Diagrammatic
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Diagrammatic Representation of Frequency Distribution
Diagrammatic representation of frequency distribution
A frequency distribution can be presented graphically in any of the following ways:1)Histogram 2)Frequency polygon 3)Smoothed frequency curve 4)Cumulative frequency curves or Ogives
Bar diagram In case of the frequency distribution of a discrete variable, we present the distribution graphically by using the Bar diagram. Consider the following example,
Number of member in a family
Frequency
1
4
2
33
3
76
4
50
5
26
6
8
7
1
Total
198
Table - 1
Histogram But in case of frequency distribution of a continuous variable, we present the distribution graphically by using Histogram. Histogram is the most common form of diagrammatic representation of a grouped frequency distribution. It consists of a set of adjoining rectangles drawn on a horizontal base line. Width of rectangles, one for each class, extends over the class boundaries (not class limits) shown on the horizontal scale.
Width of the rectangle = Width of the class
Classes with equal width
When all classes have equal width, the heights of rectangles will be proportional to the class frequencies and it is then customary to take the heights numerically equal to the class frequencies i.e. Height of each rectangle = Class Frequency
For Example:Wages (Rs.) Class Limits
Class boundaries
Frequency (number of employees)
50 ---------- 59
49.5 ----------- 59.5
8
60 ---------- 69
59.5 ----------- 69.5
10
70 ---------- 79
69.5 ----------- 79.5
16
80 ---------- 89
79.5 ----------- 89.5
14
90 ---------- 99
89.5 ----------- 99.5
10
100 ---------- 109
99.5 ----------- 109.5
5
110 ---------- 119
109.5 ----------- 119.5
2
18 16 14 12 10 8 6 4 2 0 Frequency
(number of employees) 49.5 ----------59.5 59.5 ----------69.5 69.5 ----------79.5 79.5 ----------89.5 89.5 ----------99.5 99.5 ----------109.5 109.5 -----------
Classes with unequal width
When the classes are of unequal width the rectangles will also be of unequal width, and therefore the heights must be proportional to the frequency densities.Beause then, Height of each rectangle = Frequency Density
Area of each rectangle = Width x Height = (width of
class) x (frequency density) class frequency
= (width of
class) x
----------------------------
width of the class = Class Frequency
Annual Sales(Rs.’000) Class Interval
Frequency (f)
Width of class
Frequency Density (f /w)
0 -------20
20
20
1.00
20 ------ 50
50
30
1.67
50 ------100
69
50
1.38
100 ----- 250
30
150
0.20
250 ------ 500
25
250
0.10
500 ---- 1000
19
500
0.04
Another example Weekly profit (Rs.’000)
No of shops
110 – 115
7
115 – 120
19
120 – 125
27
125 – 130
15
130 – 140
12
140 – 160
12
160 - 180
8
Since the class- intervals are unequal, frequencies must be adjusted otherwise the histogram would give a misleading picture. The adjustment is done as follows:The lowest width of the classes is 5. The frequency of the class 130 – 140 shall be divided by 2 since it’s class width is double. Similarly following this same logic the frequencies of the classes 140 – 160 and 160 – 180 must be divided by 4.
Class boundaries
Height of rectangles
110 – 115
7
115 – 120
19
120 – 125
27
125 – 130
15
130 – 140
12 ÷ 2 = 6
140 – 160
12 ÷ 4 = 3
160 - 180
8÷4=2
Histogram vs Bar Diagram
Although vertical bar chart and histogram may appear some alike, the main point of distinction between them is that
The consecutive rectangles in a histogram have no space in between but the bar diagram must have equal spaces left between the consecutive rectangles bars.
Also, the rectangles in a bar diagram must be of equal width, but those in a histogram are proportional to the widths of classes and in case of classes with unequal width, the width of the respective rectangle will also be unequal.
The histogram is after all, an area diagram, which emphasises the widths of the rectangles between the class boundaries. But in a bar diagram only the heights is all-important, the spacing and the width of bars being arbitrary.
-:Uses:The series of rectangles is a histogram give a visual representation of the relative size of various groups and the entire distribution of total frequencies among different classes becomes at once visible. The surface of the tops of rectangles also gives an idea of the nature of frequency curve for the population. The histogram may be used to find the mode graphically.
Frequency Polygon in case of discrete variable An alternative method of presenting the discrete distribution is the frequency polygon, in which the values and the corresponding frequencies are plotted as points with help of regular co-ordinates, as in Bar-diagram.
For example:Table 1, may be presented by using frequency polygon, here first we have to plot the points (1,4) , (2,33) , (3,76) ,(4,50) , (5,26) , (6,8) , (7,1) on the graph. The value preceding 1 i.e. 0 has no frequency, as has the value following 7 i.e. 8.Hence let us take two more points (0,0) and (8,0).Finally we join the successive points by line segments and get the frequency polygon.
Frequency Polygon in case of continuous variable
F
requency polygon may be used to present the frequency
distribution of a continuous variable, provided the classes are of equal width. Frequency polygon is the graphical presentation alternative to the histogram and may be looked upon as derived from histogram by joining the mid-points of the tops of consecutive rectangles. It is generally used in cases when all the classes have a common width. The two end points are joined to the base line at the mid-values of the empty classes at each end of the frequency distribution. Thus the frequency polygon has the same area as the histogram, provided the width of the classes is the same.
Another method
Another method of constructing frequency polygon is to take the mid-points of the various class-intervals and then plot the frequency corresponding to each point and to join all these points by straight lines. The figure obtained would exactly be the same as obtained by the another method. The only difference is that here we do not have to construct the histogram.
Advantages 1. The frequency polygon is particularly useful in presenting simple frequency distribution of a discrete variable. 2. It gives us an approximate idea of the shape of frequency curve. 3. By constructing the frequency polygon the value of mode can be easily ascertained. If from the apex of the polygon a perpendicular is drawn on the x axis, we get the value of the mode.
4. Moreover the frequency polygon has a special advantage over the hiatogram.The frequency polygon of several distributions may be plotted on the same graph, thereby making certain comparison possible, whereas histograms cannot be usefully employed in the same way. To compare the histograms, we must have a separate graph for each. Because of this limitation for the purposes in making a graphic comparison of frequency distribution, frequency polygon is preferred.
For example
Frequency polygon in case of continuous variable
Class Limits
Class Boundaries
Mid-value
Frequency
50 - 59
49.5 – 59.5
54.5
8
60 – 69
59.5 – 69.5
64.5
10
70 – 79
69.5 – 79.5
74.5
16
80 – 89
79.5 – 89.5
84.5
14
90 – 99
89.5 – 99.5
94.5
10
100 – 109
99.5 – 109.5
104.5
5
110 - 119
109.5 – 119.5
114.5
2
Ogive (or cumulative frequency polygon)
Ogive is the graphical presentation of a
cumulative frequency distribution and hence is also called Cumulative Frequency Polygon. When cumulative frequencies are plotted against the corresponding class boundaries and the successive points are joined by straight lines, the line diagram obtained is known as Ogive or Cumulative frequency polygon.
Types of Ogive The ogive is of “less-than” or “more-than” type according as the cumulative frequencies used are of “less-than” or “morethan” type. The “less-than” type ogive starts from the lowest class boundary on the horizontal axis and gradually rising upward ends at the highest class boundary corresponding to the cumulative frequency N i.e. total frequency. It looks like the elongated S. The “more-than” type ogive has the appearance of an elongated S, turned upside down. Unequal widths of classes in the frequency distribution do not cause any difficulty in the construction of an ogive.
Uses 1. The ogive is used to find the Median, Quartile, deciles and percentiles or the value of the variable such that its cumulative frequency is a specified number. 2. It is also useful in finding the cumulative frequency corresponding to a given value of the variable. 3. To find the number of observations which are expected to lie between two specified value of the variable.
For Example:Illustration 1:Draw histogram, frequency polygon and ogives for the following frequency distribution :Wages
50 - 59
60 – 69
70 - 79
80 – 89
90 - 99
100 – 109
110 - 119
No of employee
8
10
16
14
10
5
2
Class Limits
Class Boundaries
Mid-value
Frequency
50 - 59
49.5 – 59.5
54.5
8
60 – 69
59.5 – 69.5
64.5
10
70 – 79
69.5 – 79.5
74.5
16
80 – 89
79.5 – 89.5
84.5
14
90 – 99
89.5 – 99.5
94.5
10
100 – 109
99.5 – 109.5
104.5
5
110 - 119
109.5 – 119.5
114.5
2
Class boundary
Cumulative Cumulative frequency “less- frequency “morethan” type than” type
49.5
0
65 = N
59.5
8
57
69.5
18
47
79.5
34
31
89.5
48
17
99.5
58
7
109.5
63
2
119.5
65 = N
0
Calculation of median from ogive Class boundary 49.5 59.5 69.5 Median------79.5 89.5 99.5 109.5 119.5
Cumulative frequency “less-than” type 0 8 18 -------- N/2 = 32.5 34 48 58 63 65 = N
or
Median – 69.5 32.5 - 18 ------------------ = ---------------------79.5 – 69.5 34 - 18
Or Median = 69.5 + 10 x (14.5/16) = 69.5 + 9.06 = 78.56 (Ans)
Step Diagram
It will be noticed that the cumulative frequency
diagram for a discrete variable from a simple frequency distribution is a step diagram, the cumulative frequency remaining constant between two successive values of the variables. For example:- Consider the following table, the cumulative frequency is 0 for any value less than one, is 4 for any value greater than or equal to 1 but less than 2 ,is 37 for any value greater than or equal to 2 but less than 3 and so on.
Family size
Frequency
Cumulative frequency “less-than” type
1
4
4
2
33
37
3
76
113
4
50
163
5
26
189
6
8
197
7
1
198 = N
Total
N =198
(number of observations having value 2 or less than 2)
Frequency Curve
T
he histogram gives us only an approximate idea of the nature of distribution with limited number of observations. By drawing rectangles in a histogram, we assume that the observations in any class are uniformly distributed throughout the range of values within the class boundaries, although this may not be really true. The widths of the classes could be made smaller, but in that case some of the classes may not have any class frequency and the true pattern of the distribution in the population will not be known. If, however the number of observation is very large, the position will improve and no class is expected to be empty, even when the width of the classes are considerably small.
So if the total frequency is gradually increased indefinitely and that simultaneously the width of each class is gradually decreases that the number of classes goes on increasing, the histogram come closer and closer to a smooth curve and this limiting form of the histogram is called a frequency curve, which may be viewed as representing the true frequency distribution of the given continuous variable. The histogram shows the frequencies of the classes but the frequency curve shows the frequencies of the individual value of the variables. The frequency curve shows the probability distribution of the variable in the population and its area bounded by the ordinates at the two specified points on the horizontal axis represents the probability that a value of the variable lies between those limits. Like histogram, the frequency curve is therefore is the area diagram.
Consider the following example
15 16 17 18 19 20 21 22 23 24 25
2 3 10 8 14 15 19 16 11 20 14
26 27 28 29 30 31 32 33 34 36 37
10 7 3 9 5 8 3 6 2 1 2
39 40 43 45 46 49 50 52 57
1 3 2 1 1 1 1 1 1
Total
200
Class boundaries
Frequency(No of persons)
15-20 20-25 25-30 30-35 35-45 45-60
37 81 43 24 9 6
Total
200
90 80 70
15 -20 20 - 25 25- 30 30 - 35 35 - 45 45 - 50
60 50 40 30 20 10 0 Frequency(No of persons)
Frequency between 15 to 20 is 37 and it is assumed that observations in any classes are uniformly distributed throughout the range of values with the class boundaries i.e. . Number of observations having value 15 is approx 7.4,Number of observations having value 16 is approx 7.4 ,and so on. But in reality it is not true, the number of observations having value 15 is 2 not 7.4.
Value of variable
Frequency
15
2
16
3
17
10
18
8
19
14
16 14 12
15 16 17 18 19
10 8 6 4 2 0 Frequency
But if we keep on decreasing the width of the classes from 5 to 4, from 4 to 3, from 3 to 2 ultimately from 2 to 1, then the histogram will come closer and closer to a smooth curve and this limiting form of histogram i.e. the frequency curve will show the frequencies of individual value of the variable i.e. 15, 16,17 ……57 which is the true frequency distribution of the given continuous variable.
Problems 1)
Daily number of accidents
Frequency (Number of days)
3 4 5 6 7
5 9 11 4 1
Q 1.1) Will you choose Bar diagram or Histogram ,in order to represent this above table graphically. Draw the appropriate diagram and justify it. 1.2) Draw frequency polygon for the above table. 1.3) From the above table, calculate “less- than” cumulative frequencies and draw an appropriate cumulative frequency diagram.
Q 2.1) Draw histogram ,frequency polygon and ogives ( both “less-than” and “more-than” types) for the following frequency distribution. Class Interval (1)
Class frequency (2)
15-19 20-24 25-29 30-34 35-44 45-49
37 81 43 24 9 6
Total
200
Q 3.1) Draw histogram ,frequency polygon and ogives ( both “lessthan” and “more-than” types) for the following frequency distribution.
Size class
Frequency
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 - 100
5 11 19 21 16 10 8 6 3 1
Q 4.1) Draw histogram of the following frequency distribution. 4.2) Can we draw frequency polygon from this following distribution? If not, state the reason. Weekly profit (in 000’s Rs) 110 – 115 115 - 120 120 - 125 125 - 130 130 - 140 140 - 160 160 - 180
Number of shops 7 19 27 15 12 12 8
Q 5.1) Draw histogram ,frequency polygon and ogives ( both “less-than” and “more-than” types) for the following frequency distribution. 5.2) Determine i. the median ii. the number of companies getting between Rs. 45 crores and Rs 75 crores. Number of companies getting more than Rs. 45 crores iii. Number of companies getting more than Rs. 75 crores from the “less-than" type cumulative diagram. . .
Profits (Rs crores)
Number of companies
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 - 100
8 12 20 24 15 10 7 3 1
Q 6.1) Draw histogram ,frequency polygon and ogives ( both “less-than” and “more-than” types) for the following frequency distribution. 6.2) Determine i. the median . ii. the number of companies getting profits between Rs. 118 lakhs and Rs 148 lakhs. iii. Number of companies getting profits more than Rs. 118 lakhs . iv. Number of companies getting profits more than Rs.148 lakhs. from the “less-than" type cumulative diagram.
Yearly profits (Rs.lakh)
Number of companies
80 – 85 85 – 90 90 – 95 95 – 100 100 – 105 105 – 110 110 – 115 115 – 120 120 – 125 125 – 130 130 – 135 135 – 140 140 – 145 145 – 150 150 – 155
21 29 19 39 43 94 73 68 36 45 27 48 21 12 5
Total
580
Diagrammatic representation of frequency distribution
A frequency distribution can be presented graphically in any of the following ways:1)Histogram 2)Frequency polygon 3)Smoothed frequency curve 4)Cumulative frequency curves or Ogives
Bar diagram In case of the frequency distribution of a discrete variable, we present the distribution graphically by using the Bar diagram. Consider the following example,
Number of member in a family
Frequency
1
4
2
33
3
76
4
50
5
26
6
8
7
1
Total
198
Table - 1
Histogram But in case of frequency distribution of a continuous variable, we present the distribution graphically by using Histogram. Histogram is the most common form of diagrammatic representation of a grouped frequency distribution. It consists of a set of adjoining rectangles drawn on a horizontal base line. Width of rectangles, one for each class, extends over the class boundaries (not class limits) shown on the horizontal scale.
Width of the rectangle = Width of the class
Classes with equal width
When all classes have equal width, the heights of rectangles will be proportional to the class frequencies and it is then customary to take the heights numerically equal to the class frequencies i.e. Height of each rectangle = Class Frequency
For Example:Wages (Rs.) Class Limits
Class boundaries
Frequency (number of employees)
50 ---------- 59
49.5 ----------- 59.5
8
60 ---------- 69
59.5 ----------- 69.5
10
70 ---------- 79
69.5 ----------- 79.5
16
80 ---------- 89
79.5 ----------- 89.5
14
90 ---------- 99
89.5 ----------- 99.5
10
100 ---------- 109
99.5 ----------- 109.5
5
110 ---------- 119
109.5 ----------- 119.5
2
18 16 14 12 10 8 6 4 2 0 Frequency
(number of employees) 49.5 ----------59.5 59.5 ----------69.5 69.5 ----------79.5 79.5 ----------89.5 89.5 ----------99.5 99.5 ----------109.5 109.5 -----------
Classes with unequal width
When the classes are of unequal width the rectangles will also be of unequal width, and therefore the heights must be proportional to the frequency densities.Beause then, Height of each rectangle = Frequency Density
Area of each rectangle = Width x Height = (width of
class) x (frequency density) class frequency
= (width of
class) x
----------------------------
width of the class = Class Frequency
Annual Sales(Rs.’000) Class Interval
Frequency (f)
Width of class
Frequency Density (f /w)
0 -------20
20
20
1.00
20 ------ 50
50
30
1.67
50 ------100
69
50
1.38
100 ----- 250
30
150
0.20
250 ------ 500
25
250
0.10
500 ---- 1000
19
500
0.04
Another example Weekly profit (Rs.’000)
No of shops
110 – 115
7
115 – 120
19
120 – 125
27
125 – 130
15
130 – 140
12
140 – 160
12
160 - 180
8
Since the class- intervals are unequal, frequencies must be adjusted otherwise the histogram would give a misleading picture. The adjustment is done as follows:The lowest width of the classes is 5. The frequency of the class 130 – 140 shall be divided by 2 since it’s class width is double. Similarly following this same logic the frequencies of the classes 140 – 160 and 160 – 180 must be divided by 4.
Class boundaries
Height of rectangles
110 – 115
7
115 – 120
19
120 – 125
27
125 – 130
15
130 – 140
12 ÷ 2 = 6
140 – 160
12 ÷ 4 = 3
160 - 180
8÷4=2
Histogram vs Bar Diagram
Although vertical bar chart and histogram may appear some alike, the main point of distinction between them is that
The consecutive rectangles in a histogram have no space in between but the bar diagram must have equal spaces left between the consecutive rectangles bars.
Also, the rectangles in a bar diagram must be of equal width, but those in a histogram are proportional to the widths of classes and in case of classes with unequal width, the width of the respective rectangle will also be unequal.
The histogram is after all, an area diagram, which emphasises the widths of the rectangles between the class boundaries. But in a bar diagram only the heights is all-important, the spacing and the width of bars being arbitrary.
-:Uses:The series of rectangles is a histogram give a visual representation of the relative size of various groups and the entire distribution of total frequencies among different classes becomes at once visible. The surface of the tops of rectangles also gives an idea of the nature of frequency curve for the population. The histogram may be used to find the mode graphically.
Frequency Polygon in case of discrete variable An alternative method of presenting the discrete distribution is the frequency polygon, in which the values and the corresponding frequencies are plotted as points with help of regular co-ordinates, as in Bar-diagram.
For example:Table 1, may be presented by using frequency polygon, here first we have to plot the points (1,4) , (2,33) , (3,76) ,(4,50) , (5,26) , (6,8) , (7,1) on the graph. The value preceding 1 i.e. 0 has no frequency, as has the value following 7 i.e. 8.Hence let us take two more points (0,0) and (8,0).Finally we join the successive points by line segments and get the frequency polygon.
Frequency Polygon in case of continuous variable
F
requency polygon may be used to present the frequency
distribution of a continuous variable, provided the classes are of equal width. Frequency polygon is the graphical presentation alternative to the histogram and may be looked upon as derived from histogram by joining the mid-points of the tops of consecutive rectangles. It is generally used in cases when all the classes have a common width. The two end points are joined to the base line at the mid-values of the empty classes at each end of the frequency distribution. Thus the frequency polygon has the same area as the histogram, provided the width of the classes is the same.
Another method
Another method of constructing frequency polygon is to take the mid-points of the various class-intervals and then plot the frequency corresponding to each point and to join all these points by straight lines. The figure obtained would exactly be the same as obtained by the another method. The only difference is that here we do not have to construct the histogram.
Advantages 1. The frequency polygon is particularly useful in presenting simple frequency distribution of a discrete variable. 2. It gives us an approximate idea of the shape of frequency curve. 3. By constructing the frequency polygon the value of mode can be easily ascertained. If from the apex of the polygon a perpendicular is drawn on the x axis, we get the value of the mode.
4. Moreover the frequency polygon has a special advantage over the hiatogram.The frequency polygon of several distributions may be plotted on the same graph, thereby making certain comparison possible, whereas histograms cannot be usefully employed in the same way. To compare the histograms, we must have a separate graph for each. Because of this limitation for the purposes in making a graphic comparison of frequency distribution, frequency polygon is preferred.
For example
Frequency polygon in case of continuous variable
Class Limits
Class Boundaries
Mid-value
Frequency
50 - 59
49.5 – 59.5
54.5
8
60 – 69
59.5 – 69.5
64.5
10
70 – 79
69.5 – 79.5
74.5
16
80 – 89
79.5 – 89.5
84.5
14
90 – 99
89.5 – 99.5
94.5
10
100 – 109
99.5 – 109.5
104.5
5
110 - 119
109.5 – 119.5
114.5
2
Ogive (or cumulative frequency polygon)
Ogive is the graphical presentation of a
cumulative frequency distribution and hence is also called Cumulative Frequency Polygon. When cumulative frequencies are plotted against the corresponding class boundaries and the successive points are joined by straight lines, the line diagram obtained is known as Ogive or Cumulative frequency polygon.
Types of Ogive The ogive is of “less-than” or “more-than” type according as the cumulative frequencies used are of “less-than” or “morethan” type. The “less-than” type ogive starts from the lowest class boundary on the horizontal axis and gradually rising upward ends at the highest class boundary corresponding to the cumulative frequency N i.e. total frequency. It looks like the elongated S. The “more-than” type ogive has the appearance of an elongated S, turned upside down. Unequal widths of classes in the frequency distribution do not cause any difficulty in the construction of an ogive.
Uses 1. The ogive is used to find the Median, Quartile, deciles and percentiles or the value of the variable such that its cumulative frequency is a specified number. 2. It is also useful in finding the cumulative frequency corresponding to a given value of the variable. 3. To find the number of observations which are expected to lie between two specified value of the variable.
For Example:Illustration 1:Draw histogram, frequency polygon and ogives for the following frequency distribution :Wages
50 - 59
60 – 69
70 - 79
80 – 89
90 - 99
100 – 109
110 - 119
No of employee
8
10
16
14
10
5
2
Class Limits
Class Boundaries
Mid-value
Frequency
50 - 59
49.5 – 59.5
54.5
8
60 – 69
59.5 – 69.5
64.5
10
70 – 79
69.5 – 79.5
74.5
16
80 – 89
79.5 – 89.5
84.5
14
90 – 99
89.5 – 99.5
94.5
10
100 – 109
99.5 – 109.5
104.5
5
110 - 119
109.5 – 119.5
114.5
2
Class boundary
Cumulative Cumulative frequency “less- frequency “morethan” type than” type
49.5
0
65 = N
59.5
8
57
69.5
18
47
79.5
34
31
89.5
48
17
99.5
58
7
109.5
63
2
119.5
65 = N
0
Calculation of median from ogive Class boundary 49.5 59.5 69.5 Median------79.5 89.5 99.5 109.5 119.5
Cumulative frequency “less-than” type 0 8 18 -------- N/2 = 32.5 34 48 58 63 65 = N
or
Median – 69.5 32.5 - 18 ------------------ = ---------------------79.5 – 69.5 34 - 18
Or Median = 69.5 + 10 x (14.5/16) = 69.5 + 9.06 = 78.56 (Ans)
Step Diagram
It will be noticed that the cumulative frequency
diagram for a discrete variable from a simple frequency distribution is a step diagram, the cumulative frequency remaining constant between two successive values of the variables. For example:- Consider the following table, the cumulative frequency is 0 for any value less than one, is 4 for any value greater than or equal to 1 but less than 2 ,is 37 for any value greater than or equal to 2 but less than 3 and so on.
Family size
Frequency
Cumulative frequency “less-than” type
1
4
4
2
33
37
3
76
113
4
50
163
5
26
189
6
8
197
7
1
198 = N
Total
N =198
(number of observations having value 2 or less than 2)
Frequency Curve
T
he histogram gives us only an approximate idea of the nature of distribution with limited number of observations. By drawing rectangles in a histogram, we assume that the observations in any class are uniformly distributed throughout the range of values within the class boundaries, although this may not be really true. The widths of the classes could be made smaller, but in that case some of the classes may not have any class frequency and the true pattern of the distribution in the population will not be known. If, however the number of observation is very large, the position will improve and no class is expected to be empty, even when the width of the classes are considerably small.
So if the total frequency is gradually increased indefinitely and that simultaneously the width of each class is gradually decreases that the number of classes goes on increasing, the histogram come closer and closer to a smooth curve and this limiting form of the histogram is called a frequency curve, which may be viewed as representing the true frequency distribution of the given continuous variable. The histogram shows the frequencies of the classes but the frequency curve shows the frequencies of the individual value of the variables. The frequency curve shows the probability distribution of the variable in the population and its area bounded by the ordinates at the two specified points on the horizontal axis represents the probability that a value of the variable lies between those limits. Like histogram, the frequency curve is therefore is the area diagram.
Consider the following example
15 16 17 18 19 20 21 22 23 24 25
2 3 10 8 14 15 19 16 11 20 14
26 27 28 29 30 31 32 33 34 36 37
10 7 3 9 5 8 3 6 2 1 2
39 40 43 45 46 49 50 52 57
1 3 2 1 1 1 1 1 1
Total
200
Class boundaries
Frequency(No of persons)
15-20 20-25 25-30 30-35 35-45 45-60
37 81 43 24 9 6
Total
200
90 80 70
15 -20 20 - 25 25- 30 30 - 35 35 - 45 45 - 50
60 50 40 30 20 10 0 Frequency(No of persons)
Frequency between 15 to 20 is 37 and it is assumed that observations in any classes are uniformly distributed throughout the range of values with the class boundaries i.e. . Number of observations having value 15 is approx 7.4,Number of observations having value 16 is approx 7.4 ,and so on. But in reality it is not true, the number of observations having value 15 is 2 not 7.4.
Value of variable
Frequency
15
2
16
3
17
10
18
8
19
14
16 14 12
15 16 17 18 19
10 8 6 4 2 0 Frequency
But if we keep on decreasing the width of the classes from 5 to 4, from 4 to 3, from 3 to 2 ultimately from 2 to 1, then the histogram will come closer and closer to a smooth curve and this limiting form of histogram i.e. the frequency curve will show the frequencies of individual value of the variable i.e. 15, 16,17 ……57 which is the true frequency distribution of the given continuous variable.
Problems 1)
Daily number of accidents
Frequency (Number of days)
3 4 5 6 7
5 9 11 4 1
Q 1.1) Will you choose Bar diagram or Histogram ,in order to represent this above table graphically. Draw the appropriate diagram and justify it. 1.2) Draw frequency polygon for the above table. 1.3) From the above table, calculate “less- than” cumulative frequencies and draw an appropriate cumulative frequency diagram.
Q 2.1) Draw histogram ,frequency polygon and ogives ( both “less-than” and “more-than” types) for the following frequency distribution. Class Interval (1)
Class frequency (2)
15-19 20-24 25-29 30-34 35-44 45-49
37 81 43 24 9 6
Total
200
Q 3.1) Draw histogram ,frequency polygon and ogives ( both “lessthan” and “more-than” types) for the following frequency distribution.
Size class
Frequency
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 - 100
5 11 19 21 16 10 8 6 3 1
Q 4.1) Draw histogram of the following frequency distribution. 4.2) Can we draw frequency polygon from this following distribution? If not, state the reason. Weekly profit (in 000’s Rs) 110 – 115 115 - 120 120 - 125 125 - 130 130 - 140 140 - 160 160 - 180
Number of shops 7 19 27 15 12 12 8
Q 5.1) Draw histogram ,frequency polygon and ogives ( both “less-than” and “more-than” types) for the following frequency distribution. 5.2) Determine i. the median ii. the number of companies getting between Rs. 45 crores and Rs 75 crores. Number of companies getting more than Rs. 45 crores iii. Number of companies getting more than Rs. 75 crores from the “less-than" type cumulative diagram. . .
Profits (Rs crores)
Number of companies
10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 - 100
8 12 20 24 15 10 7 3 1
Q 6.1) Draw histogram ,frequency polygon and ogives ( both “less-than” and “more-than” types) for the following frequency distribution. 6.2) Determine i. the median . ii. the number of companies getting profits between Rs. 118 lakhs and Rs 148 lakhs. iii. Number of companies getting profits more than Rs. 118 lakhs . iv. Number of companies getting profits more than Rs.148 lakhs. from the “less-than" type cumulative diagram.
Yearly profits (Rs.lakh)
Number of companies
80 – 85 85 – 90 90 – 95 95 – 100 100 – 105 105 – 110 110 – 115 115 – 120 120 – 125 125 – 130 130 – 135 135 – 140 140 – 145 145 – 150 150 – 155
21 29 19 39 43 94 73 68 36 45 27 48 21 12 5
Total
580
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June 2020 0