Chapter 25.5 - Solved examples on Histograms

In the previous section we discussed about Histograms. In this section we will see some solved examples.
Solved example 25.13
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table (a):

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
Solution:
The class intervals in the given table are not continuous. There is a gap of '1 mm' between all the class intervals. So our first task is to make the classes continuous. Table (b) gives the required continuous classes.
A sample calculation: 
Consider the gap between 145-153 and 154-162
We will use the general method that we saw earlier:
Step 1: Gap =  lower limit of 2^nd – upper limit of 1^st = 154 - 153 = 1 
Step 2: New upper limit of 1^st  = Original upper limit  + ^Gap⁄₂ = 153 + 1/2 = 153 + 0.5 = 153.5
Step 3: New lower limit of 2^nd = Original lower limit  - ^Gap⁄₂  = 154 - 1/2 = 154 - 0.5 = 153.5
Based on table (b), the histogram is prepared as shown in fig.25.11 below:

Fig.25.11

A sample rectangle:
• Consider the third rectangle: It corresponds to the third class interval 135.5-144.5 in table (b). It's frequency in the table (b) is 9. 
• In the histogram, we see that the rectangle of this class interval has a height above the horizontal line through 8. In fact, it is half way between 8 and 10. That means, it is at 9. 
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.  
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars
Part (ii)
No. It is not correct to conclude that the maximum number of leaves are 153 mm long.
To write the reason, let us write from the beginning:
1. In this experiment, the length of each of the 40 leaves are measured.
2. It is difficult to show those 40 lengths in a table. Even if it is a 'frequency distribution table', it will be a lengthy table
3. So they adopted the 'grouped frequency distribution table'. In this, there are groups which are known as class intervals.
4. Consider the example of the class 126.5-135.5
• The leaves falling in this class may have lengths any where from 126.5 to 135.5, excluding 135.5.
• That means, several lengths are possible in this class
5. Similarly, consider the class 144.5-153.5. Several lengths are possible in this class.
■ We cannot say: 'all the 12 leaves in this class are of 153 cm length'

Solved example 25.14
The following table gives the life times of 400 neon lamps:

(i) Represent the given information with the help of a histogram. 
(ii) How many lamps have a life time of more than 700 hours?
Solution:
The required histogram is given below in fig.25.12:

Fig.25.12

A sample rectangle:
• Consider the fifth rectangle: It corresponds to the fifth class interval 700-800 in table. It's frequency in the table is 74. 
• In the histogram, we see that the rectangle of this class interval has a height just above the horizontal line through 70. The excess above the horizontal line is '4'.
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.  
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars
Part (ii)
• The number of bulbs which have a life time of 700 hours or more will be represented by the rectangles which lie on the right side of the '700 mark' on the x-axis
• So the required number is: 74 +62 +48 = 184

Solved example 25.15
A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above
Solution:
In the given table, the class intervals are not equal. So we need to find adjusted frequencies. They are tabulated in the table below:

A sample calculation:
We will use the general method that we saw earlier:
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}
• The smallest width is 1. So we will leave the first and second class intervals as such. The frequencies of those classes do not need any adjustments
• We will take the interval 10-15 as a sample
• 'n' for that interval =  ^{width of that interval}⁄_{smallest width in the problem} = 5/1 = 5
• New frequency of that interval = ^{original frequency of that interval}⁄_n  = 10/5 = 2 
The final histogram is shown below:

Fig.25.13

A sample rectangle:
• Consider the fifth rectangle: It corresponds to the fifth class interval 7-10 in the modified table. It's width is 3 and frequency is 3. 
• In the histogram, we see that the rectangle of this class interval has a height same as the horizontal line through 3.
• The width of the base on the x-axis is also 3. That is., from 7 to 10
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.

Solved example 25.16
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie
Solution:
In the given table, the class intervals are not equal. So we need to find adjusted frequencies. They are tabulated in the table below:

A sample calculation:
We will use the general method that we saw earlier:
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}
• The smallest width is 2. So we will leave the second and third class intervals as such. The frequencies of those classes do not need any adjustments
• We will take the interval 8-12 as a sample
• 'n' for that interval =  ^{width of that interval}⁄_{smallest width in the problem} = 4/2 = 2
• New frequency of that interval = ^{original frequency of that interval}⁄_n  = 16/2 = 8 

The final histogram is shown below:

Fig.25.14

A sample rectangle:
• Consider the fourth rectangle: It corresponds to the fourth class interval 8-12 in the modified table. It's width is 4 and frequency is 8. 
• In the histogram, we see that the rectangle of this class interval has a height above 5 and below 10. It is near to 10 than to 5.
• The width of the base on the x-axis is 4. That is., from 8 to 12
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars

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So we have completed the discussion on histograms. In the next section we will see 'frequency polygons'.

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Chapter 25.4 - Pictorial representation of Data - Histograms

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:

Histograms

Given below is a ‘grouped frequency distribution table’. It represents the weights of 36 students in a class.

Fig.25.5

• 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:

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

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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

• 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:

Fig.25.8

■ 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:

Fig.25.9

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:

Fig.25.10

The final histogram is shown in fig.25.11 below:

Fig.25.10 

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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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In the next section we will see some solved examples.

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Chapter 1.3 - Solved examples on Frequency distribution tables and Histograms

In the previous section we saw the method for preparing Discrete Frequency distribution tables, Grouped Frequency distribution tables, Bar graphs and Histograms. In this section we will see some solved examples.

Solved example 1.1

There are 44 students in a class. The list below shows the distance (in km) that they travel from their home to school.

6	2	7	12	1	9	2	6
5	7	3	4	1	5	4	4
5	8	6	5	2	5	9	5
11	12	1	9	2	14	4	7
9	6	6	7	3	2	6	3
4	7	9	3

Make a frequency table and answer the following questions:

1) How many students are from exactly 1 km away?

2) How many students travel more than 5 kms?

3) How many travel between 5 and 10 kms?

4) How many travel more than 10 kms?

Solution:

Fig.1.21

The frequency distribution table is shown in fig.1.21 below. It is a Discrete frequency distribution table. The distances in the raw data list are written in ascending order in the column 1.

Based on the table, we can answer the given questions.

1. The answer 3 is obtained from the first row

2. More than 5 kms means 5 km is not to be included. So we consider all the rows below the one corresponding to 5 km.  Thus we get the answer as 21 students

3. Between 5 km and 10 km means both 5 and 10 are to be included. So we consider all the rows below 4 and above 11. Thus we get the answer as 23 students.

4. More than 10 km means 10 km is not included. So we consider all the higher values than 10. That is ., row 11 and all the rows below 11. Thus we get the answer as 4.

Solved example 1.2

The weights of the members of the school health club are given below:

38	37.5	40.5	59	48	48	37.7
58	50	54.5	39	40	40.5	49
32	43	45	53	37	44	51
50.5	32.5	46	55	36	44.5	47
42.5	33

Prepare a Grouped frequency distribution table. 

Solution:
The smallest value is 32, and the largest value is 59. So we need the region between 32 and 59 on the number line. This is shown in the fig below:

The length of the required portion is 59 -32 = 27. If we decide to fix the width of class intervals to be equal to 10, then the number of class intervals will be equal to 27/10 = 2.7. So we can make utmost 3 class intervals. This is rather low. When the number of class intervals are low, the values in the raw data list will fall within a small area in the number line. It will appear like the raw data list itself. It will not serve the purpose of an ‘organised distribution’.

On the other extreme, if the class interval is very low, the values will fall within a large area in the number line. More rows will be required in the table, and it will appear like a ‘Discrete frequency distribution table’.

So instead of 10, let us choose a 'class interval' of 5. Then the number line should also show the intervals of 5. That is., number line should show 0, 5, 10, 15 etc., This is shown below:

In the above fig., we can see that
• Some portion before 32, and some portion after 59 has to be included, in order to make the intervals 'equal'. So we have to extend the red portion towards the left upto 30, and towards the right upto 60.
• Also note that, in this present problem, no values are falling in the region from 0 to 30. So that region can be avoided in our diagrams. But when such a region is avoided, a 'cut line' should be shown.

The following fig. shows these two modifications

So now we have the required region, and it is divided into 6 equal class intervals, each of width 5. We can make the frequency distribution table as shown in fig.1.25. It is a Grouped frequency distribution table. kk

Fig.1.25

The Histogram based on the table is shown in fig.1.26 below:

Fig.1.26

[It may be noted that, the Histograms are usually drawn on Graph papers. They show the units clearly. The divisions are shown with thin lines and the sub divisions are shown with thinner lines. So, for a histogram drawn on graph paper, we do not have to show the horizontal white dotted lines as in fig.1.26.]

From the above table and histogram, we can make the following conclusions:
• Only 3 students have a heavy weight of 55 to 60 kg
• Only 3 students have a light weight of 30 to 35 kg
• Most students have a weight ranging from 35 to 55 kg. And in this, the range possessed by the largest number of students is 40 to 45. It is possessed by 7 students.

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Now let us take a look at the extreme cases, when the class interval is too small or too large. First we will see the case when the class is too small. If we take the class as '2', the histogram will be as shown in the fig.1.27 below.

Fig.1.27

From the histogram, we can make the following conclusions:
• Only 2 students have heavy weights of 58 to 60 kg
• Only 3 students have light weights of 32 to 34 kg
• Most students have a weight ranging from 36 to 52 kg. And in this, the range possessed by the largest number of students is 36 to 38. It is possessed by 4 students.

We can see that this histogram gives more 'detailed' results. But it closely resembles a Bar graph, and requires more space. On many occasions, such high level of details may not be required.

Next we will see the other extreme case when the class is too large. If we take the class as '10', the histogram will be as shown in the fig.1.28 below:

Fig.1.28

From the histogram, we can make the following conclusions:
• 9 students have weights ranging from 30 to 40
• 9 students have weights ranging from 50 to 60
• 12 students have weights ranging from 40 to 50

We can see that this histogram gives less detailed results. On many occassions, such low level of details may not be sufficient. It closely resembles the 'raw data' itself because, the number of students with particular weights are distributed randomly in the three classes.

So for a particular problem, we must select the class interval according to the level of detail required. Some times the class interval that we have to use may be specified in the problem.

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Solved example 1.3
The weekly wages of 30 workers in a factory are given below:

830	835	890	810	835	836	869
845	898	890	820	860	832	833
855	845	804	808	812	840	885
835	835	836	878	840	868	890
806	840

(1) Make a Grouped frequency table with intervals as 800 – 810, 810 – 820 and so on.
(2) Draw a histogram for the frequency table made for the data in Question 3, and answer the following questions.
(i) Which interval has the maximum number of workers?
(ii) How many workers earn 860 and more?
(iii) How many workers earn less than 860?

Solution:
The class intervals are given in the question itself. So we can directly make the table. It is shown in the fig. 1.29 below:

Fig.1.29

Based on the above table, the Histogram is prepared and is shown in fig.1.30 below:

Fig.1.30

Based on the above histogram, we can answer the questions.
(i) The tallest rectangle is the one corresponding to the interval 820 - 840. It has 10 workers. So the answer is : interval 820 - 840
(ii) After the 860 mark, there are two rectangles. The first has 4 workers and the second has 5 workers. So the answer is 9. [Note that, if a worker has a wage of exactly 860, then he is included in 860 - 880, and not 840 - 860. All the values greater than 860 will fall to the right of the 860 mark on the number line. So all the rectangles to the right of the 860 mark are to be considered]
(iii) There are 3 rectangles to the left of 860. From their heights, we get the answer as 5 +10 +6 =21

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In the next section, we will discuss about Pie charts.

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Chapter 1.2 - Frequency distribution tables and histograms

In the previous section we saw how the 'equal intervals' are formed in the 'Electricity bill analysis problem'. Now we will form the frequency distribution table, based on these intervals.

The first column can be readily filled up with the 'equal intervals' as shown below:

Fig.1.16 Table for making ordered list

Now we proceed to do the tally marking. The steps are as follows:
• Take the first entry. It is 224.
• Find the interval in which 224 will fall. It is 200 - 250.
• Put a '|' mark on the second column 'in line' with this interval. This is shown in the fig.1.17 below:

Fig.1.17 Tally mark for the first entry

Repeat the process:
• Take the second entry. It is 94.
• Find the interval in which 94 will fall. It is 50 - 100.
• Put a '|' mark on the second column 'in line' with this interval. This is shown in the fig.1.18 below:

Fig.1.18 Tally mark for the second entry

In this way, all the entries in the raw data list should be marked in the second column. Then the no. of tally marks can be counted and entered in the third column. The completed table is shown below:

Fig.1.19 Completed Table

The pictorial representation of the above table is given below:

Fig.1.20 Pictorial representation

The above fig.1.20 gives a better presentation of the results of analysis. Each interval in the table is represented by a rectangle. The heights of the rectangles gives us the frequency of that interval. We can see that 
• The interval 200 – 250 has the most frequency. This means that, the 'number of houses whose bill amount falls between 200 and 250' is more in that locality.
• The number of houses which has very low bill amounts (less than 100) is less in that locality
• The number of houses which has very high bill amounts (greater than 350) is also low in that locality.
• Most of the houses have a bill amount between 150 and 350. 

Now let us see some important features of the 'methods of organising data' that we saw so far:
• Type of Table
    ♦ The type of table shown in fig.1.7 in which each of the values in the raw data list is given a unique row is called Discrete frequency distribution table. The pictorial representation of this table is called Bar graph. In the Bar graph, each of the values in the raw data list is given a unique bar. (However, repeating values will be having a single common row and a single common bar.)
     ♦ The type of table shown in fig.1.19 above, in which the values in the raw data list are grouped is called a Grouped frequency distribution table.

• Pictorial representation
     ♦ The pictorial representation of a Discrete frequency distribution table is called a Bar graph
      ♦ The pictorial representation of a Grouped frequency distribution is called a Histogram

• The intervals (shown in the first column) in a grouped frequency distribution table are called class intervals (or simply 'class'). For example, 150 – 200 in fig.1.19 is a class interval.

• Every class interval has a lower class limit and an upper class limit. For example, in the class interval 300 – 350, 300 is the lower class limit. And 350 is the upper class limit.

• The difference between the upper class limit and the lower class limit is called the width or size of the class interval. And the widths of all the class intervals in a grouped frequency distribution table should be the same.

• The class intervals should be continuous. That is., the upper class limit of one class interval should be the lower class limit of the next higher class interval. The importance of such a rule can be explained with the help of examples:

Example of a continuous class interval sequence: 0 – 20, 20 – 40, 40 – 60, . . . etc.,  
Example of a non-continuous class interval sequence: 0 – 20, 21 – 40, 41 – 60, . . . etc.,

If in the raw data list, there is a value 20.4, we cannot put it in any of the non-continuous class intervals. Because, in the class 0 - 20, no values greater than 20 can be put, and in the class 21 - 40, no values less than 21 can be put. In the continuous class interval sequence, it can be put in the interval 20 – 40. This will mean that there should be no gap between the rectangles in the Histogram. 

Based on the above, another situation can arise. What if there is a value of 20? Two class intervals, 0 – 20, and 20 – 40 can accomodate 20. In such a situation, we adopt the convention that ‘the common value will belong to the higher class’. Thus, 20 will belong to 20 – 40 and not 0 – 20. Similarly, a value of 60 if present, will belong to 60 – 80 and not 40 – 60.


Based on the above discussions, we will see some solved examples in the next section.

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