STATISTICS-Notes

Draw a histogram for given data

Solution:

1. Represent the weights on the horizontal axis using a suitable scale, such as 1 cm = 5 kg. Since the data begins at 30.5 kg rather than 0, insert a break (kink) on the x-axis immediately after the origin to accurately reflect the starting value.
2. Plot the number of students on the vertical axis using a scale that accommodates the maximum frequency observed (15 students), ensuring all data is clearly represented.
3. For each weight interval (class), draw a rectangle whose width corresponds to the class interval (e.g., 5 kg) and whose height is proportional to the number of students in that class. This visually depicts the frequency distribution of weights among students.

Draw a histogram to represent the following

Solution:

1. Represent the weights on the horizontal axis using a suitable scale, such as 1 cm = 10.
2. Plot the number of students on the vertical axis using a scale that accommodates the maximum frequency observed (15), ensuring all data is clearly represented.
3. For each weight interval (class), draw a rectangle whose width corresponds to the class interval (e.g., 10) and whose height is proportional to the number of students in that class.

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

Solution:

1. In a histogram, the area of each rectangle represents the frequency for that interval. This means that for intervals of different sizes, we must adjust the height of each rectangle so that the area accurately reflects the number of students.
2. To do this, we first identify the smallest class interval. In this case, the smallest interval is 10 marks. We'll use this interval as the reference to calculate the proportional heights for rectangles with larger or smaller intervals.
3. For intervals wider than the reference interval, the rectangle's height must be scaled down so that the area (width × height) correctly shows the frequency. If an interval is twice as wide, its rectangle must be half as high to represent the same frequency density.
4. The formula used to find the adjusted height (frequency density) for each interval is:
   
   Height = \(\left(\frac{f}{\text{class width}}\right) \times \scriptsize\text{min. class width} \)
   
   This ensures that the rectangles are comparable, since each area is proportional to its frequency.
5. Below is the table showing the calculation of rectangle heights for each class interval:

Marks Interval	(\(f\))	Class Width	Height (Rectangle)
0–20	7	20	\( \dfrac{7}{20} \times 10 = 3.5 \)
20–30	10	10	\( \dfrac{10}{10} \times 10 = 10 \)
30–40	10	10	\( \dfrac{10}{10} \times 10 = 10 \)
40–50	20	10	\( \dfrac{20}{10} \times 10 = 20 \)
50–60	20	10	\( \dfrac{20}{10} \times 10 = 20 \)
60–70	15	10	\( \dfrac{15}{10} \times 10 = 15 \)
70–100	8	30	\( \dfrac{8}{30} \times 10 \approx 2.67 \)

6. By using this method, the histogram will accurately display the distribution—even though the intervals are not all the same width. The area of each rectangle truly represents the number of students in each marks range.

Consider the marks, out of 100, obtained by 51 students of a class in a test, given in Table below.Draw a frequency polygon corresponding to this frequency distribution table.

Solution:

• To visualize the frequency distribution, we begin by drawing a histogram for the given data. Each rectangle represents a class interval, and the height of each rectangle matches the frequency of that interval.
• For clarity, we mark the mid-points at the tops of all rectangles as B, C, D, E, F, G, H, I, J, and K, respectively. These mid-points are positioned at the center of each class along the x-axis and at the frequency's value on the y-axis.
• Since the first class is 0–10, we also consider the interval just before it—which is an imaginary class from –10 to 0. By extending the x-axis into the negative direction, we locate the mid-point of this interval (at –5) and assign it a frequency of zero.
• The first segment of the frequency polygon is drawn from this new point (–5, 0) to B (the mid-point at the top of the first rectangle, 0–10). The intersection of this segment with the y-axis is marked as A.
• At the end of the data, after the last class (90–100), we add another imaginary interval (100–110). The mid-point of this interval is L (at 105, 0 frequency), providing closure to the frequency polygon.
• Connecting all marked mid-points in order—from O to A, then through B, C, D, E, F, G, H, I, J, and K, and finally to L—creates the frequency polygon, as illustrated in Figure 12.7.
• This systematic construction helps us see how data frequencies change across intervals, with the frequency polygon clearly tracing the progression and aiding comparison.