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# T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 7 - Frequency Diagrams, Histograms, Polygon and Ogive

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## T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 7 - Frequency Diagrams, Histograms, Polygon and Ogive Page/Excercise 128

Solution SAQ 1

The given distribution can be represented with the help of a bar diagram as follows:

Bar Diagram Solution SAQ 2

1. Histogram: A histogram is two dimensional diagrams. 1. Frequency Polygon: A frequency polygon is drawn by joining the mid-points of all tops of a histogram. Here, the points are joined by using a foot rule. 1. Frequency Curve: Similar to frequency polygon, a frequency curve is drawn by joining the mid-points of all tops of a histogram. But, the points are joined using a free hand. Solution SAQ 3

1. Histogram: 1. Frequency Polygon: A frequency polygon is drawn by joining the mid-points of all tops of a histogram. Here, the points are joined by using a foot rule. Solution SAQ 4

1. 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. 1. 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 ## T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 7 - Frequency Diagrams, Histograms, Polygon and Ogive Page/Excercise 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

1.
1. Frequency polygon: Presenting the frequencies in the form of rectangle and joining the mid-points 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.  1. 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 mid-points are given to draw a histogram. So these mid-points are converted into class intervals.

Procedure

Step 1: Formula to derive the class intervals corresponding to each mid-point is  Step 2: Add and subtract 5 to each mid-point 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

1. Frequency polygon using histogram By plotting the mid-points of the class intervals with their respective frequencies, the mid points are joined to draw a frequency polygon. 1. 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. ## T.R. Jain and V.K. Ohri- Statistics for Economics Solution for Class 11 Commerce Statistics for Economics Chapter 7 - Frequency Diagrams, Histograms, Polygon and Ogive Page/Excercise 155

Solution SAQ 11

1. Histogram and Frequency Polygon 1. 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 mid-points is now included in the frequency polygon. Thus, the area under frequency polygon is equal to the area under histogram.

# Text Book Solutions

CBSE XI Commerce - Statistics for Economics

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