Statistics for Engineers
Module - I
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Overview
1
Syllabus
Module - I
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Overview
1
Syllabus
2
Introduction
Module - I
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Overview
1
Syllabus
2
Introduction
3
Module 1
Measures of Central Tendancy
Measures of Dispersion
Moments
Skewness
Kurtosis
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Syllabus
1
Introduction to Statistics
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Syllabus
1
Introduction to Statistics
2
Random variables
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Syllabus
1
Introduction to Statistics
2
Random variables
3
Correlation and Regression
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Syllabus
1
Introduction to Statistics
2
Random variables
3
Correlation and Regression
4
Probability Distributions
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Syllabus
1
Introduction to Statistics
2
Random variables
3
Correlation and Regression
4
Probability Distributions
5
Hypothesis Testing - I
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Syllabus
1
Introduction to Statistics
2
Random variables
3
Correlation and Regression
4
Probability Distributions
5
Hypothesis Testing - I
6
Hypothesis Testing - II
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Syllabus
1
Introduction to Statistics
2
Random variables
3
Correlation and Regression
4
Probability Distributions
5
Hypothesis Testing - I
6
Hypothesis Testing - II
7
Reliability
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Books
Text books
1
R. E. Walpole, R. H. Mayers, S.Probability and Statistics for
Engineers and Scientists, 9th Edition, Pearson Education (2012)
2
Douglas C. Montgomery, George C. Runger, Applied Statistics and
Probability for Engineers, John Wiley & Sons, 5th Edition (2010).
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Books
Text books
1
R. E. Walpole, R. H. Mayers, S.Probability and Statistics for
Engineers and Scientists, 9th Edition, Pearson Education (2012)
2
Douglas C. Montgomery, George C. Runger, Applied Statistics and
Probability for Engineers, John Wiley & Sons, 5th Edition (2010).
Reference Books
1
E. Balagurusamy, Reliability Engineering, Tata McGraw Hill, Tenth
reprint 2010.
2
J.L Devore, Probability and Statistics , 8th Edition, Brooks/Cole,
Cengage Learning (2012)
3
R. A. Johnson, Miller & Freund’s, Probability and Statistics for
Engineers, 8th Edition, PHI (2010).
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Introduction
Data
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Introduction
Data - collection of any number of related observations
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Introduction
Data - collection of any number of related observations
Statistics deals with collection, presentation, analysis, and
interpretation of the numerical data
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Introduction
Data - collection of any number of related observations
Statistics deals with collection, presentation, analysis, and
interpretation of the numerical data
There are two major types of statistics: (a) Descriptive Statistics, (b)
Inferential Statistics
Descriptive Statistics - consist of methods of organizing and
summarizing information. (includes construction of graphs, charts,
tables and calculation of mean, median, mode, measures of variation.
Inferential Statistics - consist of methods of drawing conclusions
based on the information. (includes estimation, hypothesis testing)
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Illustration
Example
Consider the event of tossing dice. The dice is rolled 100 times. The
results are 5,4,5,6,6,6,1,1,1,1,1,2,5,5,3,2,1,4,4,4,4,2,2,6,6,6,6,2,...
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Illustration
Example
Consider the event of tossing dice. The dice is rolled 100 times. The
results are 5,4,5,6,6,6,1,1,1,1,1,2,5,5,3,2,1,4,4,4,4,2,2,6,6,6,6,2,...
Descriptive statistics is used for grouping the data to the following table
Outcome of the roll
1
2
3
4
5
6
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Frequencies
10
20
18
16
11
25
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Illustration
Example
Consider the event of tossing dice. The dice is rolled 100 times. The
results are 5,4,5,6,6,6,1,1,1,1,1,2,5,5,3,2,1,4,4,4,4,2,2,6,6,6,6,2,...
Descriptive statistics is used for grouping the data to the following table
Outcome of the roll
1
2
3
4
5
6
Frequencies
10
20
18
16
11
25
Inferential statistics can be used to verify whether the dice is a fair or not.
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Statistics
Population
is the set of observations or individuals corresponding to an entire
collection of units for which inferences are to be made.
Example: The books in our library, The students of VIT.
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Statistics
Population
is the set of observations or individuals corresponding to an entire
collection of units for which inferences are to be made.
Example: The books in our library, The students of VIT.
Sample
is a subset of a population that are actually collected in the course of an
investigation. Example: The collection of Statistics books in our library,
The students of our class.
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Statistics
Example
Let the blood types of 40 persons are as follows: O, O, A, B, A, O, A, A,
A, O, B, O, B, O, O, A, O, O, A, A, A, A, AB, A, B, A, A, O, O, A, O, O,
A, A, A, O, A, O, O, AB.
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Statistics
Example
Let the blood types of 40 persons are as follows: O, O, A, B, A, O, A, A,
A, O, B, O, B, O, O, A, O, O, A, A, A, A, AB, A, B, A, A, O, O, A, O, O,
A, A, A, O, A, O, O, AB.
Blood type
O
A
B
AB
Total
Frequencies
16
18
4
2
16
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Statistics
Example
Let the blood types of 40 persons are as follows: O, O, A, B, A, O, A, A,
A, O, B, O, B, O, O, A, O, O, A, A, A, A, AB, A, B, A, A, O, O, A, O, O,
A, A, A, O, A, O, O, AB.
Blood type
O
A
B
AB
Total
Frequencies
16
18
4
2
16
Try to graph this data (pie chart and bar graph)
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Measures of Central Tendancy
Data are classified as
Individual observations or raw data
Discrete data
Continuous data
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Measures of Central Tendancy
Data are classified as
Individual observations or raw data
Discrete data
Continuous data
The Central measures are
Mean
Median
Mode
Geometric mean
Harmonic mean
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Mean (Individual observations or raw data)
Mean (Direct method)
X̄ =
X1 +X2 +...+XN
N
=
P
Xi
N
where N is the total number of observations.
Example
Given the monthly income of 10 employees in an office,
1780,1760,1690,1750,1840,1920,1100,1810,1050,1950
Find the mean of their monthly income (or Average).
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Mean (Individual observations or raw data)
Mean (Direct method)
X̄ =
X1 +X2 +...+XN
N
=
P
Xi
N
where N is the total number of observations.
Example
Given the monthly income of 10 employees in an office,
1780,1760,1690,1750,1840,1920,1100,1810,1050,1950
Find the mean of their monthly income (or Average). Mean=1665
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Mean (Individual observations or raw data)
Mean (Direct method)
X̄ =
X1 +X2 +...+XN
N
=
P
Xi
N
where N is the total number of observations.
Example
Given the monthly income of 10 employees in an office,
1780,1760,1690,1750,1840,1920,1100,1810,1050,1950
Find the mean of their monthly income (or Average). Mean=1665
Mean (Short-cut method)
Taken an assumed mean A
Find the deviations:
d =X −A
P
X̄ = A +
d
N
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Mean (Discrete data)
Direct method
P
X̄ =
Short-cut method
fX
N
P
fd
N
where A is the assumed mean, d = X − A, N is the total number of
observations.
X̄ = A +
Problem
1. From the following data of the marks obtained by 60 students of a
class. Calculate the arithmetic mean.
Marks
No. of students
20
8
30
12
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40
20
50
10
60
6
70
4
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Mean (Discrete data)
Direct method
P
X̄ =
Short-cut method
fX
N
P
fd
N
where A is the assumed mean, d = X − A, N is the total number of
observations.
X̄ = A +
Problem
1. From the following data of the marks obtained by 60 students of a
class. Calculate the arithmetic mean.
Marks
No. of students
20
8
30
12
40
20
50
10
60
6
70
4
Arithmetic mean=41
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Mean (Continuous data)
Direct method
P
fX
)
N
P
fd
)∗h
X̄ = A + (
N
X̄ = (
Short-cut method
where A is the assumed mean, d = X −A
h , N is the total number of
observations, h is the class size, X is the mid value of each class
interval.
Example
From the following data, compute arithmetic mean
Marks
No. of students
0-10
5
10-20
10
20-30
25
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30-40
30
40-50
20
50-60
10
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Mean (Continuous data)
Direct method
P
fX
)
N
P
fd
)∗h
X̄ = A + (
N
X̄ = (
Short-cut method
where A is the assumed mean, d = X −A
h , N is the total number of
observations, h is the class size, X is the mid value of each class
interval.
Example
From the following data, compute arithmetic mean
Marks
No. of students
0-10
5
10-20
10
20-30
25
30-40
30
40-50
20
50-60
10
X̄ = 33
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Problem
Compute mean for the following data:
Marks
No. of students
0-10
5
10-30
12
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30-60
25
60-100
8
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Median
Arrange the data in ascending or descending order.
For raw data:
If N is odd, then Median is X N+1
2
If N is even, then Median is
X N +X N +1
2
2
2
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Median
Arrange the data in ascending or descending order.
For raw data:
If N is odd, then Median is X N+1
2
If N is even, then Median is
X N +X N +1
2
2
2
Discrete data:
Find the cummulative frequencies (c.f.)
Identify the Median as average of N2 th observation and N2 + 1 th
observation (if N is even) or N+1
2 th observation (if N is odd) using c.f.
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Median
Arrange the data in ascending or descending order.
For raw data:
If N is odd, then Median is X N+1
2
If N is even, then Median is
X N +X N +1
2
2
2
Discrete data:
Find the cummulative frequencies (c.f.)
Identify the Median as average of N2 th observation and N2 + 1 th
observation (if N is even) or N+1
2 th observation (if N is odd) using c.f.
Continuous data:
Find the cummulative frequencies (c.f.)
Identify the Median class as the class that contains
N
2
N
2
th observation
−c.f .
)
f
Median = L + (
∗ h where L is the lower limit of the median
class, c.f. is the cummulative frequency of the class preceding to the
median class, f is the frequency of the median class, and h is the
median class size.
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Problems
1. Compute median for the following data:
1100, 1150, 1080, 1120, 1200, 1160, 1400
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Problems
1. Compute median for the following data:
1100, 1150, 1080, 1120, 1200, 1160, 1400
Median= 1150
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Problems
1. Compute median for the following data:
1100, 1150, 1080, 1120, 1200, 1160, 1400
Median= 1150
2. Compute median for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
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800
16
2000
20
2500
6
1800
30
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Problems
1. Compute median for the following data:
1100, 1150, 1080, 1120, 1200, 1160, 1400
Median= 1150
2. Compute median for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
800
16
2000
20
2500
6
1800
30
Median = 1500
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Problems
1. Compute median for the following data:
1100, 1150, 1080, 1120, 1200, 1160, 1400
Median= 1150
2. Compute median for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
800
16
2000
20
2500
6
1800
30
Median = 1500
3. Calculate the median for the following data:
Marks
No. of students
0-10
5
10-30
12
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30-60
25
60-100
8
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Problems
1. Compute median for the following data:
1100, 1150, 1080, 1120, 1200, 1160, 1400
Median= 1150
2. Compute median for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
800
16
2000
20
2500
6
1800
30
Median = 1500
3. Calculate the median for the following data:
Marks
No. of students
0-10
5
10-30
12
30-60
25
60-100
8
Median = 39.6
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Problems
4. An incomplete frequency distribution is given below.
Variable
frequency
10-20
12
20-30
30
30-40
f1
40-50
65
50-60
f2
60-70
25
70-80
18
Total
229
Given that the median value is 46, determine the missing frequencies.
f1 = 34, f2 = 45
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Mode
Mode
The observation with maximum frequency
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Mode
Mode
The observation with maximum frequency
Discrete data:
If the frequency distribution is regular, then the observation with
maximum frequency is the mode.
For irregular frequency distribution, mode is calculated by the method
of grouping.
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Mode
Mode
The observation with maximum frequency
Discrete data:
If the frequency distribution is regular, then the observation with
maximum frequency is the mode.
For irregular frequency distribution, mode is calculated by the method
of grouping.
Continuous data:
If the frequency distribution is regular, then the Class interval with
maximum frequency is the modal class. Then Mode
1 −f0
= L + ( 2f1f−f
) ∗ h where L is the lower limit of the modal class, f1 is
0 −f2
the frequency of the modal class, f0 is the frequency of the class
preceding the modal class, f2 is the frequency of the class succeeding
the modal class, and h is the modal class size.
For irregular frequency distribution, mode is calculated by the method
of grouping.
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Problems
1. Compute mode for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
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800
16
2000
20
2500
6
1800
30
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Problems
1. Compute mode for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
800
16
2000
20
2500
6
1800
30
Mode = 1800
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Problems
1. Compute mode for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
800
16
2000
20
2500
6
1800
30
30-60
25
60-100
8
Mode = 1800
2. Calculate the mode for the following data:
Marks
No. of students
0-10
5
10-30
12
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Problems
1. Compute mode for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
800
16
2000
20
2500
6
1800
30
30-60
25
60-100
8
Mode = 1800
2. Calculate the mode for the following data:
Marks
No. of students
0-10
5
10-30
12
Modal class is 30-60. From the problem, f1 = 25, f0 = 12, f2 = 8
Mode = 43
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4. The median and mode are Rs.33.50 and Rs.34 respectively for the
following data.
Wages
frequency
0-10
4
10-20
16
20-30
f1
30-40
f2
40-50
f3
50-60
6
60-70
4
Total
230
5. Find mean, median, and mode for the following data:
CI
f
20-40
8
40-60
12
60-80
20
CI
f
80-100
30
160-180
7
100-120
40
120-140
35
140-160
18
180-200
5
Mean=108.5, Median=108.75, Mode=118.3
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Problems on Inclusive intervals
Exclusive intervals: Lower limit included but upper limit excluded
Inclusive intervals: Lower and upper limits are included
6. Calculate the mean, median, and mode for the following data:
CI
f
0-9
3
10-19
15
20-29
10
30-39
8
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40-49
3
50-59
1
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Problems on Inclusive intervals
Exclusive intervals: Lower limit included but upper limit excluded
Inclusive intervals: Lower and upper limits are included
6. Calculate the mean, median, and mode for the following data:
CI
f
0-9
3
10-19
15
20-29
10
30-39
8
40-49
3
50-59
1
7. Compute mean and median for the following data
Height (in cm)
No. of students
< 140
4
< 145
11
< 150
29
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< 155
40
< 160
46
< 165
51
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Problems on Inclusive intervals
Exclusive intervals: Lower limit included but upper limit excluded
Inclusive intervals: Lower and upper limits are included
6. Calculate the mean, median, and mode for the following data:
CI
f
0-9
3
10-19
15
20-29
10
30-39
8
40-49
3
50-59
1
7. Compute mean and median for the following data
Height (in cm)
No. of students
< 140
4
< 145
11
< 150
29
< 155
40
< 160
46
< 165
51
The empirical relationship is Mode= 3 Median - 2 Mean
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Geometric Mean
Geometric mean isPthe nth root of product of n observations.
(xi )
G.M. = Antilog ( log
)
(raw data) where N is the total number of
N
observations.
P
(xi )
)
( Discrete or Continuous data) where N
G.M. = Antilog ( fi log
N
is the total frequency.
Problem
Calculate G.M. for 125, 1462, 38, 7, 0.22, 0.08, 12.75, 0.5
G.M.=6.952
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Harmonic Mean
Harmonic mean is the reciprocal of the arithmetic mean of the reciprocal
of observations.
H.M. = PN 1
(raw data)
xi
H.M. =
PN 1
fi x
(Discrete or Continuous data)
i
Problem
Find H.M. of 2574, 475, 75, 5, 0.8, 0.08, 0.005, 0.0009.
H.M. = 0.006
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Measures of dispersion
Range
Quartile Deviation
Mean Deviation
Standard Deviation
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Range
Range is the difference between max observation and min observation
Problems
1. Find the range of 1100, 1150, 1080, 1120, 1200, 1160, 1400.
Range= 1400-1080=320
2. Find the range of the following data:
X
f
1
3
2
8
3
15
4
23
5
35
6
40
7
32
8
28
9
20
10
45
11
14
12
6
Range=12-1=11
3. Find the range of the following data:
Marks
No. of students
0-10
5
10-30
12
30-60
25
60-100
8
Range=100-0=100
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Quartile Deviation
Order the given data and find the cummulative frequencies.
First Quartile = Q1 = L + (
the first quartile class.
N
−c.f .
4
f
) ∗ h where L is the lower limit of
Second Quartile (Q2 ) coincides with Median
Third Quartile = Q3 = L + (
the third quartile class
Quartile Deviation =
3N
−c.f .
4
f
) ∗ h where L is the lower limit of
Q3 −Q1
2
Inter-quartile range = Q3 − Q1
Coefficient of Quartile Deviation =
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Q3 −Q1
Q3 +Q1
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Problems
1. Find all the quartiles, quartile deviation, and coefficient of quartile
deviation for the following data:
Marks
No. of students
0-10
5
10-30
12
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30-60
25
60-100
8
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Mean Deviation
P
For raw data, Mean deviation = N|Di | where N is the total number
of observations and Di = X − A, A is a central measure (Mean or
Median or Mode)
P
For discrete or continuous data, Mean deviation = fNi |Di | where N is
the total frequency.
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Mean Deviation
P
For raw data, Mean deviation = N|Di | where N is the total number
of observations and Di = X − A, A is a central measure (Mean or
Median or Mode)
P
For discrete or continuous data, Mean deviation = fNi |Di | where N is
the total frequency.
Problems
1. Calculate the mean deviation about median for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
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800
16
2000
20
2500
6
1800
30
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Mean Deviation
P
For raw data, Mean deviation = N|Di | where N is the total number
of observations and Di = X − A, A is a central measure (Mean or
Median or Mode)
P
For discrete or continuous data, Mean deviation = fNi |Di | where N is
the total frequency.
Problems
1. Calculate the mean deviation about median for the following data:
Income (Rs.)
No. of persons
1000
24
1500
26
800
16
2000
20
2500
6
1800
30
2. Calculate the mean deviation about median for the following data:
Marks
No. of students
0-10
5
10-30
12
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30-60
25
60-100
8
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Standard Deviation
Raw data: σ =
observations.
qP
(xi )2
N
−(
P
xi 2
N )
Discrete or Continuous data: σ =
the total frequency.
where N is the total number of
qP
fi (xi )2
N
−(
P
fi xi 2
N )
where N is
qP
P
fi (di )2
fi di 2
−
(
Discrete or Continuous data: σ = (
N
N ) ) ∗ h where
N is the total frequency and di =
xi −A
h
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Standard Deviation
Raw data: σ =
observations.
qP
(xi )2
N
−(
P
xi 2
N )
Discrete or Continuous data: σ =
the total frequency.
where N is the total number of
qP
fi (xi )2
N
−(
P
fi xi 2
N )
where N is
qP
P
fi (di )2
fi di 2
−
(
Discrete or Continuous data: σ = (
N
N ) ) ∗ h where
N is the total frequency and di =
xi −A
h
Problems
1. Calculate standard deviation for the following data:
X
f
10
3
11
12
12
18
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13
12
14
3
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Problems
2. From the following data, compute standard deviation
Marks
No. of students
0-10
5
10-20
10
20-30
25
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30-40
30
40-50
20
50-60
10
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Problems
2. From the following data, compute standard deviation
Marks
No. of students
0-10
5
10-20
10
20-30
25
30-40
30
40-50
20
50-60
10
Coefficient of Variation
C.V =
σ
x̄
∗ 100%
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Problems
1. Goals scored by two teams in a foot ball season were as follows:
No. of goals scored
No. of matches (Team A)
No. of matches (Team B)
0
15
20
1
10
10
2
7
5
3
5
4
4
3
2
5
2
1
State which team is more consistent. (A is more consistent.)
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Problems
1. Goals scored by two teams in a foot ball season were as follows:
No. of goals scored
No. of matches (Team A)
No. of matches (Team B)
0
15
20
1
10
10
2
7
5
3
5
4
4
3
2
5
2
1
State which team is more consistent. (A is more consistent.)
2. Two brands of tyres are tested with the following results
Life ( in ’000 miles)
No. of tyres (Brand X)
No. of tyres (Brand Y)
20-25
1
0
25-30
22
24
30-35
64
76
35-40
10
0
40-45
3
0
(a) Which brand of tyres have greater average life?
(b) Compare the variability and state which brand of tyres would you use
on your fleet of trucks?
X̄ = 32.1, σX = 3.441, C .VX = 10.72%, Ȳ = 31.3, σY = 2.136, C .VY =
6.824%, (a) X has greater average life (b) Y is more consistent
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Moments
Central moment
The A.M of various powers of the deviations about mean
P
(x − x̄)r
raw data
µr =
N
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Moments
Central moment
The A.M of various powers of the deviations about mean
P
(x − x̄)r
raw data
µr =
N
P
(f (x − x̄)r )
P
frequency distribution
µr =
f
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Moments
Central moment
The A.M of various powers of the deviations about mean
P
(x − x̄)r
raw data
µr =
N
P
(f (x − x̄)r )
P
frequency distribution
µr =
f
Moments about Assumed mean or Origin
0
µr =
P
(x − A)r
=
N
dr
N
P
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raw data
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Moments
Central moment
The A.M of various powers of the deviations about mean
P
(x − x̄)r
raw data
µr =
N
P
(f (x − x̄)r )
P
frequency distribution
µr =
f
Moments about Assumed mean or Origin
P r
d
(x − A)r
=
µr =
N
N
P
P
(f (x − A)r )
(fd r )
0
P
µr =
= ( P ) ∗ hr
f
f
0
P
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raw data
frequency distribution
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Calculation of central moments using moments about
origin
µ1 = 0
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Calculation of central moments using moments about
origin
µ1 = 0
0
0
µ2 = µ2 − (µ1 )2 = σ 2
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Calculation of central moments using moments about
origin
µ1 = 0
0
0
µ2 = µ2 − (µ1 )2 = σ 2
0
0
0
0
µ3 = µ3 − 3µ1 µ2 + 2(µ1 )3
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Calculation of central moments using moments about
origin
µ1 = 0
0
0
µ2 = µ2 − (µ1 )2 = σ 2
0
0
0
0
µ3 = µ3 − 3µ1 µ2 + 2(µ1 )3
0
0
0
0
0
0
µ4 = µ4 − 4µ1 µ3 + 6(µ1 )2 µ2 − 3(µ1 )4
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Calculation of central moments using moments about
origin
µ1 = 0
0
0
µ2 = µ2 − (µ1 )2 = σ 2
0
0
0
0
µ3 = µ3 − 3µ1 µ2 + 2(µ1 )3
0
0
0
0
0
0
µ4 = µ4 − 4µ1 µ3 + 6(µ1 )2 µ2 − 3(µ1 )4
1. Calculate the first four central moments from the following data :
Class Interval
Frequency
0-10
1
10-20
3
Module - I
20-30
4
30-40
2
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Calculation of central moments using moments about
origin
µ1 = 0
0
0
µ2 = µ2 − (µ1 )2 = σ 2
0
0
0
0
µ3 = µ3 − 3µ1 µ2 + 2(µ1 )3
0
0
0
0
0
0
µ4 = µ4 − 4µ1 µ3 + 6(µ1 )2 µ2 − 3(µ1 )4
1. Calculate the first four central moments from the following data :
Class Interval
Frequency
0
0
0-10
1
10-20
3
0
20-30
4
30-40
2
0
µ1 = 7, µ2 = 130, µ3 = 1900, µ4 = 37000
µ1 = 0, µ2 = 81, µ3 = −144, µ4 = 14817
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Problems
2. Calculate the first four central moments from the following data :
Marks (above)
No. of students
0
150
10
140
20
100
30
80
40
80
50
70
60
30
70
14
80
0
Table 1: Mark distribution of students
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Problems
2. Calculate the first four central moments from the following data :
Marks (above)
No. of students
0
150
10
140
20
100
30
80
40
80
50
70
60
30
70
14
80
0
Table 1: Mark distribution of students
CI
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
f
10
40
20
0
10
40
16
14
P
f = 150
0
x
5
15
25
35
45
55
65
75
d = x−35
10
-3
-2
-1
0
1
2
3
4
0
fd
-30
-80
-20
0
10
80
48
56
P
fd = 64
0
fd 2
90
160
20
0
10
160
144
224
P 2
fd = 808
fd 3
-90
-320
-20
0
10
320
432
896
P 3
fd = 1228
fd 4
810
640
20
0
10
640
1296
3584
P 4
fd = 7000
0
µ1 = 4.27, µ2 = 538.67, µ3 = 8186.67, µ4 = 466666.67
µ1 = 0, µ2 = 520.44, µ3 = 1442.02, µ4 = 384770.13
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Skewness
Lack of symmetry (measures the assymetry of the distribution)
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Skewness
Lack of symmetry (measures the assymetry of the distribution)
tells the degree and direction of departure from symmetry
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Skewness
Lack of symmetry (measures the assymetry of the distribution)
tells the degree and direction of departure from symmetry
if the distribution has longer tail towards right - positive skewness.
(Mean > Median > Mode)
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Skewness
Lack of symmetry (measures the assymetry of the distribution)
tells the degree and direction of departure from symmetry
if the distribution has longer tail towards right - positive skewness.
(Mean > Median > Mode)
if the distribution has longer tail towards left - negative skewness.
(Mode > Median > Mean)
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Skewness
Measure of skewness using moments
β1 =
µ23
,
µ32
γ1 =
p
β1
If µ3 > 0 , then positive skewness, else negative skewness.
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Kurtosis
Kurtosis is a measure of peakness.
measure of Kurtosis
β2 =
µ4
,
(µ2 )2
γ2 = β2 − 3
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Kurtosis
Kurtosis is a measure of peakness.
measure of Kurtosis
β2 =
µ4
,
(µ2 )2
γ2 = β2 − 3
If γ2 = 0, then the distribution is Normal or Mesokurtic.
If γ2 > 0, then the distribution is Leptokurtic.
If γ2 < 0, then the distribution is Platykurtic.
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Problems
1. Calculate the measure of Skewness ( using moments) and measure of
Kurtosis from the following data :
Class Interval
Frequency
0-10
1
10-20
3
Module - I
20-30
4
30-40
2
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Problems
1. Calculate the measure of Skewness ( using moments) and measure of
Kurtosis from the following data :
Class Interval
Frequency
0
0
0-10
1
10-20
3
0
20-30
4
30-40
2
0
µ1 = 7, µ2 = 130, µ3 = 1900, µ4 = 37000
µ1 = 0, µ2 = 81, µ3 = −144, µ4 = 14817
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Problems
1. Calculate the measure of Skewness ( using moments) and measure of
Kurtosis from the following data :
Class Interval
Frequency
0
0-10
1
0
10-20
3
0
20-30
4
30-40
2
0
µ1 = 7, µ2 = 130, µ3 = 1900, µ4 = 37000
µ1 = 0, µ2 = 81, µ3 = −144, µ4 = 14817
√
µ2
Measure of Skewness: β1 = µ33 = 0.039,
γ1 = β1 = 0.1975
2
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Problems
1. Calculate the measure of Skewness ( using moments) and measure of
Kurtosis from the following data :
Class Interval
Frequency
0
0-10
1
0
10-20
3
0
20-30
4
30-40
2
0
µ1 = 7, µ2 = 130, µ3 = 1900, µ4 = 37000
µ1 = 0, µ2 = 81, µ3 = −144, µ4 = 14817
√
µ2
Measure of Skewness: β1 = µ33 = 0.039,
γ1 = β1 = 0.1975
Measure of Kurtosis: β2 =
µ4
µ22
2
= 2.2583,
Module - I
γ2 = β2 − 3 = −0.7417
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Problems
1. Calculate the measure of Skewness ( using moments) and measure of
Kurtosis from the following data :
Class Interval
Frequency
0
0-10
1
0
10-20
3
0
20-30
4
30-40
2
0
µ1 = 7, µ2 = 130, µ3 = 1900, µ4 = 37000
µ1 = 0, µ2 = 81, µ3 = −144, µ4 = 14817
√
µ2
Measure of Skewness: β1 = µ33 = 0.039,
γ1 = β1 = 0.1975
Measure of Kurtosis: β2 =
µ4
µ22
2
= 2.2583,
γ2 = β2 − 3 = −0.7417
The distribution is Platykurtic and negative skewed (sign of µ3 ).
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Problems
2. The first three moments of a distribution about the value 3 of a
variable are 2,10 and 30 respectively. Obtain mean,µ2 , µ3 , and hence β1 .
Comment upon the nature of skewness. Also determine the first three
moments about zero.
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Problems
2. The first three moments of a distribution about the value 3 of a
variable are 2,10 and 30 respectively. Obtain mean,µ2 , µ3 , and hence β1 .
Comment upon the nature of skewness. Also determine the first three
moments about zero. (x̄ = 5, µ1 = 0, µ2 = 6, µ3 = −14, moments about
zero are 5, 31, 201)
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Problems
2. The first three moments of a distribution about the value 3 of a
variable are 2,10 and 30 respectively. Obtain mean,µ2 , µ3 , and hence β1 .
Comment upon the nature of skewness. Also determine the first three
moments about zero. (x̄ = 5, µ1 = 0, µ2 = 6, µ3 = −14, moments about
zero are 5, 31, 201)
3. The standard deviation of a symmetrical distribution is 3. What must
be the value of the fourth moment about the mean in order that the
distribution is mesokurtic?
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Problems
2. The first three moments of a distribution about the value 3 of a
variable are 2,10 and 30 respectively. Obtain mean,µ2 , µ3 , and hence β1 .
Comment upon the nature of skewness. Also determine the first three
moments about zero. (x̄ = 5, µ1 = 0, µ2 = 6, µ3 = −14, moments about
zero are 5, 31, 201)
3. The standard deviation of a symmetrical distribution is 3. What must
be the value of the fourth moment about the mean in order that the
distribution is mesokurtic?(µ4 = 243)
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Problems
2. The first three moments of a distribution about the value 3 of a
variable are 2,10 and 30 respectively. Obtain mean,µ2 , µ3 , and hence β1 .
Comment upon the nature of skewness. Also determine the first three
moments about zero. (x̄ = 5, µ1 = 0, µ2 = 6, µ3 = −14, moments about
zero are 5, 31, 201)
3. The standard deviation of a symmetrical distribution is 3. What must
be the value of the fourth moment about the mean in order that the
distribution is mesokurtic?(µ4 = 243)
4. The mean and SD of a set of 100 observations were worked out as 40
and 5 respectively. But by mistake a value 50 was taken in place of 40 for
one observation. Calculate the correct mean and SD.
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Problems
2. The first three moments of a distribution about the value 3 of a
variable are 2,10 and 30 respectively. Obtain mean,µ2 , µ3 , and hence β1 .
Comment upon the nature of skewness. Also determine the first three
moments about zero. (x̄ = 5, µ1 = 0, µ2 = 6, µ3 = −14, moments about
zero are 5, 31, 201)
3. The standard deviation of a symmetrical distribution is 3. What must
be the value of the fourth moment about the mean in order that the
distribution is mesokurtic?(µ4 = 243)
4. The mean and SD of a set of 100 observations were worked out as 40
and 5 respectively. But by mistake a value 50 was taken in place of 40 for
one observation. Calculate the correct mean and SD. (x̄ = 39.9, σ = 4.9)
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