`color{red} ♦` Graphical Representation of Data

The representation of data by tables has already been discussed. Now let us turn our attention to another representation of data, i.e., the graphical representation.

It is well said that one picture is better than a thousand words. Usually comparisons among the individual items are best shown by means of graphs. The representation then becomes easier to understand than the actual data.

We shall study the following graphical representations in this section.

(A) Bar graphs

(B) Histograms of uniform width, and of varying widths

(C) Frequency polygons

It is well said that one picture is better than a thousand words. Usually comparisons among the individual items are best shown by means of graphs. The representation then becomes easier to understand than the actual data.

We shall study the following graphical representations in this section.

(A) Bar graphs

(B) Histograms of uniform width, and of varying widths

(C) Frequency polygons

Previously, you've already studied and constructed bar graphs. Here we shall discuss them through a more formal approach.

Recall that a bar graph is a pictorial representation of data in which usually bars of uniform width are drawn with equal spacing between them on one axis (say, the x-axis), depicting the variable.

The values of the variable are shown on the other axis (say, the y-axis) and the heights of the bars depend on the values of the variable.

Recall that a bar graph is a pictorial representation of data in which usually bars of uniform width are drawn with equal spacing between them on one axis (say, the x-axis), depicting the variable.

The values of the variable are shown on the other axis (say, the y-axis) and the heights of the bars depend on the values of the variable.

Q 3220778611

In a particular section of Class IX, `40` students were asked about the months of their birth and the following graph was prepared for the data so obtained:

Observe the bar graph given above and answer the following questions:

(i) How many students were born in the month of November?

(ii) In which month were the maximum number of students born?

Class 9 Chapter 14 Example 5

Observe the bar graph given above and answer the following questions:

(i) How many students were born in the month of November?

(ii) In which month were the maximum number of students born?

Class 9 Chapter 14 Example 5

Note that the variable here is the ‘month of birth’, and the value of the variable is the ‘Number of students born’.

(i) `4` students were born in the month of November.

(ii) The Maximum number of students were born in the month of August. Let us now recall how a bar graph is constructed by considering the following example.

Q 3250778614

A family with a monthly income of Rs `20,000` had planned the following expenditures per month under various heads:

Draw a bar graph for the data above.

Class 9 Chapter 14 Example 6

Draw a bar graph for the data above.

Class 9 Chapter 14 Example 6

We draw the bar graph of this data in the following steps. Note that the unit

in the second column is thousand rupees. So, `‘4’` against ‘grocery’ means Rs `4000`.

1. We represent the Heads (variable) on the horizontal axis choosing any scale,

since the width of the bar is not important. But for clarity, we take equal widths

for all bars and maintain equal gaps in between. Let one Head be represented by

one unit.

2. We represent the expenditure (value) on the vertical axis. Since the maximum

expenditure is Rs 5000, we can choose the scale as `1` unit `= Rs 1000`.

3. To represent our first Head, i.e., grocery, we draw a rectangular bar with width

1 unit and height 4 units.

4. Similarly, other Heads are represented leaving a gap of 1 unit in between two

consecutive bars.

The bar graph is drawn in Fig. 14.2.

Here, you can easily visualise the relative characteristics of the data at a glance, e.g.,

the expenditure on education is more than double that of medical expenses. Therefore,

in some ways it serves as a better representation of data than the tabular form.

Continuing with the same four groups of Activity 1, represent the data by suitable bar graphs.

Let us now see how a frequency distribution table for continuous class intervals can be represented graphically.

Let us now see how a frequency distribution table for continuous class intervals can be represented graphically.

This is a form of representation like the bar graph, but it is used for continuous class intervals.

For instance, consider the frequency distribution Table 14.6, representing the weights of 36 students of a class:

Let us represent the data given above graphically as follows:

(i) We represent the weights on the horizontal axis on a suitable scale. We can choose the scale as `1 cm = 5 kg`. Also, since the first class interval is starting from `30.5` and not zero, we show it on the graph by marking a kink or a break on the axis.

(ii) We represent the number of students (frequency) on the vertical axis on a suitable scale. Since the maximum frequency is `15`, we need to choose the scale to accomodate this maximum frequency.

(iii) We now draw rectangles (or rectangular bars) of width equal to the class-size and lengths according to the frequencies of the corresponding class intervals.

For example, the rectangle for the class interval `30.5 - 35.5` will be of width `1` cm and length `4.5` cm.

(iv) In this way, we obtain the graph as shown in Fig. 14.3:

Observe that since there are no gaps in between consecutive rectangles, the resultant graph appears like a solid figure. This is called a histogram, which is a graphical representation of a grouped frequency distribution with continuous classes.

Also, unlike a bar graph, the width of the bar plays a significant role in its construction.

Here, in fact, areas of the rectangles erected are proportional to the corresponding frequencies.

However, since the widths of the rectangles are all equal, the lengths of the rectangles are proportional to the frequencies. That is why, we draw the lengths according to (iii) above.

Now, consider a situation different from the one above.

For instance, consider the frequency distribution Table 14.6, representing the weights of 36 students of a class:

Let us represent the data given above graphically as follows:

(i) We represent the weights on the horizontal axis on a suitable scale. We can choose the scale as `1 cm = 5 kg`. Also, since the first class interval is starting from `30.5` and not zero, we show it on the graph by marking a kink or a break on the axis.

(ii) We represent the number of students (frequency) on the vertical axis on a suitable scale. Since the maximum frequency is `15`, we need to choose the scale to accomodate this maximum frequency.

(iii) We now draw rectangles (or rectangular bars) of width equal to the class-size and lengths according to the frequencies of the corresponding class intervals.

For example, the rectangle for the class interval `30.5 - 35.5` will be of width `1` cm and length `4.5` cm.

(iv) In this way, we obtain the graph as shown in Fig. 14.3:

Observe that since there are no gaps in between consecutive rectangles, the resultant graph appears like a solid figure. This is called a histogram, which is a graphical representation of a grouped frequency distribution with continuous classes.

Also, unlike a bar graph, the width of the bar plays a significant role in its construction.

Here, in fact, areas of the rectangles erected are proportional to the corresponding frequencies.

However, since the widths of the rectangles are all equal, the lengths of the rectangles are proportional to the frequencies. That is why, we draw the lengths according to (iii) above.

Now, consider a situation different from the one above.

Q 3220878711

A teacher wanted to analyse the performance of two sections of students

in a mathematics test of `100` marks. Looking at their performances, she found that a

few students got under `20` marks and a few got `70` marks or above. So she decided to

group them into intervals of varying sizes as follows: `0 - 20, 20 - 30, . . ., 60 - 70,`

`70 - 100`. Then she formed the following table:

Class Chapter 14 Example 7

in a mathematics test of `100` marks. Looking at their performances, she found that a

few students got under `20` marks and a few got `70` marks or above. So she decided to

group them into intervals of varying sizes as follows: `0 - 20, 20 - 30, . . ., 60 - 70,`

`70 - 100`. Then she formed the following table:

Class Chapter 14 Example 7

Carefully examine this graphical representation. Do you think that it correctly represents

the data? No, the graph is giving us a misleading picture. As we have mentioned

earlier, the areas of the rectangles are proportional to the frequencies in a histogram.

Earlier this problem did not arise, because the widths of all the rectangles were equal.

But here, since the widths of the rectangles are varying, the histogram above does not

give a correct picture. For example, it shows a greater frequency in the interval

`70 - 100`, than in `60 - 70`, which is not the case.

So, we need to make certain modifications in the lengths of the rectangles so that

the areas are again proportional to the frequencies.

The steps to be followed are as given below:

1. Select a class interval with the minimum class size. In the example above, the

minimum class-size is `10`.

2. The lengths of the rectangles are then modified to be proportionate to the

class-size `10`.

For instance, when the class-size is `20`, the length of the rectangle is `7`. So when

the class-size is `10`, the length of the rectangle will be `7/(20) x× 10 = 3.5.`

Similarly, proceeding in this manner, we get the following table:

Since we have calculated these lengths for an interval of `10` marks in each case,

we may call these lengths as “proportion of students per `10` marks interval”.

There is yet another visual way of representing quantitative data and its frequencies. This is a polygon. To see what we mean, consider the histogram represented by Fig. 14.3.

Let us join the mid-points of the upper sides of the adjacent rectangles of this histogram by means of line segments. Let us call these mid-points B, C, D, E, F and G. When joined by line segments, we obtain the figure BCDEFG (see Fig. 14.6).

To complete the polygon, we assume that there is a class interval with frequency zero before 30.5 - 35.5, and one after 55.5 - 60.5, and their mid-points are A and H, respectively.

ABCDEFGH is the frequency polygon corresponding to the data shown in Fig. 14.3. We have shown this in Fig. 14.6.

Although, there exists no class preceding the lowest class and no class succeeding the highest class, addition of the two class intervals with zero frequency enables us to make the area of the frequency polygon the same as the area of the histogram.

Why is this so? (Hint : Use the properties of congruent triangles.)

Now, the question arises: how do we complete the polygon when there is no class preceding the first class? Let us consider such a situation.

Let us join the mid-points of the upper sides of the adjacent rectangles of this histogram by means of line segments. Let us call these mid-points B, C, D, E, F and G. When joined by line segments, we obtain the figure BCDEFG (see Fig. 14.6).

To complete the polygon, we assume that there is a class interval with frequency zero before 30.5 - 35.5, and one after 55.5 - 60.5, and their mid-points are A and H, respectively.

ABCDEFGH is the frequency polygon corresponding to the data shown in Fig. 14.3. We have shown this in Fig. 14.6.

Although, there exists no class preceding the lowest class and no class succeeding the highest class, addition of the two class intervals with zero frequency enables us to make the area of the frequency polygon the same as the area of the histogram.

Why is this so? (Hint : Use the properties of congruent triangles.)

Now, the question arises: how do we complete the polygon when there is no class preceding the first class? Let us consider such a situation.

Q 3270878716

Consider the marks, out of `100`, obtained by `51` students of a class in a test, given in Table `14.9`.

Draw a frequency polygon corresponding to this frequency distribution table.

Class 9 Chapter 14 Example 8

Draw a frequency polygon corresponding to this frequency distribution table.

Class 9 Chapter 14 Example 8

Let us first draw a histogram for this data and mark the mid-points of the

tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is

`0-10`. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative

direction and find the mid-point of the imaginary class-interval `(–10) - 0`. The first end

point, i.e., B is joined to this mid-point with zero frequency on the negative direction of

the horizontal axis. The point where this line segment meets the vertical axis is marked

as A. Let L be the mid-point of the class succeeding the last class of the given data.

Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig. 14.7.

Frequency polygons can also be drawn independently without drawing

histograms. For this, we require the mid-points of the class-intervals used in the data.

These mid-points of the class-intervals are called class-marks.

To find the class-mark of a class interval, we find the sum of the upper limit and

lower limit of a class and divide it by 2. Thus,

Class-mark ` = text( Upper limit + Lower limit)/2`

Let us consider an example.

Q 3200878718

In a city, the weekly observations made in a study on the cost of living index are given in the following table:

Draw a frequency polygon for the data above (without constructing a histogram).

Class 9 Chapter 14 Example 9

Draw a frequency polygon for the data above (without constructing a histogram).

Class 9 Chapter 14 Example 9

Since we want to draw a frequency polygon without a histogram, let us find

the class-marks of the classes given above, that is of `140 - 150, 150 - 160,....`

For `140 - 150`, the upper limit `= 150,` and the lower limit `= 140`

So, the class-mark ` = ( 150 + 140)/2 = (290)/2 = 145`

Continuing in the same manner, we find the class-marks of the other classes as well.

So, the new table obtained is as shown in the following table:

We can now draw a frequency polygon by plotting the class-marks along the horizontal

axis, the frequencies along the vertical-axis, and then plotting and joining the points

`B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6)` and `G(195, 2)` by line segments.

We should not forget to plot the point corresponding to the class-mark of the class

`130 - 140` (just before the lowest class `140 - 150`) with zero frequency, that is,

`A(135, 0)`, and the point `H (205, 0)` occurs immediately after `G(195, 2)`. So, the resultant

frequency polygon will be `ABCDEFGH` (see Fig. 14.8).

Frequency polygons are used when the data is continuous and very large. It is

very useful for comparing two different sets of data of the same nature, for example,

comparing the performance of two different sections of the same class.