PDF Measures of Dispersion Slide No 34 To 52
PDF Measures of Dispersion Slide No 34 To 52
PDF Measures of Dispersion Slide No 34 To 52
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:
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:
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:
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:
𝐍+𝟏 𝒕𝒉
Lower Quartile (Q1) = size of size of item
𝟒
𝟑(𝐍+𝟏) 𝒕𝒉
Lower Quartile (Q3) = size of size of item
𝟒
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.
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
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.
∑ 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 :
∑ 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.
100 200
200 50
300 10
∑ f = 260
INDIVIDUAL SERIES:
Example.
∑ 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 .
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.
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 :
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 d2 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
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
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
2
15 –5 ∑fd’2 ∑fd’
10–20 5 –25 25 –25 125 𝛔 =
∑f
−
∑f
×C