MEASURES OF DISPERSION

MR.
MUKESH SOLANKI
INTRODUCTION
 Measures of central tendency helps us to find a representative value
of a data. This value represents the entire data.
 Measures of central tendency have limitations that these do not
provide the complete details of information of the data.
 The actual value differ from the average value. The average value
does not revel the variability present in the data.
 For measuring the variability in the data we have to use dispersion as
statistical technique to be implemented.
MEASURES OF DISPERSION:
 Dispersion is the state of getting dispersed or spread.
 Statistical dispersion means the extent to which a numerical data
is likely to vary about an average value.
 In other words, dispersion helps to understand the distribution of
the data.
 In statistics, the measures of dispersion help to interpret the
variability of data i.e. to know how much homogenous or
heterogeneous the data is.
 In simple terms, it shows how squeezed or scattered the variable
is.
 UNDERSTANDING DISPERSION:

S. No Family A Family B Family C

 Do You notice:
1 12,000 7,000 0
 The average income of all the families
2 14,000 10,000 7,000
are same.
3 16,000 14,000 8,000
 There are considerable differences in
4 18,000 17,000 10,000
individual income.
5 ----- 20,000 50,000
 To analyse the data when the 6 ------ 22,000 --------
distribution is heterogenous, central
Total 60,000 90,000 75,000
tendency is not an appropriate measure
Income
of data analysis.
Average 15,000 15,000 15,000
 We have to use measure of dispersion. Income
OBJECTIVES OF MEASURES OF DISPERSION:
 To reveal the variability present in the data, we need to know the
dispersion (also known as spread or variation) of values.
 An average is more meaningful when expressed in the light of
dispersion.
➢ Objectives of having measures of dispersion are:
 To check the reliability of average.
 To know the extant od variability.
 To compare dispersion of two or more set of data or series.
 To facilitate statistical analysis.
CHARACTERISTICS OF A GOOD MEASURE OF DISPERSION:

➢ The main characteristics are:

 It should be easy to calculate & simple to understand.
 It should be based on all the observations of the series.
 It should be rigidly defined.
 It should not be affected by extreme values.
 It should not be unduly affected by sampling fluctuations.
 It should be capable of further mathematical treatment and statistical
analysis.
 MEASURES OF DISPERSION

Measures of
Dispersion

Absolute Relative
TYPES OF MEASURES OF DISPERSION:
 Absolute Measures
 Absolute measures of dispersion are expressed in the unit of Variable
itself.
 Like, Kilograms, Rupees, Centimeters, Marks etc.
 Relative Measures
 Relative measures of dispersion are obtained as ratios or percentages
of the average.
 These are also known as ‘Coefficient of dispersion.’
 These are pure numbers or percentages totally independent of the
units of measurements.
MEASURES OF DISPERSION:
 RANGE:
 It is the absolute measure of dispersion.
 It is simplest measure of dispersion.
 It is defined as the difference between the largest value (L) and the smallest
value in a distribution.
 Symbolically:
R=L–S
✓ Where
✓ R = Range
✓ L = Largest value in a distribution
✓ S = Smallest value in a distribution
➢ Implication: Higher value of range implies higher dispersion and vice versa.
RANGE:
 Range (R) is the difference between the largest value (L) and the
smallest value (S) in a distribution.
 Example:
 Consider the following values: 20, 30, 40, 50, 200. Calculate the Range
and Coefficient of Range
 Range = L - S
 = 200 - 20
 Range = 180
PRACTICE EXAMPLE:
➢ The following students have scored in economics as under. Find out the
range and coefficient of range:
Student’s S.No. 1 2 3 4 5 6 7
Marks Scored 20 50 75 25 80 40 60

Solution:
 Range = L - S
 = 80 – 20
 = 60
RANGE IN CASE OF DISCRETE SERIES (GROUPED DATA):
➢ Calculate the Range and Coefficient of Range:

Income (in ₹) 100 200 300 400

No. of persons 2 4 10 4

Solution:
 Range = L - S
 = 400 – 100
 = 300
 Range = ₹ 300
RANGE IN CASE OF DISCRETE SERIES (GROUPED DATA):
➢ Do it yourself:
➢ From the following data, calculate range and coefficient of range:

Marks 10 20 30 40 50 60 70

No. of students 8 12 7 30 10 5 2

Solution:
 Range = L - S
 = 70 – 10
 = 60
 Range = 60
RANGE IN CASE OF CONTINUOUS SERIES (GROUPED DATA):
➢ From the following data, calculate range and coefficient of range:

Daily wages
200-250 250-300 300-350 350-400 400-450 450-500 500-550 550-600
(in ₹)
No. of 14 28 33 30 20 15 13 7
workers
Solution:
 Range = L - S
 = 600 – 200
 = 400
 Range = ₹ 400
RANGE IN CASE OF CONTINUOUS SERIES (GROUPED DATA):
➢ Do it yourself: Find out range and coefficient of range of the following series:
➢ From the following data, calculate range and coefficient of range:

Size 5-10 10-15 15-20 20-25 25-30

Frequency 4 9 15 30 40

Solution:
 Range = L - S
 = 30 – 5
 = 25
 Range = 25
 RANGE AS A MEASURE OF DISPERSION:
 Range is unduly affected by extreme values:
 As long as the minimum and maximum values remain unaltered, any
change in other values does not affect range. Thus, it remains unchanged.
 Range cannot be calculated for open-ended frequency distribution:
 Open-ended distributions are those in which either the lower limit of the
lowest class or the upper limit of the highest class or both are not
specified.
 Range is understood and used frequently because of its simplicity. Range
is the simplest to calculate and understand.
 QUARTILES AND QUARTILE DEVIATION :
 Quartiles are the measures which divide the data into four equal parts,
each portion contains equal number of observations.
 There are three quartiles:
 The first Quartile (denoted by Q1) or lower quartile: It has 25% of the items
of the distribution below it and 75% of the items are greater than it.
 The second Quartile (denoted by Q2) or median: It has 50% of items below
it and 50% of the observations above it.
 The third Quartile (denoted by Q3) or upper Quartile: It has 75% of the
items of the distribution below it and 25% of the items above it.
 Thus, Q1 and Q3 denote the two limits within which central 50% of the
data lies.
QUARTILE IN CASE OF UNGROUPED DATA/INDIVIDUAL SERIES:

 Lower Quartile (Q1):

 It covers the lower 25% of the data.
 Formula:

𝐍+𝟏 𝒕𝒉
 Lower Quartile (Q1) = size of size of item
𝟒

 As we know quartile divide the entire distribution in 4 equal part.

 Q1 = Lower quartile
 N = number of observation
QUARTILE IN CASE OF UNGROUPED DATA/INDIVIDUAL SERIES:

 Upper Quartile (Q3):

 It covers the upper 25% of the data.
 Formula:

𝟑(𝐍+𝟏) 𝒕𝒉
 Lower Quartile (Q3) = size of size of item
𝟒

 As we know quartile divide the entire distribution in 4 equal part.

 Q3 = Upper quartile
 N = number of observation
QUARTILE IN CASE OF UNGROUPED DATA/INDIVIDUAL SERIES:
 Internal Quartile Range:
 Inter-Quartile is the absolute measure of dispersion. I.Q.R (Internal Quartile
Range) is the difference between upper quirtile and the lover quartile.
 Formula:
I.Q.R = Q3 - Q1
 Quartile deviation: It is the average if IQR. It is also known as semi inter-
quartile range.
 Symbolically:
𝑄3 −𝑄𝟏
Q.D. =
2
CALCULATION: LOWER AND UPPER QUARTILES:
 Example:
 Calculate the lower and upper quartile from the data of the marks
obtained by students in an examination. 35, 39, 41, 30, 29, 25, 48, 51,
60, 20, 70
 Solution:
 Arranging the data in an ascending order: 20, 25, 29, 30, 35, 39, 41, 48, 51, 60, 70
𝐍+𝟏 𝒕𝒉 𝟑(𝐍+𝟏) 𝒕𝒉
Q1 = Size of size of items Q3 = size of size of item
𝟒 𝟒
𝟏𝟏+𝟏 𝒕𝒉 𝟑(𝟏𝟏+𝟏) 𝒕𝒉
= items = 𝟒
items
𝟒
𝟑(𝟏𝟐)
= 12/4th item = 𝒕𝒉 𝒊𝒕𝒆𝒎
𝟒
= 3rd item = 36/4th item = 9th item
= 29 = 51
 CALCULATION: LOWER AND UPPER QUARTILES, INTER-QUARTILE RANGE:
 Example: Calculate lower and upper quartiles: 22, 26, 14, 30, 18, 11,
35, 41, 12, 32
 Solution: Arranging the data in an ascending order: 11, 12, 14, 18, 22, 26, 30, 32,
35, 41 𝟑(𝐍+𝟏) 𝒕𝒉
Q3 = size of size of item
𝐍+𝟏 𝒕𝒉 𝟒
Q1 = Size of size of items 𝟑(𝟏𝟎 + 𝟏) 𝒕𝒉
𝟒 = 𝑺𝒊𝒛𝒆 𝒐𝒇 𝟒
items
𝟏𝟎 + 𝟏 𝒕𝒉
= 𝐒𝐢𝐳𝐞 𝐨𝐟 items = Size of
𝟑(𝟏𝟏)
𝒕𝒉 𝒊𝒕𝒆𝒎
𝟒 𝟒
= 11/4th item = 33/4th item
= size of 2,75th item
= 2nd item + 0.75 (3rd item – 2nd item) = size of 8.25th item
= 12 + 0.75 (14 –12) = 8th item + 0.25 (9th item – 8th item)
= 12 + 1.5 = = 32 + 0.25 (35 –32) = 32 + 0.75
= 13.5 marks = 32.75 marks
 CALCULATION: LOWER AND UPPER QUARTILES, INTER-QUARTILE RANGE:
 Example:
𝟑(𝐍+𝟏) 𝒕𝒉
𝐍+𝟏 𝒕𝒉 Q3 = size of size of item
Q1 = Size of size of items 𝟒
𝟒 𝟑(𝟏𝟎 + 𝟏) 𝒕𝒉
𝟏𝟎 + 𝟏 𝒕𝒉 = 𝑺𝒊𝒛𝒆 𝒐𝒇 items
= 𝐒𝐢𝐳𝐞 𝐨𝐟 items 𝟒
𝟒 𝟑(𝟏𝟏)
= 11/4th item = Size of 𝒕𝒉 𝒊𝒕𝒆𝒎
𝟒
= size of 2.75th item = 33/4th item
= 2nd item + 0.75 (3rd item – 2nd item) = size of 8.25th item
= 12 + 0.75 (14 –12) = 8th item + 0.25 (9th item – 8th item)
= 12 + 1.5 = 32 + 0.25 (35 –32) = 32 + 0.75
= 13.5 marks = 32.75 marks

I.Q.R = Upper quartile (Q3) – Lower Quartile (Q1)

= 32.75 – 13.5
= 19.25
 CALCULATION: LOWER AND UPPER QUARTILES, INTER-QUARTILE RANGE &
QUARTILE DEVIATION:
 Example:
 The following data shows monthly wages of 10 workers, calculate lower and upper quartiles
monthly wages and quartile deviation:
 320, 400, 450, 530, 550, 580, 600, 610, 700, 780, 800
 Solution:
 Hint,
𝐍+𝟏 𝒕𝒉
 Q1 = Size of size of item.
𝟒
𝟑(𝐍+𝟏) 𝒕𝒉
 Q3 = size of size of item.
𝟒
 I.Q.R = 𝑄3 − 𝑄1
𝑄3 −𝑄1
 Q.D =
2
 CALCULATION: LOWER AND UPPER QUARTILES, INTER-QUARTILE RANGE &
QUARTILE DEVIATION:
 Solution:
𝐍+𝟏 𝒕𝒉 𝟑(𝐍+𝟏) 𝒕𝒉
Q1 = Size of size of item Q3 = size of size of item
𝟒 𝟒
𝟏𝟎 + 𝟏 𝒕𝒉 𝟑(𝟏𝟎 + 𝟏) 𝒕𝒉
= 𝐒𝐢𝐳𝐞 𝐨𝐟 items = 𝑺𝒊𝒛𝒆 𝒐𝒇
𝟒
items
𝟒
= 11/4th item 𝟑(𝟏𝟏)
= Size of 𝒕𝒉 𝒊𝒕𝒆𝒎
𝟒
= size of 2.75th item
= 33/4th item
= 2nd item + 0.75 (3rd item – 2nd item)
= 400 + 0.75 (450 – 400) = size of 8.25th item
= 400 + 37.5 = 8th item + 0.25 (9th item – 8th item)
= 437.5 marks = 610 + 0.25 (700 – 610) = 610 + 22.5
= 632.5 marks

I.Q.R = Q3 - Q1 Q3 − Q1
Q.D =
= 632.5 – 437.5 𝟐
= 195/2
= 195
= 97.5
QUARTILES IN GROUPED DATA:
 In case of discrete series
 Formula:
𝐍+𝟏 𝒕𝒉
 Q1 = Size of size of item.
𝟒

𝟑(𝐍+𝟏) 𝒕𝒉
 Q3 = size of size of item.
𝟒
 I.Q.R = Q3 - Q1
Q3 − Q1
 Q.D =
𝟐
DISCRETE SERIES (GROUPED DATA) :
 The frequency distribution of the number of persons and their respective
incomes (in ₹) are given below.
 (a) Calculate the lower and upper quartile values. Also, interpret
the result.
 (b) Calculate quartile deviation.

Income (in ₹) 100 200 300 400

Number of persons 2 4 10 4
 𝐍+𝟏 𝒕𝒉
 Q1= located in item.
SOLUTION: 𝟒
 21/4th item
 = 5.25 item located in
Income  Q1 = 200
(in ₹) f Cf. 𝟑(𝐍+𝟏) 𝒕𝒉
 Q3 = located in item.
𝟒

100 2 𝟑(𝟐𝟎+𝟏) 𝒕𝒉
2  = item
𝟒
200 4  = 15.75th item
6 Q3 − Q 1
 Q3 = 300
 Q.D =
300 10 16
𝟐
 I.Q.R = Q3 – Q1  = 300 − 𝟐𝟎𝟎
400 4 20 𝟐
 300 - 200  = 100/2
N = ∑f = 20  100  = 50
DISCRETE SERIES (GROUPED DATA) :
 Do it by yourself:
 Calculate Q1, Q3, Quartile Deviation and Coefficient of Quartile Deviation
from the following data.
 (a) Calculate the Q1 and Q3 values.
 (b) Calculate quartile deviation.

Marks 10 20 30 40 50 60

No. of students 4 10 20 8 6 3
 𝐍+𝟏 𝒕𝒉
 Q1= located in item.
SOLUTION: 𝟒
 52/4th item

No. of  = 13th item located in

Marks  Q1 = 20 marks
students Cf.
𝟑(𝐍+𝟏) 𝒕𝒉
10 4 4  Q3 = located in item.
𝟒

20 10 14 𝟑(𝟓𝟏+𝟏) 𝒕𝒉
 = item
𝟒
30 20 34  = 39th item
42  Q3 = 40 marks
40 8 Q3 − Q1
 Q.D =
50 6 48  I.Q.R = Q3 – Q1 𝟐
𝟐𝟎
60 3 51
 =
 40 - 20 𝟐
N = ∑f = 51  = 10
 20
 QUARTILES IN GROUPED DATA:
 In case of continuous series
 I.Q.R = Q3 - Q1
 Formula:
Q3 − Q1
𝐍 𝒕𝒉  Q.D =
 1st : Q1 = Size of size of item. 𝟐
𝟒
 Quartile class obtained by formula
𝐍
− 𝐂𝐟
 2nd : Q1 = L + 𝟒
×𝐢
𝐟
𝟑(𝐍) 𝒕𝒉
 1st : Q3 = size of size of item.
𝟒
𝐍
𝟑{ }− 𝐂𝐟
 2nd : Q3 = L + 𝟒
×𝐢
𝐟
 CONTINUOUS SERIES (GROUPED DATA) :
 Example.
 Following data relates to daily wages (in ₹) of persons working in a
factory.
 Compute the lower, upper quartiles, I.Q.R and QD.

550 – 500 – 450 – 400 – 350 – 300 – 250 – 200 –

Daily wages
600 550 500 450 400 350 300 250
No. of
7 13 15 20 30 33 28 14
workers
 SOLUTION:  Q1= located in
𝐍 𝒕𝒉
item.
𝟒
 160/4th item
 = 40th item located in the class 250-300
Daily wages No. of Workers Cumulative 𝐍
− 𝐂𝐟
(in ₹) (f ) Frequency  Q1 = L + 𝟒
×𝐢
𝐟
𝟒𝟎 −𝟏𝟒
 = 250 + 𝟐𝟖
×50

200–250 14 14  = 250 + 46.43

 = 296.43 daily wages
250–300 28 42
𝟑(𝐍) 𝒕𝒉
 Q3 = located in item.
300–350 33 75 𝟒
𝟑(𝟏𝟔𝟎) 𝒕𝒉
350–400 30 105  = 𝟒
item
 = 120th item
400–450 20 125
 Q3 = lies in the class 400 - 450
450–500 15 140 𝐍
𝟑{ }− 𝐂𝐟
 Q3 = L + 𝟒
×𝐢
𝐟
500–550 13 153 𝟏𝟐𝟎 −𝟏𝟎𝟓
 = 400 + 𝟐𝟎
× 𝟓𝟎
550–600 7 160  = 400 + 37.5
N = ∑f = 160  = 437.5 daily wages
QUARTILES IN GROUPED DATA:
 Q1 = 296.43. Q3 − Q1
 Q.D =
𝟐
 Q3 = 437.50
437.50 – 296.43
 As we know:  =
𝟐
 I.Q.R = Q3 - Q1  = 141.07/2
 = 437.50 – 296.43  = 70.54
 = 141.07
MEAN DEVIATION :

 Mean deviation is the arithmetic mean of the differences of the values

from their average.
 In case of individual series:
 Mean Deviation from Mean,
 M.D.( mean)
 Step 1: The A.M. of the values is calculated.
 step 2: Calculate the absolute deviations from mean 𝐝 = X – Mean
(ignoring arithmetic signs).
 Step 3: The A.M. of these deviations is the Mean Deviation
MEAN DEVIATION :

 In case of individual series:

 Mean Deviation from Mean,
 M.D.( mean)

∑ ΙDΙ
 𝑀𝐷 𝑥ҧ =
𝐍

 MD = Mean deviation
 IDI = Deviation from mean ignoring signs
 N = Number of observations
INDIVIDUAL SERIES :
 Example.
 Calculate the mean deviation from mean and median of the following
values. As mean is given as 6.
2, 4, 7, 8 and 9

X I d I = X - mean
2 4
4 2
7 1
8 2
9 3
Total 12
INDIVIDUAL SERIES:
 Example.
X I d I = X - mean
2 4
4 2
7 1
8 2
9 3
Total 12

∑ ΙDΙ
𝐌𝐃ത𝐱 =
𝐍
= 12/5
= 2.4
INDIVIDUAL SERIES:
 Example.
 Calculate the mean deviation from mean and median of the following
values. Calculate by median. 2, 4, 7, 8 and 9
 Median = (N+1/2) th item
 = 6/2 = 3rd item = 7
IdI = X-
X
median
2 5
4 3
7 0
8 1
9 2
Total 11
INDIVIDUAL SERIES:
 Example.
X Id I = X - mean
2 5
4 3
7 0
8 1
9 2
Total 11

∑ ΙDΙ
𝐌𝐃𝐦 =
𝐍
= 11/5
= 2.2
MEAN DEVIATION :

 In case of discrete series:

 Mean Deviation from Mean,
 M.D.( mean)

∑ f ΙDΙ
 𝑀𝐷 𝑥ҧ =
∑𝐟

 MD = Mean deviation
 IDI = Deviation from mean (X - Mean) ignoring signs
 f = frequency
DISCRETE SERIES (GROUPED DATA):
 Example.
 Plots in a housing colony come in only three sizes: 100 sq. metre, 200
sq. meters and 300 sq. metre and the number of plots are respectively
200, 50 and 10. Calculate mean deviation from mean.

Plot Size in sq. metre (X) No. of Plots (f )

100 200

200 50

300 10

∑ f = 260
INDIVIDUAL SERIES:
 Example.

Plot Size in sq.

metre (X)
No. of Plots (f ) fX IdI = X - Mean f IdI

100 200 20,000 26.92 5,384

200 50 10,000 73.08 3,654
300 10 3,000 173.08 1,730.80

260 33,000 273.08 10,768.80

∑ fX ∑ f ΙdΙ
xത = 𝐌𝐃ത𝐱 =
∑𝐟 ∑𝐟
= 33,000/260 = 10,768.80/260
= 126.92 = 41.42
DISCRETE SERIES (GROUPED DATA) BY MEDIAN:
 Example.
 The frequency distribution of the number of persons and their respective
incomes (in ₹) are given below. Calculate the Mean Deviation from
Median .

Income (in ₹) 100 200 300 400

Number of persons 2 4 10 4
DISCRETE SERIES:
 Example.
𝑵+𝟏
𝐌 = 𝒔𝒊𝒛𝒆 𝒐𝒇 𝒕𝒉
𝟐
Income (X) No. of Cumulative 𝒊𝒕𝒆𝒎
frequency IdI = f IdI = 21/2
(in ₹) persons (f ) (cf )
X - 300 = 10.5
= 300
100 2 2
200 400
100 400
200 4 6
∑ f ΙdΙ
16
0 0 𝐌𝐃𝐦 =
300 10 ∑𝐟
20 100 400 = 1,200/20
400 4
∑ f IdI =
N = ∑ f = 20 = 60
1,200
CONTINUOUS SERIES (GROUPED DATA) :
 Example.
 Calculate the mean deviation about mean.

Marks 0–10 10–20 20–30 30–40 40–50 50–60 60–70

No. of Students 5 12 15 25 8 3 2
CONTINUOUS SERIES :
 Solution.
No. of students Mid value IdI = m
Mark (X) fm f IdI
(f ) (m)
- 30.14 Mean =
0–10 5 5 25 ∑ 𝑓𝑀
25.14 125.7
10–20 12 15 180 𝛴𝑓
15.14 181.68
= 2100/70
20–30 15 25 375 77.1
5.14 = 30.14
30–40 25 35 875 121.5
4.86
40–50 8 45 360 ∑ f ΙdΙ
14.86 118.88 𝐌𝐃ത𝐱 =
165 ∑𝐟
50–60 3 55 24.86 74.58 = 769.16/70
60–70 2 65 130
34.86 69.72
∑ f = 70
∑ fm =
∑f IdI = = 10.99
2110
769.16
STANDARD DEVIATION :

 Standard Deviation is the positive square root of the mean of squared

deviations from mean.

 Suppose there are five values x1, x2, x3, x4 and x5.
 Step 1: Their mean is calculated.
 Step 2: Then deviations of the values from mean are calculated.
 Step 3: These deviations are then squared.
INDIVIDUAL SERIES DIRECT METHOD:
 Example.
 Calculate the standard deviation: 5, 10, 25, 30, 50.
∑𝐗
 Actual Mean Method: mean =
𝑁
 = 120/5 = 24

 Calculate d = X – Mean
 Take the sum of squire of d
SOLUTION: ACTUAL MEAN METHOD

d= Standard deviation 𝝈
X (X – 𝑑2
∑ 𝑑2
Mean) 𝝈 =
5 –19 361
𝑁
𝟏𝟐𝟕𝟎
10 –14 196
=
𝟓

25 +1 1
= 254
30 +6 36 = 15.937
50 +26 676
∑𝑑 2 =
0 1270
INDIVIDUAL SERIES ASSUMED MEAN METHOD:
 Example.
 Calculate the standard deviation: 5, 10, 25, 30, 50.
 Assumed Mean Method:
 A = Assumed mean from the data

 Calculate d = X – A
 Take the sum of squire of d
 ASSUMED MEAN METHOD:
FOR THE SAME VALUES, DEVIATIONS MAY BE CALCULATED FROM ANY ARBITRARY VALUE A SUCH THAT
DEVIATION = X – A. TAKE A = 25

d=X-A
𝒅𝟐 Standard deviation 𝜎 =
X (X – 25)
5 –20 400 ∑ 𝑑2 ∑𝑑
2
−
10 –15 225
𝑁 𝑁

25 0 0 =
1275
−
−5 2
5 5
30 5 25 = 255 − 1
50 25 625 = 254
∑d = – 5 ∑𝐝𝟐 = 15.937
= 1275
 DIRECT METHOD:
Standard deviation can also be calculated from the values directly, i.e.,
without taking deviations

X 𝑿𝟐 Standard deviation 𝜎 =

5 25 ∑ 𝑋2 ∑𝑋
2
−
𝑁 𝑁
10 100
4,150 120 2
= −
25 625 5 5
= 830 − 576
30 900
= 254
50 2,500 = 15.937
∑X = 120 ∑𝐗 𝟐 = 4,150
INDIVIDUAL SERIES STEP-DEVIATION METHOD:
 Example.
 Calculate the standard deviation: 5, 10, 25, 30, 50.
 Assumed Mean Method:
 A = Assumed mean from the data
 Calculate d = X – A
 Calculate d’ = d/c
 Take the sum of squire of d’
 STEP-DEVIATION METHOD:
Deviations can be calculated from an arbitrary value and then divided by a common factor (c)
𝐗−𝐴
d=X-A d =
X 𝐶 d2 Standard deviation 𝜎 =
(X – 25) = (d/5)
5 –20 –4 16 ෌ 𝒅′𝟐 ∑ 𝒅′ 𝟐
− ×C
𝑵 𝑵
10 –15 –3 9
51 −1 2
25 0 0 0 = − ×5
5 5

30 +5 1 = 10.16 × 5
1
= 15.937
50 +25 5 25
∑d = – 5 ∑d’ = –1 ∑d’2
= 51
DISCRETE SERIES (GROUPED DATA):
 Actual mean method:
∑ 𝐟𝐱
ഥ=
 Calculate mean: 𝐗
𝚺𝐟
 Find deviation from actual mean d = X – mean
 Multiply deviation with respective frequency. To find fd series.
 Apply the formula of Standard deviation 𝝈

෌ 𝑓𝑑 2
𝝈 =
∑𝑓
ACTUAL MEAN METHOD

 Calculate standard deviation by actual mean method:

Marks 5 10 15 20

No. of students 2 1 4 3

෌ 𝐟𝐝𝟐
 Apply the formula: 𝛔 =
∑𝐟
ACTUAL MEAN METHOD:
No. of ഥ = ∑ 𝐟𝐱
𝐗
students fX d= 𝚺𝐟
X (X – 14) 𝒅𝟐 fd2 = 140/10
(f )
= 14
5 2 10 –9 81 Standard deviation 𝜎 =
162
10 1 10 –4 16 ∑ 𝑓𝑑 2
16 ∑f

15 4 60 1 1 290
4 =
10

20 3 60 36 108 = 29
6
= 5.385
∑fX = ∑fd2
∑f = 10
140 = 290
ASSUMED MEAN METHOD

 Calculate standard deviation by assumed mean method:

Marks 5 10 15 20

No. of students 2 1 4 3

෌ 𝑓𝑑 2 2
∑ 𝑓𝑑
 Apply the formula: 𝛔 = −
∑f ∑f
ASSUMED MEAN METHOD:
No. of
d= 𝑨𝒔𝒔𝒖𝒎𝒆𝒅 𝑴𝒆𝒂𝒏 = 10
students fX 𝒅 𝟐 fd fd2
X (X – 10) Standard deviation 𝜎 =
(f )
5 2 10 –5 25 - 10 50 ∑ 𝑓𝑑 2 ∑ 𝑓𝑑
2
𝛔 = −
∑f ∑f
10 = A 1 10 0 0 0 0 450 40 2
= −
10 10
15 4 60 5 25 20 100 = 45 − 16
= 29
20 3 60 10 100 30 300 5.385
∑fX = ∑fd ∑fd2
∑f = 10 = 450
140 = 40
DIRECT METHOD:
No. of ഥ = ∑ 𝐟𝐱
𝐗
X students fX 𝑿𝟐 fX2 𝚺𝐟
(f ) = 140/10
14
5 2 10 25 50 Standard deviation 𝜎 =

10 1 10 100 100 ∑ 𝑓𝑋 2 ∑ 𝑓𝑋
2
𝛔 = −
∑f ∑f
15 4 60 225 900 2250 140
2
𝛔 = −
𝟏𝟎 10
20 3 60 400 1,2000 = 225 − 196
∑fX = = 29
∑f = 10 ∑fX2
140 = 2,250 = 5.385
STEP-DEVIATION MEAN METHOD:
No. of d’ = 𝑨𝒔𝒔𝒖𝒎𝒆𝒅 𝑴𝒆𝒂𝒏 = 10
d=
X students d/c 𝒅′𝟐 fd’ f.d’2 Standard deviation 𝜎 =
(X – 10)
(f ) C=5
2
∑fd’2 ∑fd’
5 2 –5 - 1 1 -2 2 𝛔 =
∑f
−
∑f
×C

2
10 = A 1 0 0 0 0 0 18 8
= − ×5
10 𝟏𝟎

15 4 5 +1 1 +4 4 = 1.8 − 0.64 × 5
= 1.16 × 5
20 3 10 +2 4 +6 12
= 1.077 × 5
∑fX = ∑fd’ ∑fd’2 = 5.385
∑f = 10 = 18
140 =+8
CONTINUOUS SERIES (GROUPED DATA):
 Actual Mean Method:
∑ 𝐟𝐦
ഥ
 Step 1: Calculate the mean of the distribution ( 𝑿 = )
∑ 𝐟
ഥ ).
 Step 2: Calculate deviations of mid values from the mean (i.e., d = m - 𝑿
 Step 3: Multiply the deviations with their corresponding frequencies to get fd
values.
 Step 4: Calculate fd2 values by multiplying f values with d2 values. Sum
up these to get ∑fd2
 Step 5: Calculate standard deviation using the following formula:

෌ 𝐟𝐝𝟐
 𝛔 =
∑𝐟
ACTUAL MEAN METHOD
 Calculate standard deviation by actual mean method:
Profits of companies
Number of companies
(₹ in lakh)

10–20 5
20–30 8
30–50 16
50–70 8
70–80 3

෌ 𝐟𝐝𝟐
 Apply the formula: 𝛔 =
∑𝐟
 ACTUAL MEAN METHOD:
∑ 𝐟𝒎
ഥ
𝐗=
d= 𝚺𝐟
CI f m fm fd fd2 = 1,620/40
m – 40.5
= 40.5
10–20 5 15 75 –25.5 –127.5 3,251.25 Standard deviation
𝜎=
20–30 8 25 200 –15.5 –124.0 1,922
30–50 16 40 640 –0.5 –8.0 4 ෌ 𝒇𝒅𝟐
∑f
50–70 8 60 480 19.5 156 3,042
𝟏𝟏,𝟕𝟗𝟎
=
70–80 3 75 225 34.5 103.5 𝟒𝟎
3,570.75
෍ 𝐟𝒎 = 𝟐𝟗𝟒. 𝟕𝟓
𝚺𝐟 = ෌ 𝑓𝑑 2 = = 17.168
40 = 1,620 0 11,790
 ASSUMED MEAN METHOD:
𝑨𝒔𝒔𝒖𝒎𝒆𝒅 𝑴𝒆𝒂𝒏 = 40
d=m–
CI f m fd Fd2 Standard deviation 𝜎 =
40
2
∑ 𝑓𝑑 2 ∑ 𝑓𝑑
10–20 5 15 –25 –125 3,125 𝛔 =
∑f
−
∑f

20–30 8 25 –15 –120 1,800 11,800 20 2

= −
40 = A 0 0 40 10
30–50 16 0
60 20 160 3,200 = 295 − 4
50–70 8
= 291
70–80 3 75 35 105 3,675 = 17.0587
∑f = 40 ෌ 𝑓𝑑 2 =
∑ 𝑓𝑑 = 20
11,800
 STEP-DEVIATION METHOD:
𝑨𝒔𝒔𝒖𝒎𝒆𝒅 𝑴𝒆𝒂𝒏 = 40
m d = m d’ = d/c Standard deviation 𝜎 =
CI f – 40 C=5 d’ 2 fd’ f.d2

2
15 –5 ∑fd’2 ∑fd’
10–20 5 –25 25 –25 125 𝛔 =
∑f
−
∑f
×C

20–30 8 25 –15 –3 9 –24 72 2

472 4
= − ×5
0 0 40 𝟒𝟎
30–50 16 40 0 0 0
=A = 11.8 − 0.01 × 5
60 +4 +32 128
50–70 8 20 16 = 11.79 × 5
147 = 3,434 × 5
70–80 3 75 35 +7 49 +21 = 17.168
∑f = ∑fd’ ෌ 𝑓𝑑 2
40 = +4 = 472