Mathematics
The following table gives the distribution of students of two sections according to the marks obtained by them:
Section A
| Marks | Frequency |
|---|---|
| 0 - 10 | 3 |
| 10 - 20 | 9 |
| 20 - 30 | 17 |
| 30 - 40 | 12 |
| 40 - 50 | 9 |
Section B
| Marks | Frequency |
|---|---|
| 0 - 10 | 5 |
| 10 - 20 | 19 |
| 20 - 30 | 15 |
| 30 - 40 | 10 |
| 40 - 50 | 1 |
Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.
Statistics
4 Likes
Answer
Frequency polygons can be drawn independently without drawing histograms. This requires the midpoints of the class intervals used in the data. The mid-points are called class-marks.
Class Mark =
Substituting the values we get,
Class mark = = 5
Similarly, proceeding in this manner, we get the following table with continuous interval and class mark:
Section A
| Marks | Class mark | Frequency |
|---|---|---|
| 0 - 10 | 5 | 3 |
| 10 - 20 | 15 | 9 |
| 20 - 30 | 25 | 17 |
| 30 - 40 | 35 | 12 |
| 40 - 50 | 45 | 9 |
Section B
| Marks | Class mark | Frequency |
|---|---|---|
| 0 - 10 | 5 | 5 |
| 10 - 20 | 15 | 19 |
| 20 - 30 | 25 | 15 |
| 30 - 40 | 35 | 10 |
| 40 - 50 | 45 | 1 |
Steps of construction of frequency polygon :
Take class marks on the x-axis.
Take frequency on the y-axis with an appropriate scale of 1 unit = 2 students.
We draw frequency polygon for section A by plotting the class-marks on x-axis and the frequencies on y-axis, mark the points A(5, 3), B(15, 9), C(25, 17), D(35, 12) and E(45, 9).
We draw a frequency polygon for section B by plotting the class-marks on x-axis and the frequencies on y-axis, marks the points U(5, 5), V(15, 19), W(25, 15), X(35,10) and Y(45, 1).
To complete the polygon, we assume that there is a class interval with frequency zero before the first interval i.e., -10 - 0, and one after the last interval i.e. (50, 60), with class marks -5 and 55 respectively. Mark points G(-5, 0) and Z(55, 0).
Join the points GABCDEZ and GUVWXYZ with the help of different line segments.
Frequency polygon GABCDEZ is formed for Section A.
Frequency polygon GUVWXYZ is formed for Section B.

From graph,
We observe that,
The performance of students of Section A is better than the performance of students of Section B, as section A shows more students securing marks between class intervals 30 – 40 and 40 – 50.
Answered By
2 Likes
Related Questions
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table :
Length (in mm) Number of leaves 118 - 126 3 127 - 135 5 136 - 144 9 145 - 153 12 154 - 162 5 163 - 171 4 172 - 180 2 (i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
The following table gives the life times of 400 neon lamps:
Life time (in hours) Number of lamps 300 - 400 14 400 - 500 56 500 - 600 60 600 - 700 86 700 - 800 74 800 - 900 62 900 - 1000 48 (i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
Number of ball Team A Team B 1 - 6 2 5 7 - 12 1 6 13 - 18 8 2 19 - 24 9 10 25 - 30 4 5 31 - 36 5 6 37 - 42 6 3 43 - 48 10 4 49 - 54 6 8 55 - 60 2 10 Represent the data of both the teams on the same graph by frequency polygons. [Hint : First make the class intervals continuous.]
A random survey of the number of children of various age groups playing in a park was found as follows:
Age (in years) Number of children 1 - 2 5 2 - 3 3 3 - 5 6 5 - 7 12 7 - 10 9 10 - 15 10 15 - 17 4 Draw a histogram to represent the data above.