MCA_UNIT-3_Computer Oriented Numerical Statistical Methods

Unit-3 FREQUAENCY DISTRIBUTION
RAI UNIVERSITY, AHMEDABAD 1
Course: MCA
Subject: Computer Oriented Numerical
Statistical Methods
Unit-3
RAI UNIVERSITY, AHMEDABAD

Unit-III- Frequaency Distribution
Sr.
No.
Name of the Topic Page
No.
1 Introduction, Collection of data, Classification of data 2
2 Introduction to frequency distribution , Class Limit, , Class
Interval,Class frequency, Class mark, Class Boundaries, Width of a
class
5
3 Frequency density, Relative frequency, Percentage frequency,
Cumulative frequency
9
4 Introduction, Arithmetic Mean, Mean forDiscrete frequency
distribution, Mean for Continuous frequency distribution,Weighted
Arithmetic Mean
11
5 Properties of A.M., Merits & De merits of A.M., 22
6 Median for raw data, Discrete frequency distribution, Continuous
frequency distribution,
23
7 Merits and demerits of Median 26
8 Mode for raw data, D.f.s., c.f.s., 27
9 Merits & demerits of mode 30
10 Introduction, Range, coefficient of range,Merit & Demerit of
Range
30
11 Quartiles, Quartiles deviations, coefficient of quartile deviations 31
12 Mean deviation and coefficient of mean deviation,Merit and
Demerit of Mean Deviation
33
13 Standard Deviation and variance for all types of frequency
distribution
38
14 Coefficient of variation 48

1.1 Introduction:
A sequence of observation, made on a set of objects included in the sample drawn
from population is known as statistical data.
(1). Ungrouped data
Data which have been arranged in a systematic order are called raw data or
ungrouped data.
(2) .Grouped data
Data presented in the form of frequency distribution is called grouped data.
1.2Collectionof Data:
The first step in any enquiry (investigation) is collection of data. The data may be
collected for the whole population or for a sample only. It is mostly collected on
sample basis. Collection of data is very difficult job. The enumerator or
investigator is the well trained personwho collects the statistical data. The
respondents (information) are the persons whom the information is collected.
1.3 ClassificationofData:
Data classification is the process oforganizing data into categories for its most
effective and efficient use.
A well-planned data classification system makes essential data easy to find and
retrieve.
Written procedures and guidelines for data classification should define what
categories and criteria the organization will use to classify data and specify the
roles and responsibilities of employees within the organization regarding data
stewardship. Once a data-classification scheme has been created, security standards
that specify appropriate handling practices for each category and storage standards
that define the data's lifecycle requirements should be addressed.
Here is an example of what a data classification scheme might look like:

Category4: Highly sensitive corporateand customer data that if disclosed could
put the organization at financial or legal risk.
Example: Employee social security numbers, customer credit card numbers
Category3: Sensitive internal data that if disclosed could negatively affect
operations.
Example: Contracts with third-party suppliers, employee reviews
Category2: Internal data that is not meant for public disclosure.
Example: Sales contest rules, organizational charts
Category1: Data that may be freely disclosed with the public.
Example: Contact information, price lists
1.3.1 Types of Data:
There are two types (sources) for the collection of data.
(1) Primary Data (2) SecondaryData
(1) Primary Data:
The primary data are the first hand information collected, compiled and published
by organization for some purpose. They are most original data in character and
have not undergone any sortof statistical treatment.
Example: Population census reports are primary data because these are collected,
complied and published by the population census organization.
(2) SecondaryData:
The secondarydata are the second hand information which are already collected by
some one (organization) for some purposeand are available for the present study.
The secondarydata are not pure in character and have undergone some treatment at
least once.
Example: Economics survey of England is secondarydata because these are
collected by more than one organization like Bureau of statistics, Board of
Revenue, the Banks etc…

1.3.2DifferencebetweenPrimary and SecondaryData:
The difference between primary and secondary data is only a change of hand. The
primary data are the first hand data information which is directly collected form
one source. They are most original data in character and have not undergone any
sort of statistical treatment while the secondarydata are obtained from some other
sources or agencies. They are not pure in character and have undergone some
treatment at least once.
For Example: Supposewe interested to find the average age of MS students. We
collect the age’s data by two methods; either by directly collecting from each
student himself personally or getting their ages from the university record. The
data collected by the direct personal investigation is called primary data and the
data obtained from the university record is called secondary data.
1.3.3 Types of Classification:
1. Geographical Classification:
i.e. Area wise, e.g. cities, districts, etc.
2. Chronological Classification:
i.e. on the basis of time
3. Qualitative Classification:
i.e. according to some attributes
4. Quantitative Classification:
i.e. in terms of magnitudes
Quantitative classification refers to the classification of data according to some
characteristics that can be measured, such as height, weight, income, sales, profits
etc.
In this type of classification , there are two elements namely
(i). the variable
(ii). The frequency: Frequency is how often something occurs.

2.1Definition:frequency Distribution
By counting frequencies we can make a Frequency Distribution table.
There are two types of frequency distribution:
(a). Discrete frequency distribution.
(b). Continuous frequency Distribution.
Example: Newspapers
These are the numbers of newspapers sold at a local shop over the last 10 days:
22, 20, 18, 23, 20, 25, 22, 20, 18, 20
Let us count how many of each number there is:
Here Table-A is a example of discrete frequency distribution
Table-B is a example of continuous frequency distribution.
TABLE-A
Papers Sold Frequency
18 2
19 0
20 4
21 0
22 2
23 1
24 0
25 1
It is also possible to group the values. Here they are in grouped :
Table-B

15-19 2
20-24 7
25-29 1
2.2ClassLimit:
Class limits are the smallest and largest observations (data, events etc) in each
class. Therefore, each class has two limits: a lower and upper.
Example for, In the above Table-B
For the first class 15-19
The Lower class limit is =15
The Upper class limit is = 19
2.3Class Interval:
The difference between the upper and lower limit of a class is known as class
interval of that class.
Table-B
15-19 2
20-24 7
25-29 1
For example: ForTable-B in the class 15-19 the class interval = 4
(i.e. 19-15)
A simple formula to obtain the estimate of appropriate class interval,
i.e. 𝑖 =
𝐿−𝑆
𝑘
where 𝐿 = largest item, S= Smallest item, K = no. of classes

2.3.1 Example:If the salary of 100 employees in a commercialundertaking
varied between Rs.10,000and Rs. 30,000and we want to form 10 classes,then
Find the class interval.
Solution: 𝑖 =
𝐿−𝑆
𝑘
 Here 𝐿 = 30,000, 𝑆 = 10,000, 𝑘 = 10
 𝑖 =
30,000−10,000
10
=
20,000
10
= 2000
2.4 Class Frequency:
The number of observations correspondingto a particular class is known as the
frequency of that class or the class frequency.
Table-B
15-19 2
20-24 7
25-29 1
For example,
In the Table-B the class 20-24 has frequency 7.
i.e there are 7 days in which no. of papers sold is between 20-24.
If we add together frequencies of all individual classes, we obtain the total
frequency.
The total frequency of Table-B = 2+7+1=10
2.5 Class mark or class mid point:
It is the value lying half-way between the lower and upper class limits of a class
interval.

∴ 𝑐𝑙𝑎𝑠𝑠 𝑚𝑎𝑟𝑘 =
𝑈𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠+𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑎𝑠𝑠
2
2.6 Class Boundaries:
Class Boundaries are the midpoints between the upper class limit of a class and the
lower class limit of the next class in the sequence. Therefore, each class has an
upper and lower class boundary.
For Example: Table-B
15-19 2
20-24 7
25-29 1
For the first class in table-B , 15 – 19
The lower class boundary is the midpoint between 14 and 15, that is 14.5
The upper class boundary is the midpoint between 19and 20, that is 19.5
2.7 Width of a Class:
Difference between two consecutive lower class limits
Difference between two consecutive upper class limits
Example for, Table-B
15-19 2
20-24 7
25-29 1
Difference between two consecutive lower class limits
20-15 = 5

Difference between two consecutive upper class limits
24-19 = 5
∴Class width=5
3.1Frequency Density:
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
𝑐𝑙𝑎𝑠𝑠 𝑤𝑖𝑑𝑡ℎ
Frequency density is use to draw histogram .
The following table shows the ages of 25 children on a school bus:
Age Frequency
5-10 6
11-15 15
16-17 4
> 17 0
To draw the histogram we need frequency density:
For class 5-10
=
6
6
= 1
For class 11-15
=
15
5
= 3

3.2 Relative frequency:
Relative frequency is the ratio of the number of times an event occurs to the
number of occasions on which it might occurin the same period.
In other words, how often something happens divided by all outcomes.
Example: if your team has won 9 games from a total of 12 games played:
* the Frequency of winning is 9
* the Relative Frequency of winning is
9
12
= 75%
3.3 Percentagefrequency:
Percentage frequency that means calculate percentage of given frequency.
Percentage frequency=
100×𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝑔𝑖𝑣𝑒𝑛 𝑐𝑙𝑎𝑠𝑠
𝑡𝑜𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
15-19 2
20-24 7
25-29 1
Forclass 15-19 frequency is = 2
∴ Percentage frequency for class 15-19 =
100×2
10
= 20%

3.4Cumulative frequency:
The total of a frequency and all frequencies so far in a frequency distribution.
It is the 'running total' of frequencies.
4.1Introduction:
A measure of central tendency is a single value that attempts to describe a set of
data by identifying the central position within that set of data. Measures of central
tendency are sometimes called measures of central location. They are also classed
as summary statistics. The mean (often called the average) is most likely the
measure of central tendency that you are most familiar with, but there are others,
such as the median and the mode.
The mean, median and mode are all valid measures of central tendency.
4.2Arithmetic Mean:
The most popular and widely used measure of representing the entire data by one
value is what most laymen call an average and what the statisticians call the

arithmetic mean. Its value is obtained by adding together all the items and by
dividing this total by the number of items.
Arithmetic mean may either be
i. Simple arithmetic mean, or
ii. Weighted arithmetic mean
First of all we have to discuss about Simple Arithmetic Mean.
There are two method for finding simple arithmetic Mean :
1. Direct Method
2. Short-cut Method
4.2.1 DirectMethod for finding Arithmetic Mean :
𝑿̅ =
𝑿 𝟏+𝑿 𝟐+𝑿 𝟑+⋯……+𝑿 𝒏
𝑵
=
∑ 𝑿
𝑵
Here 𝑋̅ = Arithmetic Means
∑ 𝑋 = Sum of all the values of the variable 𝑋
𝑁 = Number of the Observations
Steps for finding Arithmetic mean :
1. Add together all the values of the variable 𝑋 and obtain the total
i.e., ∑ 𝑋
2. Divide this total by the number of observations, i.e., 𝑁
Example-The following table data is the monthly income (in Rs.)of 10
employees in an office:
14780,15760,26690,27750,24840,24920,16100,17810,27050,26950
Calculate the arithmetic mean of incomes.
Solution: Here 𝑁 = 10
 ∴ 𝑋̅ =
𝑋1+𝑋2+𝑋3+⋯……+𝑋 𝑛
𝑁
=
∑ 𝑋
𝑁

 ∴ 𝑋̅ =
14780+15760+26690+27750+24840+24920+16100+17810+27050+26950
10
 ∴ 𝑋̅ =
2,22,650
10
 ∴ 𝑋̅ = 22,265
 Hence the Average Income is Rs.22,265
4.2.2 Short-cut Method for finding Arithmetic Mean:
The arithmetic mean can be calculated by using what is known as an arbitrary
Origin. When deviations are taking from an arbitrary origin, the formula for
calculating arithmetic mean is
𝑋̅ = 𝐴 +
∑ 𝑑
𝑁
Where A is the assumed mean and d is the deviation of items from assumed mean,
i.e.,𝑑 = (𝑋 − 𝐴)
Steps for finding Arithmetic mean by shortcut Method
1. Take an assumed mean.
2. Take the deviations of items from the assumed mean and denote these
deviations by 𝑑.
3. Obtain the sum of these deviations, i.e., ∑ 𝑑
4. Apply the formula 𝑋̅ = 𝐴 +
∑ 𝑑
𝑁
Example-- The following table data is the monthly income (in Rs.)of 10
employees in an office:
14780,15760,26690,27750,24840,24920,16100,17810,27050,26950
Calculate the arithmetic mean of incomes by using short cut method.
Solution:
Supposeassumed mean 𝐴 = 22000

Employees Income(Rs.) ( 𝑿 − 𝟐𝟐𝟎𝟎𝟎) = 𝒅
1 14,780 -7220
2 15,760 -6240
3 26,690 +4690
4 27,750 +5750
5 24,840 +2840
6 24,920 +2920
7 16,100 -5900
8 17,810 -4190
9 27,050 +5050
10 26,950 +4950
N=10 ∑ 𝑑 = 2650
 𝑋̅ = 𝐴 +
∑ 𝑑
𝑁
 𝐻𝑒𝑟𝑒 𝐴 = 22000,∑ 𝑑 = 2650, 𝑁 = 10
 𝑋̅ = 22000 +
2650
10
= 22,265
 Hence the average income is Rs. 22,265.
4.3 Calculationof Arithmetic Mean-DiscretefrequencyDistribution:
In discrete series arithmetic mean may be computed by applying
1. Direct Method
2. Short-Cut Method
4.3.1DirectMethod:
The formula for computing mean is

𝑋̅ =
∑ 𝑓𝑥
𝑁
Where 𝑓 = frequency
𝑋 = The variable in Question
𝑁 = Total number of Observations i.e. ∑ 𝑓
Steps for finding Arithmetic Mean for Discrete frequency Distribution:
1. Multiply the frequency of each row with the variable and obtain the
total ∑ 𝑓𝑋
2. Divide the total obtained by step(i) by the number of observations ,
i.e. Total frequency
Example-- From the following data of the marks obtained by 60 students of a
class calculatesthe arithmetic mean by Direct Method:
Marks No. of Students
20 8
30 12
40 20
50 10
60 6
70 4
Solution:
Let the marks be denoted by 𝑋 and the number of students by 𝑓.
Calculation of Arithmetic Mean
Marks(𝑿) No. of
Students(𝒇)
𝒇𝑿
20 8 160

30 12 360
40 20 800
50 10 500
60 6 360
70 4 280
𝑵 = 𝟔𝟎 ∑𝒇𝒙 = 𝟐𝟒𝟔𝟎
 𝑋̅ =
∑ 𝑓𝑋
𝑁
=
2460
60
= 41
Hence, the average marks=41
4.3.2 Short-Cut Method:
According to this method, 𝑋̅ = 𝐴 +
∑ 𝑓𝑑
𝑁
Where 𝐴 = Assumed mean ;
𝑑 = ( 𝑋 − 𝐴);
𝑁 = Total number of observations i.e., ∑ 𝑓.
Steps for finding Arithmetic Mean for Discrete frequency Distribution:
1. Take an assumed mean.
2. Take the deviations of the variable X from the assumed mean and
denote the deviations by 𝑑.
3. Multiply these deviations with the respective frequency and take the
total ∑ 𝑓𝑑.
4. Divide the total obtained in third step by the total frequency.
Example--From the following data of the marks obtained by 60 students of a
class, calculatethe arithmetic mean by Shot-cut Method:
Marks No. of Students
20 8

30 12
40 20
50 10
60 6
70 4
Solution:
Supposeassumed mean A= 40
Calculation of Arithmetic Mean by short-cut Method
Marks(𝑿) No. of
Students(𝒇)
( 𝑿 − 𝟒𝟎) = 𝒅 𝒇𝒅
20 8 -20 -160
30 12 -10 -120
40 20 0 0
50 10 +10 +100
60 6 +20 +120
70 4 +30 +120
𝑁 = 60 ∑ 𝑓𝑑 = 60
 𝑋̅ =
∑ 𝑓𝑑
𝑁
= 40 +
60
60
= 40 + 1 = 41
Hence the Arithmetic mean by Shortcut method is =60
 Note:We can Observe that value of Arithmetic mean does not change in
both the method . so, we can use any one for finding arithmetic mean.
4.4 Calculationof Arithmetic Mean – Continuous Frequency Distribution

In continuous frequency distribution arithmetic mean may be computed by
applying any of the following methods:
1. Direct Method
2. Short-Cut Method
4.4.1 DirectMethod: (Meanof Continuous frequency distribution)
When direct method is used
𝑿̅ =
∑ 𝒇𝒎
𝑵
Where, m = mid-point of various classes
𝑓 = Frequency of each class
𝑁 = The total frequency
Steps for finding Arithmetic mean by DirectMethod (C.F.D):
1. Obtain the mid-point of each class and denote it by m.
2. Multiply these midpoints by the respective frequency of each class
and obtain the total ∑ 𝑓𝑚
3. Divide the total obtained in step(i) by the sum of the frequency,i.e.,N.
Example--.from the following data compute arithmetic mean by direct
method:
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of Students 5 10 25 30 20 10
Solution:
Calculation for Arithmetic Mean
Marks(𝑿) Mid-points
(𝒎)
No. of Students
(𝒇)
𝒇𝒎
0-10 5 5 25
10-20 15 10 150

20-30 25 25 625
30-40 35 30 1050
40-50 45 20 900
50-60 55 10 550
𝑵 = 𝟏𝟎𝟎 ∑ 𝒇𝒎 = 𝟑, 𝟑𝟎𝟎
 𝑋̅ =
∑ 𝑓𝑚
𝑁
=
3300
100
= 33
Hence The value of Arithmetic mean is 33.
4.4.2 Short-Cut Method: (Meanof Continuous frequency distribution)
When short-cut method is used, arithmetic mean is computed by applying the
following formula:
𝑋̅ = 𝐴 +
∑ 𝑓𝑑
𝑁
Where A= Assumed mean
𝑑 = Deviations of mid-points from assumed mean i.e.,(m-A)
𝑁 = Total number of observations
Steps for finding Arithmetic Mean by Short-Cut Method: (C.F.D)
1. Take an assumed mean
2. From the mid-point of each class deductthe assumed mean.
3. Multiply the respective frequencies of each class by these deviations
and obtain the total ∑ 𝑓𝑑.
4. Apply the formula 𝑋̅ = 𝐴 +
∑ 𝑓𝑑
𝑁
Example --from the following data compute arithmetic mean by Short-Cut
method:
Marks 0-10 10-20 20-30 30-40 40-50 50-60
No. of Students 5 10 25 30 20 10

Solution:
Assumed that Arithmetic Mean =35
N= 100
Calculation of Arithmetic mean by short-cut method
Marks(𝑿) Mid-points
(𝒎)
No. of Students
(𝒇)
( 𝒎 − 𝟑𝟓)
= 𝒅
𝒇𝒅
0-10 5 5 -30 -150
10-20 15 10 -20 -200
20-30 25 25 -10 -250
30-40 35 30 0 0
40-50 45 20 +10 +200
50-60 55 10 +20 +200
𝑵 = 𝟏𝟎𝟎 ∑ 𝒇𝒅 = −𝟐𝟎𝟎
 𝑋̅ = 𝐴 +
∑ 𝑓𝑑
𝑁
= 35 −
200
100
= 35 − 2 = 33
Hence the value of Arithmetic mean by Short-Cut method for continuous
frequency distribution is 33.
Example-- The mean marks of 100 students were found to be 40. Later on it
was discoveredthat a score of 53 was misread as 83.Findthe correctmean
corresponding to correctScore.
Solution:
 We are given 𝑁 = 100, 𝑋̅ = 40
 Since 𝑋̅ =
∑ 𝑋
𝑁
 ∑ 𝑋 = 𝑁𝑋̅ = 100× 40 = 4000
 But this is not correct ∑ 𝑋

 Correct ∑ 𝑋 = Incorrect ∑ 𝑋 − Wrong item + Correct item
 Correct ∑ 𝑋 = 4000− 83 + 53 = 3970
 ∴ Correct 𝑋̅ =
correct ∑ 𝑋
N
=
3970
100
= 39.7
Hence the Correct mean=39.7
4.5 Weightedarithmetic mean
One of the limitations of the arithmetic mean discussed above is that it gives equal
importance to all the items. But there are cases where the relative importance of the
different items.The formula for computing weighted arithmetic mean is:
𝑋 𝑤
̅̅̅̅ =
∑ 𝑊𝑋
∑ 𝑊
 Where 𝑋 𝑤
̅̅̅̅ represents the weighted arithmetic mean; X represents the
variable values.
W represents the weights attached to variable values.
Steps for finding Weighted mean:
1. Multiply the weights by the variable X and obtain the total ∑ 𝑊𝑋
2. Divide this total by the sum of the weights, i.e.,∑ 𝑊.
Example-- Calculate weightedaverage ofthe following data:
Course BA BSc MA MCA MBA
%of Pass 70 65 75 90 99
No of
Students
20 30 30 50 40
Solution:
 Weighted Average =
∑ 𝑊𝑋
∑ 𝑊
 Weighted Average =13300 / 170 = 76.47
Calculation for weighted Average (Mean)

% of Pass X No of Student W XW
70 20 1400
65 30 1950
75 30 2250
90 50 4500
80 40 3200
X=380 W=170 13300
Example--A train runs 25 miles at a speedof 30 m.p.h., another50 miles at a
speedof 40 m.p.h, then due to repairs a track travels for 6 minutes at a speed
of 10 m.p.h. what is the average speedin miles per hour?
Solution:
Time taken in covering 25 miles at a speed of 30 m.p.h=50 minutes. Time taken in
covering 50 miles at a speed of 40 m.p.h = 75 minutes. Distance covered in 6
minutes at a speed of 10 m.p.h = 1 mile. Time taken in covering 24 miles at a
speed of 24 m.p.h = 60 minute Therefore, taking the time taken as weights we have
the weighted mean as
Speed (in m.p.h)
(𝑿)
Time taken
(in min) W
𝑾𝑿
30 50 1500
40 75 3000
10 6 60
24 60 1440
∑ 𝑊 = 191 ∑ 𝑊 𝑋 = 6000
 ∴ Weighted arithmetic mean =𝑋 𝑤
̅̅̅̅ =
∑ 𝑊𝑋
∑ 𝑊
=
6000
191
= 31.41 𝑚. 𝑝. ℎ.

Hence value of weighted arithmetic mean is 31.41 𝑚. 𝑝. ℎ.
5.1 MathematicalProperties Of Arithmetic Mean:
The following are a few important mathematical properties of the arithmetic mean:
1. The sum of the deviations of the items from the arithmetic mean (taking
signs into account) is always zero. i.e. ∑(𝑋 − 𝑋̅) = 0
2. The sum of the squared deviations of the items from arithmetic mean is
minimum, that is, less than the sum of the squared deviations of the items
from any other value.
3. It is clear that ∑(𝑋 − 𝑋̅)2
is greater. This Property that the sum of the
squares of items is least from the mean is of immense use regression
analysis which shall be discussed later.
5.2 Merits of arithmetic mean:
Arithmetic mean is the simplest measurement of central tendency of a group. It is
extensively used because:
 It is easy to calculate and easy to understand.
 It is based on all the observations.
 It is rigidly defined.
 It provides good basis of comparison.
 It can be used for further analysis and algebraic treatment.
5.3 Demerits of the arithmetic mean:
 It is affected by the extreme values.
 It may lead to a wrong conclusion.
 It is unrealistic.
 Arithmetic mean cannot be obtained even if single observation is missing
 It cannot be identified observation or graphic method
6.1Median:

The median is the middle score for a set of data that has been arranged in order of
magnitude.
If the number of events are even then the average of two middle are taken.
The median is better for describing the typical value.
Example:
In order to calculate the median, supposewe have the data below:
65 55 89 56 35 14 56 55 87 45 92
We first need to rearrange that data into order of magnitude (smallest first):
14 35 45 55 55 56 56 65 87 89 92
Our median mark is the middle mark - in this case, 56 (highlighted in bold).
The median by definition refers to the middle value in a distribution. The median is
that value of the series which divides the group into two equal parts, one part
comprising all values greater than the median value and the other part comprising
all the values smaller than the median value.
6.2 Steps for Calculationof Median – Individual Observations
1. Arrange the data in ascending or descending order of magnitude.
(Both arrangements would give the same answer)
2. Apply the formula Median = 𝑀 =
𝑁+1
2
Example-- Find the median for the following data:
5, 15, 10, 15, 5, 10, 10, 20, 25 and 15.
Solution:
First of all we have to arrange all the Observations in ascending order
5, 5, 10, 10, 10, 15, 15, 15, 20, 25
Here by Observation we can say that 𝑁 = 10
𝐻𝑒𝑛𝑐𝑒, Median 𝑀 =
𝑁+1
2
=
10+1
2
= 5.5th item =
10+15
2
= 12.5
Example-- Find the median for the following data:

25900,26950,27020,27200,28280
Solution:
First of all we have to arrange all the Observations in ascending order
25900, 26950, 27020, 27200, 28280
Here by Observation we can say that 𝑁 = 5
𝐻𝑒𝑛𝑐𝑒, Median 𝑀 =
𝑁+1
2
=
5+1
2
=
6
2
= 3𝑟𝑑 item =27020
6.3 Steps for Calculationof median of continuous frequency distribution:
1. Determine the particular class in which the value of median lies.
2. Use
𝑁
2
as the rank of the median and not
𝑁+1
2
.
3. After ascertaining the class in which median lies, the following formula is
used for determining the exact value of median
𝑀𝑒𝑑𝑖𝑎𝑛 = 𝐿 +
𝑁
2
− 𝑐. 𝑓.
𝑓
× 𝑖
Where,
𝐿 = Lower limit of the median class i.e., the class in which the middle item of
the distribution lies.
𝑐. 𝑓. = Cumulative frequency of the class preceding the median class or sum of the
Frequency of the frequencies of all classes lower than the median class.
𝑓 = Simple frequency of the Median class.
𝑖 = The class interval of the Median class.
Example--: Find the median of the following data.
Cost 10-20 20-30 30-40 40-50 50-60
Items in a
group
4 5 3 6 3
Solution:

Calculation for Median
Cost Number of items in the group Cumulative frequency
10-20 4 4
20-30 5 9
30-40 3 12
40-50 6 18
50-60 3 21
 Here N=21 ⇒
𝑁
2
= 10.5
 The median class is 30-40.
From Formula,
𝑀𝑒𝑑𝑖𝑎𝑛 = 𝐿 + (
𝑁
2
− 𝑐𝑓
𝑓
) × 𝑖
 Here, L=30, 𝑖 = 10, 𝑐𝑓 = 9
 𝑀𝑒𝑑𝑖𝑎𝑛 = 30+
(10.5−9)
12
× 10 = 30 + 1.25 = 31.25
7.1 MathematicalProperty of Median:
1. The sum of the deviations of the items from median, ignoring signs, is the
least.
For example, the median of 4,6,8,10,12 is 8. The deviations from 8 ignoring signs
are 4, 2, 0, 2, 4 and the total is 12.This total is smaller than the one obtained if
deviations are taken from any other value. Thus if deviations are taken from 7,
values ignoring signs would be 3, 1, 1, 3, 5 and the total 13.
7.2 Merits of Median:
 It is based on all the observations.
 It is rigidly defined.

 It eliminates the impact of extreme values.
 It can be used for further analysis and algebraic treatment.
 Median can be found out just by inspection in some cases.
7.3Demerits Of Median:
 It simply ignores the extreme values.
 It may lead to a wrong conclusion. When distribution of observations is
Irregular.
 The median is estimated in continuous case.
8.1 Mode:
The value of the variable which occurs mostfrequently in a distribution is called
the mode. Mode = 3 Median – 2 Mean
8.2 Calculationof Mode – Individual Operations.
For determining mode count the number of times the various values repeat
themselves and the value occurring maximum number of times is the mdal value
.The more often the modal value appears relatively,the more valuable the measure
is an average to represent data.
Example--The following is the number of problems that Ms. Matty assigned
for homework on 10 different days. What is the mode?
8, 11, 9, 14, 9, 15, 18, 6, 9, 10
Solution:
Ordering the data from least to greatest, we get:
6, 8, 9, 9, 9, 10, 11, 14, 15, 18
The scorewhich occurs mostoften is 9.
Therefore, the mode is 9.

Example-2 In a crashtest, 11 cars were testedto determine what impact speed
was required to obtain minimal bumper damage. Find the mode of the speeds
given in miles per hour below.
24, 15, 18, 20, 18, 22, 24, 26, 18, 26, 24
Solution:
15, 18, 18, 18, 20, 22, 24, 24, 24, 26, 26
Since both 18 and 24 occurthree times, the modes are 18 and 24 miles per hour.
This data set is bimodal.
Example-3 A marathon race was completed by 5 participants. What is the
mode of these times given in hours?
2.7 hr, 8.3 hr, 3.5 hr, 5.1 hr, 4.9 hr
Solution:
2.7, 3.5, 4.9, 5.1, 8.3
Since each value occurs only once in the data set, there is no mode for this set of
data.
8.2Steps for Calculationof Mode For Continuous frequency Distribution:
1. Constructthe table

2. Find the Modal class
3. Find out Mode class by using N / 2
4. Apply the formula
𝑀𝑜𝑑𝑒 = 𝐿 +
(𝑓1 − 𝑓0)
2𝑓1 − 𝑓0 − 𝑓2
× 𝑖
Where, L = Lower limit of the Modal class
𝑓1Frequency of the Modal class
𝑓0 = Frequency of the class preceding the Modal class
𝑓2 =Frequency of the class succeeding Modal class
i = Class interval of modal class
Example--: Calculate mode of the following data:
Marks 10-20 20-30 30-40 40-50 50-60
F 5 20 25 15 5
Solution:
Construction of the table to find Mode:
Marks F
10-20 5
20-30 20
30-40 25
40-50 15
50-60 5
Modal class is 30-40 since highest frequency occurs here i.e. frequency of that
class is =25
𝑀𝑜𝑑𝑒 = 𝐿 +
(𝑓1 − 𝑓0)
2𝑓1 − 𝑓0 − 𝑓2
× 𝑖
Where, L = Lower limit of the Modal class = 30

𝑓1 = Frequency of the Modal class = 25
𝑓0 = Frequency of the class preceding the Modal class = 20
𝑓2 =Frequency of the class succeeding Modal class =15
i = Class interval of modal class =10
 𝑀𝑜𝑑𝑒 = 30+
(25−20)
(2×25)−20−15
× 10
 𝑀𝑜𝑑𝑒 = 30+
50
15
 𝑀𝑜𝑑𝑒 = 33.33
9.1Merits of Mode:
 It eliminates the impact of extreme values
 It can be identified by using graphical method
9.2Demerits of Mode
 It is not suitable for further mathematical treatments.
 It may lead to a wrong conclusion. When bimodal distribution.
 It is difficult to compute in some cases.
 Mode is influenced by length of the class interval.
10.1 Introduction
Averages give us information of concentration of the observations about the central
part of the distribution. But they fail to give anything further about the data.
According to George Simpson and Fritz Kafka, “An average does not tell the full
Story. It is hardly fully representative of a mass, unless we know the manner in
which the individual items scatter around it.
“Dispersion is the measure of the variations of the items.’’- A.L Bowley
10.2 RANGE
The range is the difference between two extreme values of the given observations
Range = Largest value – Smallest value

10.3 Co-efficientof Range
Co- efficient of Range =
𝐿𝑎𝑟𝑔𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒−𝑆𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒
𝐿𝑎𝑟𝑔𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒+𝑆𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒
=
𝐿−𝑆
𝐿+𝑆
Example-- Find the Co-efficient of range of Marks of 10 students from the
following
65,35,48,99,56,88,78,20,66,53
Solution:
 Range = L – S
 Range = 99 -20
 Range = 79
 Co- efficient of Range =
99−20
99+20
=
79
109
= 0.72
10.4 Merits:
1. It is easy to compute and understand.
2. It gives an idea about the distribution immediately.
10.5 Demerits:
1. Calculation range depends only on the basis of extreme items, hence it is
not reliable.
2. It is not applied to open end cases
3. Not suitable for mathematical treatments.
11.1Quartile Deviation:
The range which includes the middle 50 per cent of the distribution.That is one
quarter of the observations at the lower end, another quarter of the observations at
the upper end of the distribution are excluded in computing the interquartile
range.In the other words, interquartile range represents the difference between the
third quartile and the first quartile.
Symbolically,

𝐼𝑛𝑡𝑒𝑟𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝑟𝑎𝑛𝑔𝑒 = 𝑄3 − 𝑄1
Very often the interquartile range is reduced to the form of the semi-interquartile
range or quartile deviation by dividing it by 2.
𝑄𝑢𝑎𝑟𝑡𝑖𝑙𝑒 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝑄. 𝐷. =
𝑄3 − 𝑄1
2
11.2 Coefficientof Quartile Deviation:
Quartile deviation is an absolute measure of dispersion .The relative measure
corresponding to this measure, called the coefficient of quartile deviation, is
calculated as follows.
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑄. 𝐷. =
𝑄3 − 𝑄1
2
𝑄3 + 𝑄1
2
=
𝑄3 − 𝑄1
𝑄3 + 𝑄1
Example--find out the value of quartile deviation and its coefficientfrom the
following data:
Roll No. 1 2 3 4 5 6 7
Marks 20 28 40 12 30 15 50
Solution:
 Marks arranged in ascending order: 12 15 20 28 30 40 50
 𝑄1 = Size of
𝑁+1
4
th item =
7+1
4
= 2𝑛𝑑 item
 Size of 2nd item is 15. Thus 𝑄1 = 15
 𝑄3 = size of 3 (
𝑁+1
4
) 𝑡ℎ item = Size of (
3×8
4
) 𝑡ℎ item =6th item
 Size of 6th item is 40.Thus 𝑄3 = 40
𝑄. 𝐷. =
𝑄3 − 𝑄1
2
=
40 − 15
2
= 12.
 Now we have to find coefficient of Q.D.
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑄. 𝐷. =
𝑄3−𝑄1
𝑄3+𝑄1
=
40−15
40+15
=
25
55
= 0.455

Example--Compute coefficientnofnquartile deviation from the following data:
Marks 10 20 30 40 50 60
No. of Students 4 7 15 8 7 2
Solution:
Calculation of coefficient of quartile deviation
Marks Frequency c.f.
10 4 4
20 7 11
30 15 26
40 8 34
50 7 41
60 2 43
 𝑄1 = size of
𝑁+1
4
𝑡ℎ item =
43+1
4
= 11th item
 Size of 11th item is 20. Thus 𝑄1 = 20
 𝑄3 = Size of 3 (
𝑁+1
4
) 𝑡ℎ item =
3×44
4
= 33rd item
 Size of 33rd item is 40. Thus, 𝑄3 = 40
𝑄. 𝐷. =
𝑄3−𝑄1
2
=
40−20
2
= 10
 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑄. 𝐷. =
𝑄3−𝑄1
𝑄3+𝑄1
=
40−20
40+20
= 0.333
12.1MeanDeviation:
Mean deviation is the arithmetic mean of the difference of a series computed from
any measure of central tendency i.e., Deviations from Mean or Mode or Median.
All the deviation’s absolute values are considered.
The mean deviation is also known as the average deviation. It is the average
difference between the items in a distribution and the median or mean of that

series. Theoritically there is an advantage in taking the deviations from median
because the sum of deviations of items from median is minimum when signs are
ignored.
12.2Computation of MeanDeviation-Individual Observations:
If 𝑋1, 𝑋2, 𝑋3, …… 𝑋 𝑁 are 𝑁 given observations then the deviation about an average
𝐴 is given by
𝑀. 𝐷. =
1
𝑁
∑| 𝑋 − 𝐴| =
1
𝑁
∑| 𝐷| 𝑜𝑟
∑| 𝐷|
𝑁
Where | 𝐷| = | 𝑋 − 𝐴|.
Read as mod (X-A) is the modulus value or absolute value of the deviation
ignoring plus and minus signs.
12.2 Steps for Computation of mean deviation: (Indiviadual Observations)
1. Compute the median of the series.
2. The deviations of items from median ignoring ± signs and denote these
deviations by | 𝐷|.
3. Obtain the total of these deviations,ie. ∑| 𝐷|.
4. Divide the total obntained in step (3) by the total number of observations.
12.3Coefficientof MeanDeviation:
The relative measure corresponding to the mean deviation called the coefficient of
mean deviation is obtained by dividing mean deviation by the particular average
used in computing mean deviation. Thus if mean deviation has been computed
from median, the coefficient of mean deviation shall be obtained by dividing mean
deviation by mean, median or mode.
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑀. 𝐷. =
𝑀.𝐷.
𝑀𝑒𝑎𝑛 ,𝑀𝑒𝑑𝑖𝑎𝑛 𝑜𝑟 𝑀𝑜𝑑𝑒
Example--Calculate mean deviation and coefficient of mean deviation from
the following data:

100 200 300 400 500 600 700
Solution:
Calculation for Mean Deviation
𝑿 | 𝑫| = | 𝑿 − 𝑨|=| 𝑿 − 𝟒𝟎𝟎|
100 300
200 200
300 100
400 0
500 100
600 200
700 300
∑| 𝑫| = 𝟏𝟐𝟎𝟎
 Arithmetic Mean=A =
∑ 𝑋𝑖
𝑁
=
2800
7
= 400
 Mean Deviation= 𝑀. 𝐷. =
1
𝑁
∑| 𝑋 − 𝐴| =
1
𝑁
∑| 𝐷|
𝑀. 𝐷. =
1200
7
= 171.42
 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑀. 𝐷. =
𝑀.𝐷.
𝑀𝑒𝑎𝑛
=
171.42
400
= 0.4285
 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑀. 𝐷 = 0.43
12.2.2 Calculationof MeanDeviation-(Discrete frequency distribution):
In discrete series the formula for calculating mean deviation is
𝑀. 𝐷. =
∑ 𝑓| 𝐷|
𝑁
Where | 𝐷| denote deviation from median ignoring signs.

Steps for calculation:
1. Calculate the median of the series.
2. Take the deviations of the items from median ignoring signs and denote
them by | 𝐷|.
3. Divide the total obtained in step (ii) by the number of observations.This
gives us the value of mean deviation.
Example--Claculate Meandeviation from the following series
X 10 11 12 13 14
f 3 12 18 12 3
Solution:
Calculation of Mean Deviation
X 𝒇 | 𝑫| 𝒇| 𝑫| 𝒄. 𝒇.
10 3 2 6 3
11 12 1 12 15
12 18 0 0 33
13 12 1 12 45
14 3 2 6 48
𝑁 = 48 ∑ 𝑓| 𝐷|=36
 𝑀. 𝐷.=
∑ 𝑓| 𝐷|
𝑁
 Median = Size of
𝑁+1
2
𝑡ℎ item =
48+1
2
= 24.5𝑡ℎ item
 Size of 24.5th item is 12, hence Median = 12
 𝑀. 𝐷. =
∑ 𝑓| 𝐷|
𝑁
=
36
48
= 0.75
12.2.3 Calculationof MeanDeviation– Continuous Frequency distribution

For calculating mean deviation in continuous series the procedureremains the
same as discussed above.Theonly difference is that here we have to obtain the mid
point of the various classes and take deviations of these points from median. The
formula is same, i.e.,
𝑀. 𝐷. =
∑ 𝑓| 𝐷|
𝑁
Example-- Calculate the mean deviation from mean for the following data:
Class Interval 2-4 4-6 6-8 8-10
frequency 6 8 4 2
Solution:
Calculation for Mean Deviation
Class Mid
value(m)
Frequency 𝒇𝒎 | 𝑫|
= | 𝑿 − 𝑨|
𝒇| 𝑫|
2-4 3 6 18 2.2 13.2
4-6 5 8 40 0.2 1.6
6-8 7 4 28 1.8 7.2
8-10 9 2 18 3.8 7.6
Total 20 104 27.6
 𝐴 =
∑ 𝑓𝑚
∑ 𝑓
=
104
20
= 5.2
 𝑀. 𝐷. =
∑ 𝑓| 𝐷|
𝑁
=
27.6
20
= 1.48 (ℎ𝑒𝑟𝑒 𝑁 = ∑ 𝑓 = 20)
 𝑀. 𝐷. = 1.48
12.4Merits of MeanDeviation:
1. It is simple to understand and easy to calculate
2. The computation process is based on all items of the series

3. It is less affected by the extreme items.
4. This measure is flexible, Since it can be calculated from mean, meadian, or
mode.
5. This measure is rigidly defined.
12.5Demerits of Mean Deviation:
1. This measure is not a very accurate measure of dispersion.
2. Not suitable for further mathematical calculation.
3. It is rarely used.
4. Absolute values are considered, mathematically unsound and illogical.
13.1 Standard Deviation:
The famous statistician karl pearsonintroduced the conceptof standard deviation
in 18
This is the most accepted measure of dispersion and also widely used in many
statistical applications. Standard deviation is also referred as root-mean square
deviation or Mean square error. It gives accurate results.
The standard deviation is also denoted by the greek letter ( 𝜎).
13.2Variance:
The term variance was used to describe the square of standard deviation by
R.A.Fisher in 1913.
The conceptof variance is highly important in advanced work where it is possible
to split the total into several parts,each attributable to one of the factors causing
variation in their original series.
Variance is defined as follows
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 =
∑( 𝑋−𝐴)2
𝑁
Thus variance is nothing but the square of the standard deviation
𝑖. 𝑒., 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝜎2
𝑜𝑟 𝜎 = √ 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒

In a frequency distribution where deviations are taken from assumed mean
variance may directly be computed as follows
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = {
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
} × 𝑖2
𝑤ℎ𝑒𝑟𝑒, 𝑑 =
𝑋−𝐴
𝑖
and 𝑖=class interval
13.2. CalculationOfStandard Deviation- Individual Observation
There are two method of calculating standard deviation in an individual
observation:
(i) Direct Method – Deviation taken from actual mean
(ii) Short- cut Method – Deviation taken from assumed mean
13.2.1 (i) DirectMethod:
The following are the steps:
1. Find out actual mean of the given observations.
2. Compute deviation of each observation from the mean (𝑋 − 𝑀𝑒𝑎𝑛).
3. Square the deviations and find out the sum i.e. ( 𝑋 − 𝐴)2
4. Divide the total by the number of observations and take square rootof the
quotient,the value is standard deviation.
𝜎 = √
∑( 𝑋 − 𝐴)2
𝑁
Example-- Calculate the standard deviation from the following data:
𝟏𝟓, 𝟏𝟐, 𝟏𝟕, 𝟏𝟎, 𝟐𝟏, 𝟏𝟖, 𝟏𝟏, 𝟏𝟔
Solution:
Calculation of S.D. from Mean
Values (𝑿) (𝑿 − 𝑨) ( 𝑿 − 𝑨) 𝟐

15 0 0
12 -3 9
17 2 4
10 -5 25
21 6 36
18 3 9
11 -4 16
16 1 1
𝑿 = 𝟏𝟐𝟎 ( 𝑿 − 𝑨) 𝟐
= 𝟏𝟎𝟎
 𝐴 =
∑ 𝑋𝑖
𝑁
=
120
8
= 15
 S.D.=𝜎 = √
∑( 𝑋−𝐴)2
𝑁
= √
100
8
 𝜎 = 3.53
 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝜎2
= 12.46
13.2.2(ii) Short-Cut Method - Deviation takenfrom assumedmean
This method is used when arithmetic mean is fractional value. A deviation from
fractional value leads to tedious task. To save calculation time,we apply this
method the formula is
𝜎 = √
∑ 𝑑2
𝑁
− (
∑ 𝑑
𝑁
)
2
Where, d = deviations from assumed mean =(𝑋 − 𝐴)
𝑁 = Number of Observations
Steps for calculations:
1. Take the deviations of the items from an assumed mean, i.e. obtain
( 𝑋 − 𝐴). Denote these deviations by d.
2. Take the total of these deviations. i.e., obtain ∑ 𝑑.

3. Square these deviations and obtain the total ∑ 𝑑2
.
4. Substitute the values of ∑ 𝑑2
,∑ 𝑑 𝑎𝑛𝑑 𝑁 in the above formula.
Example--Bloodserum cholesterollevels of10 persons are as under:
240, 260, 290,245, 255, 288,272, 263, 277,251
Calculate standard deviation and variance with the help of assumed
mean.
Solution:
Calculation of Standard Deviation
𝑿 𝒅 = (𝑿 − 𝟐𝟔𝟒) 𝒅 𝟐
240 -24 576
260 -4 16
290 +26 676
245 -19 361
255 -9 81
288 +24 576
272 +8 64
263 -1 1
277 +13 169
251 -13 169
∑ 𝑿 =2641 ∑ 𝑑 = +1 ∑ 𝑑2
= 2689
 𝜎 = √
∑ 𝑑2
𝑁
− (
∑ 𝑑
𝑁
)
2
 Now ∑ 𝑑2
= 2689,∑ 𝑑 = +1, 𝑁 = 10

 ∴ 𝜎 = √2689
10
− (
1
10
)
2
 ∴ 𝜎 = √268.9− 0.01
 ∴ 𝜎 = 16.398
 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 𝜎2
= (16.398)2
= 268.894
13.3 Calculationof Standard Deviation– Discrete frequency distribution
For calculating standard deviation in discrete series, any of the following methods
may be applied:
1. Actual Mean Method
2. Assumed Mean Method
3. Step deviation method
13.3.1 ActualMeanMethod:
When this method is applied, deviations are taken from the actual mean,The
formula is applied
𝜎 = √
∑ 𝑓( 𝑋 − 𝐴)2
𝑁
However in practice this method is rarely used becauseif the actual mean is in
fraction the calculations take a lot of time.
(1) Compute mean of the observations
(2) Compute deviation from the mean 𝑑 = ( 𝑋 − 𝐴)
(3) Square the deviations d2
and multiply these values with respective frequencies
f i.e., fd2
(4) Sum the products 𝑓𝑑2
and apply the formula
𝜎 = √
∑ 𝑓𝑑2
𝑁
= √
∑ 𝑓( 𝑋 − 𝐴)2
𝑁

Example-- Compute standard deviation and variance from the following data
Marks 10 20 30 40 50
Frequency 2 8 10 8 2
Solution:
Construct the table to compute the standard deviation
Marks (X) 𝒇 𝒇𝑿 𝒅 = 𝑿 − 𝟑𝟎 𝒅 𝟐 𝒇𝒅 𝟐
10 2 20 -20 400 800
20 8 160 -10 100 800
30 10 300 0 0 0
40 8 320 10 100 800
50 2 100 20 400 800
𝑓 = 30 𝑓𝑋 = 900 𝑓𝑑2
= 3200
 Mean = 𝐴 =
∑ 𝑋𝑖
𝑁
=
900
30
= 30
 𝜎 = √
∑ 𝑓𝑑2
𝑁
= √
∑ 𝑓( 𝑋−𝐴)2
𝑁
 𝜎 = √
3200
30
 𝜎 = √106.66 = 10.325
= 106.66
13.3.2 AssumedMeanMethod:
When this method is used, the following formula is applied:
𝜎 = √
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
Where 𝑑 = (𝑋 − 𝐴)
1. Assume any one of the given value as assumed mean A

2. Compute deviation from the assumed mean (𝑑 = 𝑋 − 𝐴).
3. Multiply these deviations by its frequencies 𝑓𝑑.
4. Square the deviations (𝑑2
) and multiply these values with respective
Frequencies ( 𝑓) i.e., 𝑓𝑑2
5. Sum the products 𝑓𝑑2
and apply the formula 𝜎 = √
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
Example-- Compute standard deviation and variance from the following data
Marks 10 20 30 40 50
Frequency 2 8 10 8 2
Solution: Constructthe table to compute the standard deviation
Marks (X) 𝒇 𝒅 = 𝑿 − 𝟐𝟎 𝒅 𝟐 𝒇𝒅 𝒇𝒅 𝟐
10 2 -10 100 -20 200
20 8 0 0 0 0
30 10 10 100 100 1000
40 8 20 400 160 3200
50 2 30 900 60 1800
𝑓 = 30 𝑓𝑑 = 300 𝑓𝑑2
= 6200
 𝜎 = √
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
 𝑁𝑜𝑤, here ∑ 𝑓𝑑2
= 6200,∑ 𝑓𝑑 = 300, 𝑁 = ∑ 𝑓 = 30
 𝜎 = √6200
30
− (
300
30
)
2
 𝜎 = √206.66 − 100 = √106.66
 𝜎 = 10.325

= 106.66
13.3.3StepdeviationMethod:
When this method is used we take deviations of midpoints from an assumed mean
and divide these deviations by the width of class interval,i.e. ′𝑖′
. In case class
intervals are unequal. We divide the deviations of midpoints by the lowest
common factor and use ‘c’ instead of ′𝑖′
in the formula for calculating standard
deviation.
The formula for calculating standard deviation is:
𝜎 = √
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
× 𝑖
Where 𝑑=
𝑋−𝐴
𝑖
and 𝑖 = class interval
Take a common factor and divide that item by all deviations
1. Assume any one of the given value as assumed mean A
2. Compute deviation from the assumed mean (𝑑 =
𝑋−𝐴
𝑖
).
3. Multiply these deviations by its frequencies 𝑓𝑑.
4. Square the deviations (𝑑2
) and multiply these values with respective
Frequencies ( 𝑓) i.e., 𝑓𝑑2
5. Sum the products 𝑓𝑑2
and apply the formula.
Example-- The annual salaries ofa group of employees are given in the
following table:
Salaries 45000 50000 55000 60000 65000 70000 75000 80000
Number of
persons
3 5 8 7 9 7 4 7
Calculate the standard deviation of the salaries.

Solution:
Calculation of standard deviation
Salaries (X) No. of
persons(𝒇)
𝒅 =
𝑿 − 𝟔𝟎𝟎𝟎𝟎
𝟓𝟎𝟎𝟎
𝒇𝒅 𝒇𝒅 𝟐
45000 3 -3 -9 27
50000 5 -2 -10 20
55000 8 -1 -8 8
60000 7 0 0 0
65000 9 +1 +9 9
70000 7 +2 +14 28
75000 4 +3 +12 36
80000 7 +4 +28 112
𝑵 = 𝟓𝟎 ∑ 𝒇𝒅 =36 ∑ 𝒇𝒅 𝟐
= 𝟐𝟒𝟎
 𝜎 = √
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
× 𝑖
 Now here ∑ 𝑓𝑑 = 36, ∑ 𝑓𝑑2
= 240 ,𝑁 = 50, 𝑖 = 5000
 ∴ 𝜎 = √240
50
− (
36
50
)
2
× 5000
 ∴ 𝜎 = √4.8− 0.5184× 5000
 ∴ 𝜎 = 10346.01
13.4 Calculationof Standard Deviation– Continuous Series.
In continuous series any of the methods discussed above for discrete frequency
distribution can be used. However, in practice it is the step deviation method that is
most used.
The formula is,

𝜎 = √
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
× 𝑖
Where 𝑑 =
𝑚−𝐴
𝑖
, 𝑖 = 𝑐𝑙𝑎𝑠𝑠 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
1. Find the mid point of various classes.
2. Take the deviations of these mid points from an assumed mean and denote
thrse deviations by 𝑑.
3. Wherever possible take a common factor and denote this column by 𝑑.
4. Multiply the frequencies of each class with these deviations and obtain ∑ 𝑓𝑑.
5. Square the deviations and multiply them with the respective frequencies of
each class and obtain ∑ 𝑓𝑑2
.
Thus the only difference in procedurein case of continuous series is to find mid-
points of the various classes.
Example--Calculate Standard deviation from the following data:
Age 20-25 25-30 30-35 35-40 40-45 45-50
No. of Persons 170 110 80 45 40 35
Solution:
Take assumed average = 32.5
Calculation of Standard deviation
Age Mid
point(m)
No. of
Persons (f)
𝒅 =
𝒎 − 𝟑𝟐. 𝟓
𝟓
𝒇𝒅 𝒇𝒅 𝟐
20-25 22.5 170 -2 -340 680
25-30 27.5 110 -1 -110 110
30-35 32.5 80 0 0 0
35-40 37.5 45 +1 +45 45

40-45 42.5 40 +2 +80 160
45-50 47.5 35 +3 +105 315
𝑵 = 𝟒𝟖𝟎 ∑ 𝒇𝒅 =-220 ∑ 𝒇𝒅 𝟐
=1310
 𝜎 = √
∑ 𝑓𝑑2
𝑁
− (
∑ 𝑓𝑑
𝑁
)
2
× 𝑖
 𝜎 = √1310
480
− (
−220
480
)
2
× 5
 𝜎 = √2.279− 0.21 × 5
 𝜎 = √2.519× 5
 𝜎 = 1.587× 5 = 7.936
14.1Coefficientof Variation:
The standard deviation discussed aboveis an absolute measure of dispersion. The
corresponding relative measure is known as the Coefficient of variation.
This measure developed by karl pearson is the most commonly used measure of
relative variation. It is used in such problems where we cant to compare the
variability of two or more than two series.
That series for which the coefficient of variation is greater is said to be more
variable or conversely less consistent,less uniform,less srable .
On the other hand , the series for which coefficient of variation is less is said to be
less variable or more consistent,more uniform,more stable .
Coefficient of variation is denoted by C.V. and is obtained as follows:
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝐶. 𝑉. =
𝜎
𝐴
× 100
Where, 𝜎 = standard deviation, 𝐴 = 𝑎𝑟𝑖𝑡ℎ𝑚𝑒𝑡𝑖𝑐 𝑀𝑒𝑎𝑛

Example--From the prices of shares of 𝑿 and 𝒀 below find out which is more
stable in value:
𝑿 35 54 52 53 56 58 52 50 51 49
𝒀 108 107 105 105 106 107 104 103 104 101
Solution:
In order to find out which shares are more stable, we have to compare coefficient
of variations.
Calculation of Coefficient of Variation
𝑿 𝒙 = (𝑿 − 𝑨) 𝒙 𝟐 𝒀 𝒚 = (𝒀 − 𝑨) 𝒚 𝟐
35 -16 256 108 +3 9
54 +3 9 107 +2 4
52 +1 1 105 0 0
53 +2 4 105 0 0
56 +5 25 106 +1 1
58 +7 49 107 +2 4
52 +1 1 104 -1 1
50 -1 1 103 -2 4
51 0 0 104 -1 1
49 -2 4 101 -4 16
∑ 𝑿 =510 ∑ 𝒙 =0 ∑ 𝒙 𝟐
=350 ∑ 𝒀 =1050 ∑ 𝒚=0 ∑ 𝒚 𝟐
=40
Coefficientof variation 𝑿:
 𝐶. 𝑉. =
𝜎
𝐴
× 100
 Here 𝐴 =
∑ 𝑋
𝑁
=
510
10
= 51

 𝜎 = √
∑ 𝑥2
𝑁
= √
350
10
= 5.916
 ∴ 𝐶. 𝑉. =
𝜎
𝐴
× 100 =
5.196
51
× 100
 ∴ 𝐶. 𝑉. = 11.6
Coefficientof variation 𝒀:
 𝐶. 𝑉. =
𝜎
𝐴
× 100
 Here 𝐴 =
∑ 𝑦
𝑁
=
1050
10
= 105
 𝜎 = √
∑ 𝑦2
𝑁
= √
40
10
= 2
 ∴ 𝐶. 𝑉. =
𝜎
𝐴
× 100 =
2
105
× 100
 ∴ 𝐶. 𝑉. = 1.905
Since Coefficient of variation is much less in case of shares 𝑌, Hence they are
more stable in value.
EXERCISE
Q-1 Evaluate the following Questions:
(1). Find the Mean, Median and Mode of the following data.
Cost 10-20 20-30 30-40 40-50 50-60
Items in a
group
4 5 3 6 3
(2).Find the Mean, Median and Mode of the following distribution:
Class
Interval
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 5 9 8 12 28 20 12 11
(3).On his first 5 biology tests, Bob received the following scores: 72, 86, 92, 63,
and 77. What test scoremust Bob earn on his sixth test so that his average (mean
score)for all six tests will be 80? Show how you arrived at your answer.
Q-2 Evaluate the following Questions:

(1). Calculate quartile deviation and it coefficient from the following data:
Wages in Rupees per
Day
Less then
35
35-
37
38-40 41-43 Over 43
Number of Wage
earners
14 62 99 18 7
(2). Calculate Mean Deviation for the following data
Size 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 7 12 18 25 16 14 8
(3). The owner of a restaurant is interested in how much people spend at the
restaurant. He examines 10 randomly selected receipts for parties of four and write
down the following data: 44, 50, 38, 96, 42, 47, 40,39, 46, 50.
Find mean, standard deviation and variance.
15. Referenc Book andWebsite Name:
1. http://www.emathzone.com/tutorials/basic-statistics/collection-of-statistical-
data.html
2. http://wizznotes.com/mathematics/statistics/class-limits-boundaries-and-
intervals
3. http://tistats.com/definitions/class-width/
4. http://mathespk.blogspot.in/2011/10/frequency-density.html
5. http://www.mathsisfun.com/definitions/relative-frequency.html
6. http://www.mathsisfun.com/definitions/cumulative-frequency.html
7. http://www.mathgoodies.com/lessons/vol8/mode.html
8. Statistical Methods by S.P.Gupta.

MCA_UNIT-3_Computer Oriented Numerical Statistical Methods

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