Standard deviation, s = − 2 Standard error of the mean =

𝑛 𝑛 𝑛

690 106

= − 2

2.53

20 20 =

20

= 34.5 − 28.09

= 6.41 = 0.5561

s = 2.53