This document provides an overview of a probability and statistics course, including the textbook, topics covered, evaluation scheme, and sample chapter content. The course covers topics such as descriptive statistics, probability theory, random variables, sampling distributions, and inferential statistics. Evaluation is based on a midterm exam, report, activities, and final exam. The sample chapter discusses descriptive statistics, data types, measures of central tendency (mean, median, mode), measures of dispersion (range, variance, standard deviation), and data representation methods.

1. Probability and Statistics
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2. Text book
• Probability & Statistics for Engineers &
Scientists, Ronald E. Walpole, 9th edition 2012,
Pearson
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3. Brief list of Course topics
1. Introduction to statistics and data analysis.
2. Introduction to probability theory.
3. Random variables and probability distributions.
4. Mathematical Expectation
5. Some discrete probability distribution.
6. Some continuous probability distribution.
7. Functions of Random Variables
8. Fundamental sampling distributions and data
descriptions.
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4. Evaluation Scheme
• Midterm Exam 40
• Report
• Activities
• Final exam 40
• Total 100 =100%
60 = 60%
40 = 40%
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5. Chapter 1
Introduction to statistics and
data analysis
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6. What is Statistics?
• Statistics is the science of collecting,
organizing, summarizing, and analyzing
information to draw conclusions or answer
questions.
• Statistics is a way to get information from data.
It is the science of uncertainty.
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7. Steps of Statistical Practice
• Preparation: Set clearly defined goals, questions of
interests for the investigation
• Data collection: Make a plan of which data to collect
and how to collect it
• Data analysis: Apply appropriate statistical methods
to extract information from the data
• Data interpretation: Interpret the information and
draw conclusions
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8. Statistical Methods
• Descriptive statistics include the collection,
presentation and description of numerical data .
• Inferential statistics include making inference,
decisions by the appropriate statistical methods by
using the collected data.
• Model building includes developing prediction
equations to understand a complex system.
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9. Descriptive Statistics
• Descriptive statistics involves the arrangement,
summary, and presentation of data, to enable
meaningful interpretation, and to support decision
making.
• Descriptive statistics methods make use of
– graphical techniques
– numerical descriptive measures.
• The methods presented apply both to
– the entire population
– the sample
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10. Basic Definitions
• Population: The collection of all items of interest in a
particular study.
• Variable: A characteristic of interest about each
element of a population or sample.
• Statistic: A descriptive measure of a sample
• Sample: A set of data drawn from the population;
a subset of the population available for observation
• Parameter: A descriptive measure of the population,
e.g., mean
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11. Collecting Data
• Target Population: The population about
which we want to draw inferences.
• Sampled Population: The actual population
from which the sample has been taken.
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12. Types of Variables
Qualitative Quantitative
Discrete Continuous
Ordinal Non Ordinal
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13. Types of data - examples
Numerical data
Age - income
55 75000
42 68000
. .
. .
Weight gain
+10
+5
.
.
Nominal
Person Marital status
1 married
2 single
3 single
. .
. .
Computer Brand
1 IBM
2 Dell
3 IBM
. .
. .
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Types of data - examples
Numerical data
Age - income
55 75000
42 68000
. .
. .
Nominal data
A descriptive statistic
for nominal data is
the proportion
of data that falls into
each category.
IBM Dell Compaq Other Total
25 11 8 6 50
50% 22% 16% 12%
Weight gain
+10
+5
.
.
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15. Types of Variables
Qualitative Quantitative
Discrete Continuous
Ordinal Non Ordinal
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16. Types of Variables
•Qualitative variables (what, which type…)
measure a quality or characteristic on each
experimental unit. (categorical data)
•Examples:
•Hair color (black, brown, blonde…)
•Make of car (Dodge, Honda, Ford…)
•Gender (male, female)
•State of birth (Iowa, Arizona,….)
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17. Types of Variables
•Quantitative variables (How big, how
many) measure a numerical quantity on each
experimental unit. (denoted by x)
Discrete if it can assume only a finite or
countable number of values.
Continuous if it can assume the infinitely
many values corresponding to the points
on a line interval.
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18. Graphing Qualitative Variables
• Use a data distribution to describe:
– What values of the variable have been measured
– How often each value has occurred
• “How often” can be measured 3 ways:
–Frequency
–Relative frequency = Frequency/n
–Percent frequency = Relative frequency* 100
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19. Example
• A bag contains 25 colored balls:
• Raw Data:
• Statistical Table:
Color Tally Frequency Relative
Frequency
Percent
Red 3 3/25 = .12 12%
Blue 6 6/25 = .24 24%
Green 4 4/25 = .16 16%
Orange 5 5/25 = .20 20%
Brown 3 3/25 = .12 12%
Yellow 4 4/25 = .16 16%
m
m
m
m m
m
m m
m
m
m
m m
m
m m
m
m
m
m
m
m
m
m
m
m
m
m
m
m m
m m
m m m
m m m
m m
m m
m
m
m
m
m m
m
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20. Graphs
Bar Chart
Pie Chart
Angle=
Relative Frequency times 360
Color
Frequency
Green
Orange
Blue
Red
Yellow
Brown
6
5
4
3
2
1
0
16.0%
Green
20.0%
Orange
24.0%
Blue
12.0%
Red
16.0%
Yellow
12.0%
Brown
Pareto Chart
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21. A sample of 30 persons who often consume donuts
were asked what variety of donuts was their favourite.
The responses from these 30 persons were as follows:
glazed filled other plain glazed other
frosted filled filled glazed other frosted
glazed plain other glazed glazed filled
frosted plain other other frosted filled
filled other frosted glazed glazed filled
Construct a frequency distribution table for these data.
Example
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22. Solution
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23. Solution
s
frequencie
all
of
Sum
category
that
of
Frequency
category
a
of
frequency
lative
Re 
Calculating Percentage Frequency
Percentage Frequency = (Relative frequency) * 100
Relative Frequency and Percentage Distributions
23

24. Graphical Presentation of Qualitative Data
A graph made of bars whose heights represent the
frequencies of respective categories is called a bar
graph.
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25. Graphical Presentation of Qualitative Data
A circle divided into portions that represent the
relative frequencies or percentages of a population
or a sample belonging to different categories is
called a pie chart.
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26. Calculating Angle Sizes for the Pie Chart
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27. Pie chart for the percentage distribution
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28. Scatter Plot
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29. Scatter Plot
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30. Dot Plot
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Draw the dot plot for the following data, then calculate
the mean, median, and mode
0.86, 0.49, 0.46, 0.52, 0.62, 0.79, 0.75, 0.47, 0.26, 0.43
Mean Calculation:-
65
.
5
.....
6
5
4
3
2
1











x
x
x
x
x
x
x
x
x n
565
.
0
10
65
.
5




n
x
x

31. Dot Plot
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Rearrange the data
n=10 Median Order =5&6
0.26, 0.43, 0.46, 0.47, 0.49, 0.52, 0.62, 0.75, 0.79, 0.86
Median =( 0.49+0.52)/2 = 0.505
Mode No Mode
Median Calculation:-

32. Stem and Leaf Displays
In a stem-and-leaf display of quantitative data, each
value is divided into two portions – a stem and a leaf.
The leaves for each stem are shown separately in a
display.
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33. Example
The following are the scores of 30 college students
on a statistics test:
Construct a stem-and-leaf display.
75
69
83
52
72
84
80
81
77
96
61
64
65
76
71
79
86
87
71
79
72
87
68
92
93
50
57
95
92
98
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34. Solution
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35. Solution
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36. Solution
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37. Example
The following data give the monthly rents paid by a
sample of 30 households selected from a small town.
Construct a stem-and-leaf display for these data.
880
1210
1151
1081
985
630
721
1231
1175
1075
932
952
1023
850
1100
775
825
1140
1235
1000
750
750
915
1140
965
1191
1370
960
1035
1280
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38. Solution
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39. Example
39
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Construct a stem-and-leaf display for the given data
39

40. Solution
40
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41. Solution
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42. Mean
The mean for ungrouped data is obtained by dividing the
sum of all values by the number of values in the data set. Thus,
Mean for population data:
Mean for sample data:
Where
is the sum of all values; N is the population size;
n is the sample size;
is the population mean;
is the sample mean.
N
x



n
x
x


x

x
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43. Mean
1. Most common measure of central tendency
2. Acts as ‘balance point’
3. Affected by extreme values (‘outliers’)
4. Denoted where
x
x
n
x x x
n
i
i
n
n
 
  


1 1 2
…
x
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44. Example
Find the mean of cash donations made by these
eight Persons.
63
,
26
,
315
,
21
,
63
,
110
,
199
,
319
8
7
6
5
4
3
2
1 x
x
x
x
x
x
x
x
x 








million
n
x
x 5
.
139
$
5
.
139
8
1116





1116
63
26
315
21
63
110
199
319 








Solution
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45. Example
Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7
x
x
n
x x x x x x
i
i
n
 
    

    



1 1 2 3 4 5 6
6
10 3 4 9 8 9 11 7 6 3 7 7
6
8 30
. . . . . .
.
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46. Median
1. Measure of central tendency
2. Middle value in ordered sequence
• If n is odd, middle value of sequence
• If n is even, average of 2 middle values
3. Position of median in sequence
4. Not affected by extreme values
Order 

n 1
2
Order 
n
2
n
2
, +1
n is odd
n is even
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47. Median Example
• Raw Data: 24.1 22.6 21.5 23.7 22.6
• Ordered: 21.5 22.6 22.6 23.7 24.1
• Position: 1 2 3 4 5
Positioning Point
Median






n 1
2
5 1
2
3 0
22 6
.
.
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48. Median Example
• Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7
• Ordered: 4.9 6.3 7.7 8.9 10.3 11.7
• Position: 1 2 3 4 5 6
Positioning Point
Median
  



n
2
6
2
3 4
7 7 8 9
2
8 30
,
. .
.
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49. Mode
1. Measure of central tendency
2. Value that occurs most often
3. Not affected by extreme values
4. May be no mode or several modes
5. May be used for quantitative or qualitative
data
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50. Mode Example
• No Mode
Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7
• One Mode
Raw Data: 6.3 4.9 8.9 6.3 4.9 4.9
• More Than 1 Mode
Raw Data: 21 28 28 41 43 43
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51. Range
1. Measure of dispersion
2. Difference between largest & smallest
observations
Range= R = Max Value – Min Value
3. Ignores how data are distributed
7 8 9 10 7 8 9 10
Range = 10 – 7 = 3 Range = 10 – 7 = 3
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52. Variance & Standard Deviation
1. Measures of dispersion
2. Most common measures
3. Consider how data are distributed
4 6 10 12
x = 8.3
4. Show variation about mean (x or μ)
8
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53. Standard Notation
Measure Sample Population
Mean x 
Standard
Deviation s 
Variance s 2
 2
Size n N
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54. Variance and Standard Deviation
Basic Formulas for the Variance and Standard Deviation for
Ungrouped Data
where σ² is the population variance, s² is the sample
variance, σ is the population standard deviation, and s
is the sample standard deviation.
   
   
1
and
1
and
2
2
2
2
2
2














n
x
x
s
N
x
n
x
x
s
N
x




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55. Variance and Standard Deviation
Short-cut Formulas for the Variance and Standard Deviation for
Ungrouped Data
where σ² is the population variance, s² is the sample variance, σ
is the population standard deviation, and s is the sample
standard deviation.
 
 
 
 
)
1
(
and
)
1
(
and
2
2
2
2
2
2
2
2
2
2










 
 
 
 
n
n
x
x
n
s
N
N
x
x
n
n
x
x
n
s
N
N
x
x


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56. Variance Example
Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7
s
x x
n
x
x
n
s
i
i
n
i
i
n
2
2
1 1
2
2 2 2
1
8 3
10 3 8 3 4 9 8 3 7 7 8 3
6 1
6 368



 

     


 
 
( )
( ) ( ) ( )
where .
. . . . . .
.
…
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57. Variance Example
Raw Data: 10.3 4.9 8.9 11.7 6.3 7.7
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  8
.
49
x   18
.
445
2
x
n= 6
 
)
1
(
2
2
2



 
n
n
x
x
n
s
368
.
6
5
*
6
)
8
.
49
(
18
.
445
*
6 2
2



s
523
.
2
368
.
6 

s

58. Summary of Variation Measures
Measure Formula Description
Range XMax – XMin Total Spread
Standard Deviation
(Sample)
Dispersion about
Sample Mean
Standard Deviation
(Population)
Dispersion about
Population Mean
Variance
(Sample)
Squared Dispersion
about Sample Mean
xi  x
 2
i1
n

n 1
xi  µx
 2
i1
n

N
xi  x
 2
i1
n

n 1
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59. A box and whisker plot also called a box plot
displays the five number summary of a set of data.
The five number summary is
• The minimum value
• First quartile (Q1)
• Median,
• Third quartile (Q3)
• The maximum value
Box-and-whisker Plot
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60. Lower quartile Upper quartile
Median
minimum maximum
Box-and-whisker Plot
In a box plot, we draw a box from the first quartile to
the third quartile. A vertical line goes through the box
at the median. The whiskers go from each quartile to
the minimum or maximum.
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61. Construct a box-and-whisker plot for the following
data.
Example
85,92,78,88,90,88,89
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62. 1. Order the test scores from least to greatest
2. Find the median of the test scores.
78, 85,88, 88, 89, 90,92
88
3. Find Find the quartiles.
 The first quartile (Q1) is the median of the data
points to the left of the median.
Solution
85
 The third quartile (Q3) is the median of the data
points to the right of the median
90
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63. 4. Complete the five-number summary by
finding the min and the max.
Solution
Min = 78
Max = 92
Q1 Q3
Median
Min Maz
78 85 88 88 89 90 92
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64. Use the given data to make a box-and-whisker plot.
31, 23, 33, 35, 26, 24, 31, 29
Example
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65. Order the data from least to greatest. Then find the
minimum, lower quartile, median, upper quartile, and
maximum.
minimum: 23
maximum: 35
lower quartile: = 25
24 + 26
2
upper quartile: = 32
31 + 33
2
median: = 30
29 + 31
2
23 24 26 29 31 31 33 35
Solution
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66. Draw a number line and plot a point above each value.
Draw the box and whiskers.
23 24 26 29 31 31 33 35
22 24 26 28 30 32 34 36 38
Solution
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67. Frequency Histograms
• Divide the range of the data into 5-12
subintervals of equal length.
• Calculate the approximate width of the
subinterval as Range/number of subintervals.
• Round the approximate width up to a
convenient value.
• Sturges rule K= 1+3.3log (N).
• Create a statistical table including the
subintervals, their frequencies and relative
frequencies.
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68. Example
• The following are balances (in $) of 100 accounts
receivable taken from the ledger of XYZ Store.
31 38 41 52 59 46 74 69 93 60
69 83 78 74 77 35 79 80 71 65
56 69 34 33 92 37 60 43 51 61
74 68 83 49 34 71 58 83 94 66
78 48 34 50 68 65 64 95 92 81
77 84 41 40 38 38 60 67 50 86
76 99 38 94 48 70 80 95 98 42
55 49 54 60 62 70 88 94 85 51
59 68 51 87 53 57 54 46 46 76
69 64 61 63 78 55 66 73 75 64
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69. • Using 7 equal intervals with the lowest starting at
30, compute the mean, and the variance using short-
cut method.
• calculate mode and median (analytically and
graphically)
• Estimate the value below which 75% of the values
fall.
Example
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70. Solution
• Determine the Min value
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71. Example
• The following are balances (in $) of 100 accounts
receivable taken from the ledger of XYZ Store.
31 38 41 52 59 46 74 69 93 60
69 83 78 74 77 35 79 80 71 65
56 69 34 33 92 37 60 43 51 61
74 68 83 49 34 71 58 83 94 66
78 48 34 50 68 65 64 95 92 81
77 84 41 40 38 38 60 67 50 86
76 99 38 94 48 70 80 95 98 42
55 49 54 60 62 70 88 94 85 51
59 68 51 87 53 57 54 46 46 76
69 64 61 63 78 55 66 73 75 64
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72. Solution
• Determine the Min value = 31
• Determine the Max value =
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73. Example
• The following are balances (in $) of 100 accounts
receivable taken from the ledger of XYZ Store.
31 38 41 52 59 46 74 69 93 60
69 83 78 74 77 35 79 80 71 65
56 69 34 33 92 37 60 43 51 61
74 68 83 49 34 71 58 83 94 66
78 48 34 50 68 65 64 95 92 81
77 84 41 40 38 38 60 67 50 86
76 99 38 94 48 70 80 95 98 42
55 49 54 60 62 70 88 94 85 51
59 68 51 87 53 57 54 46 46 76
69 64 61 63 78 55 66 73 75 64
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74. Solution
• Determine the Min value = 31
• Determine the Max value = 99
• Calculate the range = Max – Min
• But the starting point is given 30
• use Min = 30
• Range = 99 – 30 = 69
• Interval Length C = Range / No. of intervals
C= 69 / 7 = 9.85 =10
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75. Solution
L.L U.L
30
75
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76. Solution
L.L U.L
30
40
C=10
76
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77. Solution
L.L U.L
30
40
50
C=10
C=10
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78. Solution
L.L U.L
30
40
50
60
70
80
90
78
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79. Solution
L.L U.L
30 39
40 49
50 59
60 69
70 79
80 89
90 99
Classes
79
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80. Solution
L.L U.L f
30 39
40 49
50 59
60 69
70 79
80 89
90 99
80
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81. Example
• The following are balances (in $) of 100 accounts
receivable taken from the ledger of XYZ Store.
31 38 41 52 59 46 74 69 93 60
69 83 78 74 77 35 79 80 71 65
56 69 34 33 92 37 60 43 51 61
74 68 83 49 34 71 58 83 94 66
78 48 34 50 68 65 64 95 92 81
77 84 41 40 38 38 60 67 50 86
76 99 38 94 48 70 80 95 98 42
55 49 54 60 62 70 88 94 85 51
59 68 51 87 53 57 54 46 46 76
69 64 61 63 78 55 66 73 75 64
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82. Solution
L.L U.L f
30 39 11
40 49
50 59
60 69
70 79
80 89
90 99
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83. Example
• The following are balances (in $) of 100 accounts
receivable taken from the ledger of XYZ Store.
31 38 41 52 59 46 74 69 93 60
69 83 78 74 77 35 79 80 71 65
56 69 34 33 92 37 60 43 51 61
74 68 83 49 34 71 58 83 94 66
78 48 34 50 68 65 64 95 92 81
77 84 41 40 38 38 60 67 50 86
76 99 38 94 48 70 80 95 98 42
55 49 54 60 62 70 88 94 85 51
59 68 51 87 53 57 54 46 46 76
69 64 61 63 78 55 66 73 75 64
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84. Solution
L.L U.L f
30 39 11
40 49 12
50 59
60 69
70 79
80 89
90 99
84
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85. Solution
L.L U.L f
30 39 11
40 49 12
50 59 16
60 69 23
70 79 17
80 89 11
90 99 10
100
85
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86. Solution
L.L U.L f F f relative X
30 39 11 11 0.11 34.5
40 49 12 23 0.12 44.5
50 59 16 39 0.16 54.5
60 69 23 62 0.23 64.5
70 79 17 79 0.17 74.5
80 89 11 90 0.11 84.5
90 99 10 100 0.1 94.5
100 1
class mark (X) or Mid Point = (LL+UL )/2
X1 = (30+39)/2 =34.5
C=10
C=10
F = Cumulative Frequency
f = Relative Frequency
86
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87. Solution
L.L U.L f F f relative X f*X f*(x - x )2 f*X2
30 39 11 11 0.11 34.5 379.5 9637.76 13092.75
40 49 12 23 0.12 44.5 534 4609.92 23763
50 59 16 39 0.16 54.5 872 1474.56 47524
60 69 23 62 0.23 64.5 1483.5 3.68 95685.75
70 79 17 79 0.17 74.5 1266.5 1838.72 94354.25
80 89 11 90 0.11 84.5 929.5 4577.76 78542.75
90 99 10 100 0.1 94.5 945 9241.6 89302.5
100 1 6410 31384 442265
Mean (X ) = 64.1
n
X
f
X i
i


*
87
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88. Solution
L.L U.L f F f relative X f*X f*(x-x )2 f*X2
30 39 11 11 0.11 34.5 379.5 9637.76 13092.75
40 49 12 23 0.12 44.5 534 4609.92 23763
50 59 16 39 0.16 54.5 872 1474.56 47524
60 69 23 62 0.23 64.5 1483.5 3.68 95685.75
70 79 17 79 0.17 74.5 1266.5 1838.72 94354.25
80 89 11 90 0.11 84.5 929.5 4577.76 78542.75
90 99 10 100 0.1 94.5 945 9241.6 89302.5
100 1 6410 31384 442265
Mean (X ) = 64.1
Variance (S2 ) 317.010101
S.D (s) 17.80477748
C.V 0.277765639
1
)
( 2
2





n
X
X
f
S
i
i
)
1
(
)
( 2
2
2



 
n
n
X
f
X
f
n
S
i
i
i
i
X
s
CV 
88
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89. histogram
89
0
5
10
15
20
25
30 39 40 49
50 59
Dr Yehya Mesalam

90. histogram
90
0
5
10
15
20
25
30 39 40 49
50 59
Dr Yehya Mesalam

91. histogram
91
0
5
10
15
20
25
30 39 40 49
50 59
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92. histogram
X3 X7
44.5
C
92
X1 X2
34.5
54.5 64.5 74.5 84.5 94.5
Take scale every 1 Cm = 10 $ or degree as given in your data
1 Cm
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93. histogram
X3 X7
C
93
X1 X2
C
2
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94. Solution
L.L U.L L. B U. B
30 39 29.5 39.5
40 49 39.5 49.5
50 59 49.5 59.5
60 69 59.5 69.5
70 79 69.5 79.5
80 89 79.5 89.5
90 99 89.5 99.5
94
L.L & U.L is the Lower Limit & upper limit for the class
L.B & U.B is the Lower boundary& upper boundary for the class
The graph of histogram
must be on the boundaries
not on the limits
L.B for class i=
L.L i +U.L i-1
2
U.B for class i=
L.L i+1 + U.L i
2
L.B 1 = (30+29)/2 =29.5
U.B 1 = (39+40)/2 =39.5
L.B 2 = (40+39)/2 =39.5
Then L.B i = U.B i-1 or U.B i = L.B i+1
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95. histogram
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
C
95
X1 X2
29.5
39.5
49.5
59.5
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96. histogram
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
fr
0.05
0.20
96
Relative
frequency
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97. Polygon
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
97
Dr Yehya Mesalam

98. Polygon
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
98
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99. Polygon
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
99
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100. Polygon
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
10
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101. Polygon
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
10
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102. histogram
0
5
10
15
20
25
24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5
frequency
class mark
Mode
fr
0.05
0.20
10
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103. Median
Class
limit
fi F
30-39 11 11
40-49 12 23
50-59 16 39
60-69 23 62
70-79 17 79
80-89 11 90
90-99 10 100
100
med
med
med
f
F
n
C
L
X
1
~
2
*




60-0.5= 59.5
10
Dr Yehya Mesalam

104. Median
Class
limit
fi Less
than F
30-39 11 11
40-49 12 23
50-59 16 39
60-69 23 62
70-79 17 79
80-89 11 90
90-99 10 100
100
med
med
med
f
F
n
C
L
X
1
~
2
*




60-0.5= 59.5
10
Dr Yehya Mesalam

105. Median
Class
limit
fi Less
than F
30-39 11 11
40-49 12 23
50-59 16 39
60-69 23 62
70-79 17 79
80-89 11 90
90-99 10 100
100
med
med
med
f
F
n
C
L
X
1
~
2
*




60-0.5= 59.5
10
Dr Yehya Mesalam

106. Median
Class
limit
fi Less
than F
30-39 11 11
40-49 12 23
50-59 16 39
60-69 23 62
70-79 17 79
80-89 11 90
90-99 10 100
100
med
med
med
f
F
n
C
L
X
1
~
2
*




60-0.5= 59.5
Median =59.5+10(50-39)/23 =
64.28
10
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107. Mode
Class
limit
fi
30-39 11
40-49 12
50-59 16
60-69 23
70-79 17
80-89 11
90-99 10
100
2
1
1
mod
^
*





 C
L
X
60-0.5= 59.5
10
Dr Yehya Mesalam

108. Mode
Class
limit
fi
30-39 11
40-49 12
50-59 16
60-69 23
70-79 17
80-89 11
90-99 10
100
2
1
1
mod
^
*





 C
L
X
60-0.5= 59.5
7
16
23
1 



10
Dr Yehya Mesalam

109. Mode
Class
limit
fi
30-39 11
40-49 12
50-59 16
60-69 23
70-79 17
80-89 11
90-99 10
100
2
1
1
mod
^
*





 C
L
X
60-0.5= 59.5
7
16
23
1 



6
17
23
2 



10
Dr Yehya Mesalam

110. Mode
Class
limit
fi
30-39 11
40-49 12
50-59 16
60-69 23
70-79 17
80-89 11
90-99 10
100
Mode=59.5+10(7/13) = 64.88
2
1
1
mod
^
*





 C
L
X
60-0.5= 59.5
7
16
23
1 



6
17
23
2 



11
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111. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
29.5 0
39.5
49.5
59.5
69.5
79.5
89.5
99.5
11
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112. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
29.5 0
39.5
11
49.5
59.5
69.5
79.5
89.5
99.5
11
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113. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
29.5 0
39.5 11
49.5 23
59.5 39
69.5 62
79.5 79
89.5 90
99.5 100
11
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114. Solution
L.L U.L f F
30 39 11 11
40 49 12 23
50 59 16 39
60 69 23 62
70 79 17 79
80 89 11 90
90 99 10 100
100
11
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115. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
More
Than
29.5 0 100
39.5 11
49.5 23
59.5 39
69.5 62
79.5 79
89.5 90
99.5 100
11
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116. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
More
Than
29.5 0 100
39.5 11 89
49.5 23
59.5 39
69.5 62
79.5 79
89.5 90
99.5 100
11
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117. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
More
Than
29.5 0 100
39.5 11 89
49.5 23 77
59.5 39 61
69.5 62 38
79.5 79 21
89.5 90 10
99.5 100 0
M than =n- L than
M than +L than =n
11
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118. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
More
Than
29.5 0 100
39.5 11 89
49.5 23 77
59.5 39 61
69.5 62 38
79.5 79 21
89.5 90 10
99.5 100 0
M than +L than =n
11
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119. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
More
Than
29.5 0 100
39.5 11 89
49.5 23 77
59.5 39 61
69.5 62 38
79.5 79 21
89.5 90 10
99.5 100 0
0
10
20
30
40
50
60
70
80
90
100
110
29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5
Cum.
Frequency
Lower Boundary
O-Gives
More Than
Less Than
11
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120. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
More
Than
29.5 0 100
39.5 11 89
49.5 23 77
59.5 39 61
69.5 62 38
79.5 79 21
89.5 90 10
99.5 100 0
0
10
20
30
40
50
60
70
80
90
100
110
29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5
Cum.
Frequency
Lower Boundary
O-Gives
More Than
Less Than
12
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121. O-Gives ( Less Than & More than)
Lower
Boundary
Less
Than
More
Than
29.5 0 100
39.5 11 89
49.5 23 77
59.5 39 61
69.5 62 38
79.5 79 21
89.5 90 10
99.5 100 0
0
10
20
30
40
50
60
70
80
90
100
110
29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5
Cum.
Frequency
Lower Boundary
O-Gives
More Than
Less Than
Mediam at n=50
Median
12
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122. O-Gives ( Less Than & More than)
0
10
20
30
40
50
60
70
80
90
100
110
29.5 39.5 49.5 59.5 69.5 79.5 89.5 99.5
Cum.
Frequency
Lower Boundary
O-Gives
More Than
Less Than
12
Estimate the value below which 75% of the values fall.
n= 100 100%
? 75%
Then at frequency value =75
draw horizontal line cuts
Less Than and More Than
then determine the required
value
75% of the sample obtained
more ( above) the value 51
75% of the sample obtained
less (blew)the value 77 51
77
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123. Short Cut Method
n
d
f
C
X
X i
i


 0
)
1
(
)
( 2
2
2
2



 
n
n
d
f
d
f
n
C
S
i
i
i
i
12
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124. L.L U.L f F f relative d f*d f *d2
30 39 11 11 0.11 -3 -33
40 49 12 23 0.12 -2 -24
50 59 16 39 0.16 -1 -16
60 69 23 62 0.23 0 0
70 79 17 79 0.17 1 17
80 89 11 90 0.11 2 22
90 99 10 100 0.1 3 30
Sum 100 1 -4
Mean (X ) = 64.1
Variance (S2 ) 317.010101
S.D (s) 17.80477748
C.V 0.277765639
X
s
CV 
12
Short Cut Method
n
d
f
C
X
X i
i


 0
X = 64.5+ 10 (-4/100)=64.1
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125. L.L U.L f F f relative d f*d f*d2
30 39 11 11 0.11 -3 -33 99
40 49 12 23 0.12 -2 -24 48
50 59 16 39 0.16 -1 -16 16
60 69 23 62 0.23 0 0 0
70 79 17 79 0.17 1 17 17
80 89 11 90 0.11 2 22 44
90 99 10 100 0.1 3 30 90
Sum
100 1 -4 314
Mean (X ) = 64.1
Variance (S2 ) 317.010101
S.D (s) 17.80477748
C.V 0.277765639
125
Short Cut Method
S2 = 102 *[(100*314-(-4)2 )/(100*99)]
=317.010101
)
1
(
)
( 2
2
2
2



 
n
n
d
f
d
f
n
C
S
i
i
i
i
Dr Yehya Mesalam

126. L.L U.L f F f relative d f*d f *d2
30 39 11 11 0.11 0 0
40 49 12 23 0.12 1 12
50 59 16 39 0.16 2 32
60 69 23 62 0.23 3 69
70 79 17 79 0.17 4 68
80 89 11 90 0.11 5 55
90 99 10 100 0.1 6 60
Sum 100 1 296
Mean (X ) = 64.1
Variance (S2 ) 317.010101
S.D (s) 17.80477748
C.V 0.277765639
126
Short Cut Method
n
d
f
C
X
X i
i


 0
X = 34.5+ 10 (296/100)=64.1
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127. Example
Complete the following table, and then find the mean, mode,
median, variance and CV.
Draw the Histogram, Frequency polygon, Relative frequency
histogram
Class limits
50 - 8
- 10
- 34 0
- 14
- 10
-
- 119 65 16
U
L X
X  i
x i
f F i
d i
i d
f i
i d
f 2
C
120
C
127
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128. Complete the following table, and then find the mean, mode,
median, variance and CV.
Draw the Histogram, Frequency polygon, Relative frequency
histogram
Class limits
50 - 8
- 10
- 34 0
- 14
- 10
-
- 119 65 16
U
L X
X  i
x i
f F i
d i
i d
f i
i d
f 2
C
120= 50+7C Then C=10
128
Solution
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129. Complete the following table, and then find the mean, mode,
median, variance and CV.
Draw the Histogram, Frequency polygon, Relative frequency
histogram
Class limits
50 - 59 54.5 8 -2
60 - 69 64.5 10 -1
70 - 79 74.5 34 0
80 - 89 84.5 1 14
90 - 99 94.5 10 2
100 - 109 104.5 3
110 - 119 114.5 65 4 16
U
L X
X  i
x i
f F i
d i
i d
f i
i d
f 2
129
Solution
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130. Complete the following table, and then find the mean, mode,
median, variance and CV.
Draw the Histogram, Frequency polygon, Relative frequency
histogram
Class limits
50 - 59 54.5 8 -2
60 - 69 64.5 10 -1
70 - 79 74.5 34 0
80 - 89 84.5 14 1 14
90 - 99 94.5 10 2
100 - 109 104.5 3
110 - 119 114.5 1 65 4 16
U
L X
X  i
x i
f F i
d i
i d
f i
i d
f 2
130
Solution
Dr Yehya Mesalam

131. Solution
Complete the following table, and then find the mean, mode,
median, variance and CV.
Draw the Histogram, Frequency polygon, Relative frequency
histogram
Class limits
50 - 59 54.5 8 8 -2
60 - 69 64.5 10 18 -1
70 - 79 74.5 16 34 0
80 - 89 84.5 14 48 1 14
90 - 99 94.5 10 58 2
100 - 109 104.5 6 64 3
110 - 119 114.5 1 65 4 16
U
L X
X  i
x i
f F i
d i
i d
f i
i d
f 2
131
Dr Yehya Mesalam

132. Solution
Complete the following table, and then find the mean, mode,
median, variance and CV.
Draw the Histogram, Frequency polygon, Relative frequency
histogram
Class limits
50 - 59 54.5 8 8 -2 -16 32
60 - 69 64.5 10 18 -1 -10 10
70 - 79 74.5 16 34 0 0 0
80 - 89 84.5 14 48 1 14 14
90 - 99 94.5 10 58 2 20 40
100 - 109 104.5 6 64 3 18 54
110 - 119 114.5 1 65 4 4 16
U
L X
X  i
x i
f F i
d i
i d
f i
i d
f 2
30 166
132
Dr Yehya Mesalam

133. Solution
Complete the following table, and then find the mean, mode,
median, variance and CV.
Draw the Histogram, Frequency polygon, Relative frequency
histogram
Class limits
50 - 59 54.5 8 8 -2 -16 32
60 - 69 64.5 10 18 -1 -10 10
70 - 79 74.5 16 34 0 0 0
80 - 89 84.5 14 48 1 14 14
90 - 99 94.5 10 58 2 20 40
100 - 109 104.5 6 64 3 18 54
110 - 119 114.5 1 65 4 4 16
U
L X
X  i
x i
f F i
d i
i d
f i
i d
f 2
30 166
133
Dr Yehya Mesalam

134. Short Cut Method
n
d
f
C
X
X i
i


 0
)
1
(
)
( 2
2
2
2



 
n
n
d
f
d
f
n
C
S
i
i
i
i
Mean = 74.5 + 10 ( 30/65) = 79.11
Variance = (10)2 [65*166 – (30)2 ] / [65*64 ] = 237.74
S.D = (237.74) 0.5 =15.41
134
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135. Shape
1. Describes how data are distributed
2. Measures of Shape
• Skew = Symmetry
Right-Skewed
Left-Skewed Symmetric
Mean = Median
Mean Median Median Mean
135
Dr Yehya Mesalam

136. Moment
n
X
X
f
m
K
i
i
K
 

)
(
0
)
(
1 



n
X
X
f
m
i
i
n
X
X
f
m
i
i
 

2
2
)
(
n
X
X
f
m
i
i
 

3
3
)
(
n
X
X
f
m
i
i
 

4
4
)
(
n
X
f
m
K
i
i
K


/
X
n
X
f
m
i
i



/
1
n
X
f
m
i
i


2
/
2
n
X
f
m
i
i


3
/
3
n
X
f
m
i
i


4
/
4
About the Origin About the Mean
136
Dr Yehya Mesalam

137. Moment
• Coefficient of Skewness 1

3
2
3
1
m
m


0
1 
 0
1 

0
1 

Normal Distribution Skewness to Right
Skewness to Left
Right-Skewed
Median Mean
Left-Skewed
Mean Median
Symmetric
Mean = Median
137
See page 38
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138. Moment
• Coefficient of Kurtosis 2

2
2
4
2
m
m


3
2 

3
2 

3
2 

Normal Distribution
Leptokurtic Platykurtic
Symmetric
138
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139. Example
• From the given graph, complete the following tables, draw the histogram
and polygon, determine the mode and median graphically, and calculate
the mean, median, mode, variance, standard deviation, and coefficient
of variation
Class limits
i
x i
f r
f
139
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140. Solution
•
Class
limits
d fd
12 - 16 14 6 6/80 -3 -18
17 - 21 19 8 8/80 -2 -16
22 - 26 24 14 14/80 -1 -14
27 - 31 29 24 24/80 0 0
32 - 36 34 14 14/80 1 14
37 - 41 39 8 8/80 2 16
52 - 46 44 6 6/80 3 18
sum 80 1 0
i
x i
f r
f
X = 29+ 5(0/80)=29
140
Dr Yehya Mesalam

141. Example
• From the given graph, complete the following tables, draw the histogram
and polygon, determine the mode and median graphically, and calculate
the mean, median, mode, variance, standard deviation, and coefficient
of variation
Class limits
i
x i
f r
f
141
Dr Yehya Mesalam

142. Example
• Complete the table, compute the mean, variance, , and mode and
median analytical and graphical
Class limit Frequency
Relative
frequency
Boundaries
Cumulative
frequency
? - ? ? ? More than ? 100
20 - ? ? ? More than 19.95 92
? - ? 17 ? More than ? ?
? - ? ? ? More than ? 46
? - ? ? 0.12 More than 37.95 ?
? - ? ? ? More than ? 5
? ? More than ? ?
142
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143. Solution
•
Class limit Frequency
Relative
frequency
Boundaries
Cumulative
frequency
14 - 19.9 8 0.08 More than 13.95 100
20 - 25.9 29 0.29 More than 19.95 92
26 - 31.9 17 0.17 More than 25.95 63
32 - 37.9 29 0.29 More than 31.95 46
38 - 43.9 12 0.12 More than 37.95 17
44 - 49.9 5 0.05 More than 43.95 5
100 1 More than 49.95 0
143
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144. Solution
L.L U.L f d F d f d2
14 19.9 8 -3 -24 72
20 25.9 29 -2 -58 116
26 31.9 17 -1 -17 17
32 37.9 29 0 0 0
38 43.9 12 1 12 12
44 49.9 5 2 10 20
Sum 100 -77 237
mean X 30.33
Variance S2 64.62181818
S.D 8.038769693
144
Dr Yehya Mesalam

145. Solution
8
29
17
29
12
5
0
5
10
15
20
25
30
35
16.95 22.95 28.95 34.95 40.95 46.95
frequency
class mark
viscosity
Mode
Mode
145
Dr Yehya Mesalam

146. Solution
100
92
63
46
17 5
0
0
8
37
54
83
95
100
0
10
20
30
40
50
60
70
80
90
100
110
13.96 19.96 25.96 31.96 37.96 43.96 49.96
Cum.
Frequency
Lower Boundary
O-Gives
More Than
Less Than
Mediam at n=50
146
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147. 147