Examples (Revised) - Chapter 14 - Statistics - Ncert Solutions class 9 - Maths

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Chapter 12 - Statistics | NCERT Solutions for Class 9 Maths

Example 1 :

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?

Solution :

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.
Example 2 :

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

Draw a bar graph for the data above. 

Solution :

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 ₹ 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 ₹ 5000 , we can choose the scale as 1 unit $=₹ 1000$.
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. 12.2.

Example 3 :

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:

A histogram for this table was prepared by a student as shown in Fig. 12.4. 

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 $\frac{7}{20} \times 10=3.5$.
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".
So, the correct histogram with varying width is given in Fig. 12.5.

(C) Frequency Polygon

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. 12.3. 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. 12.6). 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 $\mathrm{A}$ and $\mathrm{H}$, respectively. ABCDEFGH is the frequency polygon corresponding to the data shown in Fig. 12.3. We have shown this in Fig. 12.6.

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.

Example 4 :

Consider the marks, out of 100 , obtained by 51 students of a class in a test, given in Table 12.5.

Draw a frequency polygon corresponding to this frequency distribution table.
Solution : 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 $\mathrm{A}$. Let $\mathrm{L}$ 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. 12.7.

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,
$
\text { Class-mark }=\frac{\text { Upper limit }+ \text { Lower limit }}{2}
$

Let us consider an example.
Example 5 :

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).
Solution : 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 $140-150,150-160, \ldots$.
For $140-150$, the upper limit $=150$, and the lower limit $=140$
So, the class-mark $=\frac{150+140}{2}=\frac{290}{2}=145$.
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 $\mathrm{B}(145,5), \mathrm{C}(155,10), \mathrm{D}(165,20), \mathrm{E}(175,9), \mathrm{F}(185,6)$ and $\mathrm{G}(195,2)$ 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 $\mathrm{ABCDEFGH}$ (see Fig. 12.8).

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.