In the previous section we completed the discussion on Bar graphs. We saw that bar graphs correspond to frequency distribution tables. In a similar way, Histograms correspond to Grouped frequency distribution tables. We can write this:
• Bar graphs → Frequency distribution tables
• Histograms → Grouped Frequency distribution tables
We have learned some basics about histograms in earlier classes. Details here. In the following discussion, we will see more details:
• We see that the class-intervals are: 30.5-35.5, 35.5-40.5, ..., 55.5-60.5
• There must be one rectangle (a rectangle can be considered as a ‘bar with a width’) for each class-interval
• The base of all the rectangles should rest on the x-axis
• For histograms, we cannot give ‘any convenient’ width for the rectangles. There must be a fixed scale. In this example, while drawing on a fresh graph paper, the scale '1 cm = 5 kg' will be convenient.
• Once we fix this scale, we can calculate the width that should be given to the rectangles.
• The class-interval 30.5-35.5 has a class width of 5 kg. So, according to the scale ‘1 cm = 5kg’, the width of the rectangle will be 1 cm
• All the class-intervals have the same width of 5 kg. So all the rectangles will be 1 cm wide
• The class-intervals are continuous. That is., each class-interval begins at the exact point where it’s preceding class ends.
• So there are no gaps between class-intervals
• Thus, in the histogram, there will not be any gaps between the rectangles
• Just as in the bar graph, in histograms also, the height of the bars are taken from the frequency column of the table.
• To mark the heights, we need to fix a suitable scale for the y-axis also.
• In this example, while drawing on the graph paper, the scale '1 cm = 5 numbers' will be convenient.
• The completed histogram is shown below in fig.25.6:
• Note that, since there is no gap between the individual rectangles, the whole graph appears like a solid fig.
• Consider the areas of the various rectangles.
♦ The rectangle which contain greater frequency will be having greater area
♦ The rectangle which contain lesser frequency will be having lesser area
• But the widths of all the rectangles are same. Height is the variable. So we can write this:
♦ The rectangle which contain greater frequency will be having greater height
♦ The rectangle which contain lesser frequency will be having lesser height
Let us see another case:
A teacher takes maths classes for 2 classes in STD IX of a school. There are a total of 90 students in the 2 classes. She wanted to make an assessment of their performance in an exam. So she tabulated the marks obtained by the students in various class intervals:
• Looking at the table, we can see that:
♦ the first interval has a width of 20
♦ the last interval has a width of 30
♦ All other intervals in between have width 10
• Why would she use different widths like this?
♦ Ans: She found that, only a very few students scored less than 20.
So if she gives two intervals 0-10 and 10-20, the number of students in each of those two intervals will be still less. So she decided to put those two intervals together
♦ Similarly, she found that, only a very few students scored greater than 70.
So if she gives three intervals 70-80, 80-90 and 90-100, the number of students in each of those three intervals will be still less. So she decided to put those three intervals together
• Thus we get the table shown in fig.25.7 above. Such a table may give a rough assessment about the students' performance. But when we draw a histogram, the results will go wrong.
Let us draw a histogram based on the table in fig.25.7 and find it's drawbacks:
■ Consider the first rectangle. It has width 20 and height 7. So it's area = 20×7 = 140 sq.units
• Consider the second rectangle. It has width 10 and height 10. So it's area = 10×10 = 100 sq.units
• The first rectangle has greater area. And second rectangle has lesser area.
• We have seen above that, the area of rectangles are proportional to it's frequency
• This would mean that, the first rectangle has greater frequency than the second
• But this is not the reality. The frequency 7 of first rectangle is less than the frequency 10 of the second rectangle.
■ Consider the last rectangle. It has width 30 and height 8. So it's area = 30×8 = 240 sq.units
• Consider the second last rectangle. It has width 10 and height 15. So it's area = 10×15 = 150 sq.units
• The last rectangle has greater area. And second last rectangle has lesser area.
• We have seen above that, the area of rectangles are proportional to it's frequency
• This would mean that, the last rectangle has greater frequency than the second last
• But this is not the reality. The frequency 8 of last rectangle is less than the frequency 15 of the second last rectangle.
■ So the histogram is giving misleading results. We have to adjust the heights of the rectangles. Let us see how this can be done:
■ Consider the first rectangle in fig.25.8. We want to reduce it's height. For that, the following procedure is adopted:
1. Divide the rectangle into equal rectangles
2. Width of each of those equal rectangles should be equal to the smallest class width in the given problem
3. The smallest class width in the given problem is 10
4. So divide the first rectangle into equal rectangles, each of width 10
5. How many such equal rectangles will we get?
The answer is: width of rectangle⁄smallest width = 20⁄10 = 2
6. So the first rectangle is divided into 2 equal rectangles of width 10 each. This is shown by red dotted lines in fig.25.9 below:
7. Now distribute the frequency uniformly among the 2 new rectangles. That means, the frequency associated with the first large rectangle in fig.25.8 is now divided equally among the 2 new rectangles in fig.25.9
8. So the frequency of each of the new rectangles in fig.25.9 = 7⁄2 = 3.5
9. So height of the 2 new equal rectangles in fig.25.9 = 3.5 units
■ Consider the last rectangle in fig.25.8. We want to reduce it's height. For that, the same procedure is adopted:
1. Divide the rectangle into equal rectangles
2. Width of each of those equal rectangles should be equal to the smallest class width in the given problem
3. The smallest class width in the given problem is 10
4. So divide the last rectangle into equal rectangles, each of width 10
5. How many such equal rectangles will we get?
The answer is: width of rectangle⁄smallest width = 30⁄10 = 3
6. So the last rectangle is divided into 3 equal rectangles of width 10 each. This is shown by green dotted lines in fig.25.9 above
7. Now distribute the frequency uniformly among the 3 new rectangles. That means, the frequency associated with the last large rectangle in fig.25.8 is now divided equally among the 3 new rectangles in fig.25.9
8. So the frequency of each of the new rectangles in fig.25.9 = 8⁄3 = 2.67
9. So height of the 3 new equal rectangles in fig.25.9 = 2.67 units
The dashed lines in fig.25.9 above shows the 'equal division'. They need not be shown in the final histogram. But we do need to make a modified table as shown below:
The final histogram is shown in fig.25.11 below:
Now we will write the steps in general so that we can apply it to all such cases:
1. Find the smallest width in the problem
2. Leave all the intervals with the smallest width as such
3. Take the rectangles with the higher widths
4. Divide each of them into equal rectangles
5. Width of all those equal rectangles should be the 'smallest width' in the problem
6. So number (n) of 'equal rectangles' in each large rectangle = width⁄smallest width
7. Divide the frequency uniformly among the new 'equal rectangles'
8. So new frequency = original frequency⁄n
Once we understand the above basic steps, the steps can be further condensed:
1. Find the smallest width in the problem
2. Leave all the intervals with the smallest width as such
3. New frequency of each of the other intervals is given by:
original frequency of that interval⁄n
Where n = width of that interval⁄smallest width in the problem
In the next section we will see some solved examples.
• Bar graphs → Frequency distribution tables
• Histograms → Grouped Frequency distribution tables
We have learned some basics about histograms in earlier classes. Details here. In the following discussion, we will see more details:
Histograms
Given below is a ‘grouped frequency distribution table’. It represents the weights of 36 students in a class. |
| Fig.25.5 |
• There must be one rectangle (a rectangle can be considered as a ‘bar with a width’) for each class-interval
• The base of all the rectangles should rest on the x-axis
• For histograms, we cannot give ‘any convenient’ width for the rectangles. There must be a fixed scale. In this example, while drawing on a fresh graph paper, the scale '1 cm = 5 kg' will be convenient.
• Once we fix this scale, we can calculate the width that should be given to the rectangles.
• The class-interval 30.5-35.5 has a class width of 5 kg. So, according to the scale ‘1 cm = 5kg’, the width of the rectangle will be 1 cm
• All the class-intervals have the same width of 5 kg. So all the rectangles will be 1 cm wide
• The class-intervals are continuous. That is., each class-interval begins at the exact point where it’s preceding class ends.
• So there are no gaps between class-intervals
• Thus, in the histogram, there will not be any gaps between the rectangles
• Just as in the bar graph, in histograms also, the height of the bars are taken from the frequency column of the table.
• To mark the heights, we need to fix a suitable scale for the y-axis also.
• In this example, while drawing on the graph paper, the scale '1 cm = 5 numbers' will be convenient.
• The completed histogram is shown below in fig.25.6:
 |
| Fig.25.6 |
• Consider the areas of the various rectangles.
♦ The rectangle which contain greater frequency will be having greater area
♦ The rectangle which contain lesser frequency will be having lesser area
• But the widths of all the rectangles are same. Height is the variable. So we can write this:
♦ The rectangle which contain greater frequency will be having greater height
♦ The rectangle which contain lesser frequency will be having lesser height
Let us see another case:
A teacher takes maths classes for 2 classes in STD IX of a school. There are a total of 90 students in the 2 classes. She wanted to make an assessment of their performance in an exam. So she tabulated the marks obtained by the students in various class intervals:
 |
| Fig.25.7 |
♦ the first interval has a width of 20
♦ the last interval has a width of 30
♦ All other intervals in between have width 10
• Why would she use different widths like this?
♦ Ans: She found that, only a very few students scored less than 20.
So if she gives two intervals 0-10 and 10-20, the number of students in each of those two intervals will be still less. So she decided to put those two intervals together
♦ Similarly, she found that, only a very few students scored greater than 70.
So if she gives three intervals 70-80, 80-90 and 90-100, the number of students in each of those three intervals will be still less. So she decided to put those three intervals together
• Thus we get the table shown in fig.25.7 above. Such a table may give a rough assessment about the students' performance. But when we draw a histogram, the results will go wrong.
Let us draw a histogram based on the table in fig.25.7 and find it's drawbacks:
 |
| Fig.25.8 |
• Consider the second rectangle. It has width 10 and height 10. So it's area = 10×10 = 100 sq.units
• The first rectangle has greater area. And second rectangle has lesser area.
• We have seen above that, the area of rectangles are proportional to it's frequency
• This would mean that, the first rectangle has greater frequency than the second
• But this is not the reality. The frequency 7 of first rectangle is less than the frequency 10 of the second rectangle.
■ Consider the last rectangle. It has width 30 and height 8. So it's area = 30×8 = 240 sq.units
• Consider the second last rectangle. It has width 10 and height 15. So it's area = 10×15 = 150 sq.units
• The last rectangle has greater area. And second last rectangle has lesser area.
• We have seen above that, the area of rectangles are proportional to it's frequency
• This would mean that, the last rectangle has greater frequency than the second last
• But this is not the reality. The frequency 8 of last rectangle is less than the frequency 15 of the second last rectangle.
■ So the histogram is giving misleading results. We have to adjust the heights of the rectangles. Let us see how this can be done:
■ Consider the first rectangle in fig.25.8. We want to reduce it's height. For that, the following procedure is adopted:
1. Divide the rectangle into equal rectangles
2. Width of each of those equal rectangles should be equal to the smallest class width in the given problem
3. The smallest class width in the given problem is 10
4. So divide the first rectangle into equal rectangles, each of width 10
5. How many such equal rectangles will we get?
The answer is: width of rectangle⁄smallest width = 20⁄10 = 2
6. So the first rectangle is divided into 2 equal rectangles of width 10 each. This is shown by red dotted lines in fig.25.9 below:
 |
| Fig.25.9 |
8. So the frequency of each of the new rectangles in fig.25.9 = 7⁄2 = 3.5
9. So height of the 2 new equal rectangles in fig.25.9 = 3.5 units
■ Consider the last rectangle in fig.25.8. We want to reduce it's height. For that, the same procedure is adopted:
1. Divide the rectangle into equal rectangles
2. Width of each of those equal rectangles should be equal to the smallest class width in the given problem
3. The smallest class width in the given problem is 10
4. So divide the last rectangle into equal rectangles, each of width 10
5. How many such equal rectangles will we get?
The answer is: width of rectangle⁄smallest width = 30⁄10 = 3
6. So the last rectangle is divided into 3 equal rectangles of width 10 each. This is shown by green dotted lines in fig.25.9 above
7. Now distribute the frequency uniformly among the 3 new rectangles. That means, the frequency associated with the last large rectangle in fig.25.8 is now divided equally among the 3 new rectangles in fig.25.9
8. So the frequency of each of the new rectangles in fig.25.9 = 8⁄3 = 2.67
9. So height of the 3 new equal rectangles in fig.25.9 = 2.67 units
The dashed lines in fig.25.9 above shows the 'equal division'. They need not be shown in the final histogram. But we do need to make a modified table as shown below:
 |
| Fig.25.10 |
 |
| Fig.25.10 |
Now we will write the steps in general so that we can apply it to all such cases:
1. Find the smallest width in the problem
2. Leave all the intervals with the smallest width as such
3. Take the rectangles with the higher widths
4. Divide each of them into equal rectangles
5. Width of all those equal rectangles should be the 'smallest width' in the problem
6. So number (n) of 'equal rectangles' in each large rectangle = width⁄smallest width
7. Divide the frequency uniformly among the new 'equal rectangles'
8. So new frequency = original frequency⁄n
Once we understand the above basic steps, the steps can be further condensed:
1. Find the smallest width in the problem
2. Leave all the intervals with the smallest width as such
3. New frequency of each of the other intervals is given by:
original frequency of that interval⁄n
Where n = width of that interval⁄smallest width in the problem
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