Class 11commerce TR JAIN AND VK OHRI Solutions Statistics for Economics Chapter 7  Frequency Diagrams, Histograms, Polygon and Ogive
Frequency Diagrams, Histograms, Polygon and Ogive Exercise 128
Solution SAQ 1
The given distribution can be represented with the help of a bar diagram as follows:
Bar Diagram
Solution SAQ 2
 Histogram: A histogram is two dimensional diagrams.
 Frequency Polygon: A frequency polygon is drawn by joining the midpoints of all tops of a histogram. Here, the points are joined by using a foot rule.
 Frequency Curve: Similar to frequency polygon, a frequency curve is drawn by joining the midpoints of all tops of a histogram. But, the points are joined using a free hand.
Solution SAQ 3
 Histogram:
 Frequency Polygon: A frequency polygon is drawn by joining the midpoints of all tops of a histogram. Here, the points are joined by using a foot rule.
Solution SAQ 4
 Less than ogive curves: In this method, frequencies are cumulated and presented in a graph corresponding to upper limits of the classes in a frequency distribution. Firstly, all the data are converted into less than cumulative frequency distribution as follows
Marks 
Cumulative Frequency 
Less than 5 
7 
Less than 10 
7 + 10 = 17 
Less than 15 
17 + 20 = 37 
Less than 20 
37 + 13 = 50 
Less than 25 
50 + 12 = 62 
Less than 30 
62 + 19 = 81 
Less than 35 
81 + 14 = 95 
Less than 40 
95 + 9 = 104 
This curve is drawn by plotting cumulative frequencies against the upper limit of the class intervals. And these points are joined to obtain the less than ogive curve.
 More than ogive curves: In this method, frequencies are cumulated and presented in a graph corresponding to lower limits of the classes in a frequency distribution. Firstly, all the data are converted into more than cumulative frequency distribution as follows
Marks 
Cumulative Frequency 
More than 0 
104 
More than 5 
104  7 = 97 
More than 10 
97  10 = 87 
More than 15 
87  20 = 67 
More than 20 
67  13 = 54 
More than 25 
54  12 = 42 
More than 30 
42  19 = 23 
More than 35 
23  14 = 9 
More than 40 
9  9 =0 
This curve is drawn by plotting cumulative frequencies against the lower limit of the class intervals. And these points are joined to obtain more than ogive curve.
Solution SAQ 5
Ogive curve is a smooth curve presented by plotting the frequency data on a graph. This curve represents the frequencies corresponding to lower limits or upper limits in the distribution of data.
Less than ogive curves: In this method, frequencies are cumulated and presented in a graph corresponding to upper limits of the classes in a frequency distribution. Firstly, all the data are converted into less than cumulative frequency distribution as follows
Capital (in lakh) 
Cumulative Frequency 
Less than 10 
2 
Less than20 
2 + 3 = 5 
Less than 30 
5 + 7 = 12 
Less than 40 
12 + 11 = 23 
Less than 50 
23 + 15 = 38 
Less than 60 
38 + 7 = 45 
Less than 70 
45 + 23 = 68 
This curve is drawn by plotting cumulative frequencies against the upper limit of the class intervals. And these points are joined to obtain the less than ogive curve.
Solution SAQ 6
Histogram
Solution SAQ 7
Histogram
Frequency Diagrams, Histograms, Polygon and Ogive Exercise 129
Solution SAQ 8
i. The frequency distribution of the given data is as follows:
Class Interval (Marks) 
Frequency (No. of Student) 
20  29 
2 
30  39 
5 
40  49 
8 
50  59 
6 
60  69 
4 

∑f = 25 
 Frequency polygon: Presenting the frequencies in the form of rectangle and joining the midpoints of the tops of the consecutive rectangles is known as frequency polygon. It is an alternative to histogram which is derived from histogram itself. However, frequency polygon can be drawn even without presenting the histogram.
Firstly, the mid points of the respective class intervals are calculated and presented graphically against their respective frequencies. Here, the points are joined by using a foot rule to obtain the frequency polygon curve.
 Less than ogive curves: In this method, frequencies are cumulated and presented in a graph corresponding to upper limits of the classes in a frequency distribution. Firstly, all the data are converted into less than cumulative frequency distribution as follows
Less than Ogive 
Cumulative Frequency 
Less than 29 
2 
Less than 39 
2 + 5 = 7 
Less than 49 
7 + 8 = 15 
Less than 59 
15+6=21 
Less than 69 
21+4=25 
This curve is drawn by plotting cumulative frequencies against the upper limit of the class intervals. And these points are joined to obtain the less than ogive curve.
Solution SAQ 9
Only midpoints are given to draw a histogram. So these midpoints are converted into class intervals.
Procedure
Step 1: Formula to derive the class intervals corresponding to each midpoint is
Step 2: Add and subtract 5 to each midpoint to get the following class intervals.
Mid point 
Class Interval 
Size 
115 
110  120 
6 
125 
120  130 
55 
135 
130  140 
48 
145 
140  150 
72 
155 
150  160 
116 
165 
160  170 
60 
175 
170  180 
38 
185 
180  190 
22 
195 
190  200 
3 
Solution SAQ 10
i. Frequency polygon: Frequency polygon can be drawn even without presenting the histogram. Firstly, the mid points of the respective class intervals are calculated and presented graphically against their respective frequencies. Here, the points are joined by using a foot rule to obtain the frequency polygon curve.
ii. Ogive curve: It is a smooth curve presented by plotting cumulative frequency data on a graph. This can be drawn by understanding the frequencies corresponding to lower limits and upper limits in the distribution of data.
Firstly, all the data are converted into more than and less than cumulative frequency distribution as follows
Marks 
Cumulative Frequency 
Less than 10 
3 
Less than 20 
3 + 10 = 13 
Less than 30 
13 + 14 = 27 
Less than 40 
27 + 10 = 37 
Less than 50 
37 + 3 =40 
Marks 
Cumulative Frequency 
More than 0 
40 
More than 10 
40  3 = 37 
More than 20 
37 + 10 = 27 
More than 30 
27 + 14 = 13 
More than 40 
13 + 10 = 3 
This less than curve is drawn by plotting cumulative frequencies against the upper limit of the class intervals. And these points are joined to obtain the less than ogive curve. And more than curve is drawn by plotting cumulative frequencies against the lower limit of the class intervals. And these points are joined to obtain more than ogive curve.
Solution SAQ 12
 Frequency polygon using histogram
By plotting the midpoints of the class intervals with their respective frequencies, the mid points are joined to draw a frequency polygon.
 Frequency polygon without using histogram
Solution SAQ 13
If the class intervals are equal but the series are inclusive, then inclusive series are converted into an exclusive series.
Step1: Apply the formula to convert into exclusive series
Step 2: Add and subtract 0.5 to each class intervals.
Weight 
Frequency 
29.5  34.5 
3 
34.5  39.5 
5 
39.5  44.5 
12 
44.5  49.5 
18 
49.5  54.5 
14 
54.5  59.5 
6 
59.5  64.5 
2 
i. Less than ogive curves: In this method, frequencies are cumulated and presented in a graph corresponding to upper limits of the classes in a frequency distribution. Firstly, all the data are converted into less than cumulative frequency distribution as follows
Weight 
Cumulative Frequency 
Less than 34.5 
3 
Less than 39.5 
3 + 5 = 8 
Less than 44.5 
8 + 12 = 20 
Less than 49.5 
20 + 18 = 38 
Less than 54.5 
38 + 14 = 52 
Less than 59.5 
52 + 6 = 58 
Less than 64.5 
58 + 2 = 60 
This curve is drawn by plotting cumulative frequencies against the upper limit of the class intervals. And these points are joined to obtain the less than ogive curve.
ii. More than ogive curves: In this method, frequencies are cumulated and presented in a graph corresponding to lower limits of the classes in a frequency distribution. Firstly, all the data are converted into more than cumulative frequency distribution as follows
Weight 
Cumulative Frequency 
More than 0 
60 
More than 34.5 
60  3 = 57 
More than 39.5 
57  5 = 52 
More than 44.5 
52  12 = 40 
More than 49.5 
40  18 = 22 
More than 54.5 
22  14 = 8 
More than 59.5 
8  6 =2 
More than 64.5 
2  2 = 0 
This curve is drawn by plotting cumulative frequencies against the lower limit of the class intervals. And these points are joined to obtain more than ogive curve.
Frequency Diagrams, Histograms, Polygon and Ogive Exercise 155
Solution SAQ 11
 Histogram and Frequency Polygon
 The first class interval is extended to the left side by half the size of class interval which denotes the initial point of the frequency polygon. And the last class interval is extended to the right side by half the size of the class interval which denotes the final point of the frequency polygon. This implies that the area not included at the time of joining the midpoints is now included in the frequency polygon. Thus, the area under frequency polygon is equal to the area under histogram.