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Following data represents Classes Frequency (f) c.f. r.f. % freq

the plant height (cm) of a
sample of 30 plants. 86–90 6 6 0.200 20.0
87 91 89 91–95 4 10 0.133 13.3
88 89 91 96–100 10 20 0.333 33.3
87 92 90 101–105 6 26 0.200 20.0
98 95 97

Frequency Distribution & 96

96
98
100
98
100
101
99
102
106–110
111–115
Total 30
3
1
29
30
0.100
0.033
1.000
10.0

100.0
3.3

Histogram 99
103
106
101
107
107
105
105
112 12
Histogram

10
10
8

Frequency
Muhammad Usman Frequency 6
6 6

distribution 4
4
3
& 2 1
Histogram 0
85.5–90.5 90.5–95.5 95.5–100.5 100.5–105.5 105.5–110.5 110.5–115.5
Class Boundries

Frequency Distribution Some definitions

Class Limits
• Tabular arrangement of data in which various items are arranged
• The class limits are defined as the number or the values of the variables which are
into classes or groups and the number of items falling in each class used to separate two classes. Sometimes classes are taken as 20--25, 25--30 etc In
is stated. such a case, these class limits means " 20 but less than 25", "25 but less than 30" etc
• The number of observations falling in a particular class is referred Class marks or midpoints
to as class frequency "f". • The class mark or the midpoint is that value which divides a class into two equal parts.
It is obtained by dividing the sum of lower and upper class limits or class boundaries
• Data presented in the form of a frequency distribution is also called of a class by 2.
grouped data. Class interval
• The difference between either two successive lower class limits or two successive
upper class limits OR
• The difference between two successive midpoints.
• denoted by "h".

Construction of a frequency distribution Example

1. Decide the number of classes: K=1+3.3 log(n) or 𝑛 • The following data represents the height of 30 wheat plants taken from the
2. Determine the range of variation of the data i.e, R= Max – Min experimental area. Construct a frequency distribution and appropriate
𝑹 graphs to explain the distribution of data:
3. Determine the approximate size of class interval 𝒉=𝑲
4. Decide where to locate the class limits 87 91 89 88 89 91 87 92 90 98 95
5. Distribute the data into appropriate classes 97 96 100 101 96 98 99 98 100 102 99
101 105 103 107 105 106 107 112

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Construction of a frequency distribution Frequency Distribution

• Decide the number of classes: Classes Class Boundaries Tally Freq (f) c.f. r.f. % freq Cumulative %
freq
K=1+3.3 log(n)=5.87 or 𝑛=5.47  6 Classes
86–90 85.5–90.5 6 6 0.200 20.0 20.00
• Determine the range of variation of the data i.e,
91–95 90.5–95.5 4 10 0.133 13.3 33.3
R= Max – Min = 112 – 87 = 25
96–100 95.5–100.5 10 20 0.333 33.3 66.6
• Determine the approximate size of class interval
101–105 100.5–105.5 6 26 0.200 20.0 86.6
𝑹
𝒉 = 𝑲 = 25/6 = 4.17  5 Class Interval 106–110 105.5–110.5 3 29 0.100 10.0 96.6

• Decide where to locate the class limits  86-90, 91-95, … 111–115 110.5–115.5 1 30 0.033 3.3 100.0

• Distribute the data into appropriate classes Total 30 1.000 100.0

Class Boundaries Histogram

Histogram of Height of 30 Students
• Class Boundaries 12
• Subtract any Upper Class Limit from its Subsequent Lower Class limit and 10
10
divide the difference with 2, you will get the Continuity correction factor
• Subtract this factor from all Lower Class Limits and add it to all Upper Class 8
Frequency

limits. 6
6 6

4
4 3
For example, 91-90 = ½ =0.05 or 96-95 = ½ =0.05
2 1

0
85.5–90.5 90.5–95.5 95.5–100.5 100.5–105.5 105.5–110.5 110.5–115.5
Class Boundries

Frequency Polygon Cumulative Frequency Polygon / Ogive

• Frequency polygons are a graphical device for understanding the shapes • A cumulative frequency polygon is a plot of the cumulative
of distributions. They serve the same purpose as histograms, but are frequency against the upper class boundary with the points joined by a line
especially helpful for comparing sets of data. segment.
• Mid Points vs Frequency Frequency Polygon • Upper Class Boundaries vs Cumulative Frequency
12
Cumulative Frequency Polygon / Ogive
35
10
30
Cumulative Frequency

8
Frequency

25
6 20
15
4
10
2 5
0 0
88 93 98 103 108 113 90.5 95.5 100.5 105.5 110.5 115.5
Mid Points Upper Class Boundaries

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Stem & Leaf Display Example

• A relatively small data set can be represented by stem and leaf display. Represent the following data by Stem and Leaf display by
• In addition to information on the number of observations falling in the (i) taking 10 unit as the width of the class
various classes, it displays details of what those observations actually are. (ii) taking 5 unit as the width of the class
• Each number in the data set is divided into two parts, a Stem and a Leaf. A 32 45 38 41 49 36 52 56 51 62 63 59
stem is the leading digit(s) of each number and is used in sorting, while a leaf 68
Steam Leaf *indicate 0—4
is the rest of the number or the trailing digit(s) and shown in display. 3* 2
Steam Leaf .indicate 5—9
3. 8 6
3 2 8 6 4* 1
4. 5 9 * and . are called placeholder
4 5 1 9
5* 2 1
5 2 6 1 9 5. 6 9
6* 2 3
6 2 3 8
6. 8

Example
Use the data below to make a stem- Stem Leaf
and-leaf plot by taking 10 as a unit.
7 0589
85 115 126 92 104 8 4558
85 116 100 121 123 9 022379
79 90 110 129 108
10 0478
107 78 131 114 92
131 88 97 99 116
11 04566
93 84 75 70 132 12 1369
13 112
7 0589
These values are 70, 75, 78 and 79