Frequency Polygon
A frequency polygon is a line graph that represents a frequency distribution. It is drawn by plotting points at the class marks (mid-points) of each class interval against their frequencies, and then joining these points with straight lines.
A frequency polygon can be drawn directly from a frequency table, or it can be drawn over a histogram by connecting the mid-points of the top of each bar.
Frequency polygons are useful for comparing two or more distributions on the same graph. This topic is part of NCERT Class 9 Mathematics, Chapter: Statistics.
What is Frequency Polygon?
Definition: A frequency polygon is a closed line graph obtained by plotting class marks (mid-values) on the x-axis and corresponding frequencies on the y-axis, then joining the plotted points with straight line segments.
Class mark formula:
Class mark = (Lower limit + Upper limit) / 2
Important:
- Each plotted point is at (class mark, frequency).
- The polygon is closed by extending it to the mid-points of two imaginary classes (one before the first class and one after the last class) with frequency 0.
- This ensures the polygon touches the x-axis at both ends, enclosing the same area as the histogram.
- Frequency polygons are ideal for comparing distributions because multiple polygons can be drawn on the same axes.
Frequency Polygon Formula
Key Formulas:
1. Class mark (mid-point):
Class mark = (Lower limit + Upper limit) / 2
2. Steps for plotting:
- Find the class mark for each class interval.
- Plot points: (class mark, frequency).
- Add two extra points at frequency 0: one before the first class, one after the last class.
- Join all points with straight lines.
3. Imaginary classes:
- Before first class: class mark = first class mark − class width
- After last class: class mark = last class mark + class width
- Both have frequency = 0
4. From histogram:
- Mark the mid-point of the top of each bar.
- Connect mid-points with straight lines.
- Close the polygon by extending to the x-axis at both ends.
Derivation and Proof
Steps to Draw a Frequency Polygon:
Method 1: Directly from frequency table
- Calculate the class mark for each class: (lower + upper) / 2
- Add an imaginary class before the first class with frequency 0.
- Add an imaginary class after the last class with frequency 0.
- Plot the points: (class mark, frequency) for all classes including the imaginary ones.
- Join the points with straight line segments.
- The resulting closed figure is the frequency polygon.
Method 2: From a histogram
- Draw the histogram for the given data.
- Mark the mid-point of the top edge of each bar.
- Connect these mid-points with straight lines.
- To close the polygon, connect the first and last mid-points to the x-axis at the class marks of the imaginary classes.
Why close the polygon?
- The area under the frequency polygon equals the area under the histogram.
- The two triangles added at the ends (going down to the x-axis) compensate for the triangles cut off from the histogram bars.
Types and Properties
Types of frequency polygon problems:
1. Drawing from a frequency table
- Given a grouped frequency distribution, calculate class marks and plot.
- Most common type in Class 9.
2. Drawing from a histogram
- Overlay the polygon on an existing histogram.
- Connect mid-points of bar tops.
3. Comparing distributions
- Draw two or more frequency polygons on the same axes.
- Useful for comparing boys vs girls, two classes, two years, etc.
4. Reading a frequency polygon
- Given a frequency polygon, determine the frequency table.
- Read off the coordinates of each point.
5. Frequency polygon with unequal class widths
- Use frequency density on the y-axis instead of frequency.
- Plot (class mark, frequency density).
Solved Examples
Example 1: Example 1: Drawing a frequency polygon from a table
Problem: Draw a frequency polygon for the following data.
| Marks | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Students | 4 | 8 | 14 | 10 | 4 |
Solution:
Step 1: Find class marks
| Class | Class Mark | Frequency |
|---|---|---|
| (-10)–0 (imaginary) | −5 | 0 |
| 0–10 | 5 | 4 |
| 10–20 | 15 | 8 |
| 20–30 | 25 | 14 |
| 30–40 | 35 | 10 |
| 40–50 | 45 | 4 |
| 50–60 (imaginary) | 55 | 0 |
Step 2: Plot points (−5, 0), (5, 4), (15, 8), (25, 14), (35, 10), (45, 4), (55, 0).
Step 3: Join the points with straight lines to form the polygon.
Example 2: Example 2: Frequency polygon from histogram
Problem: A histogram has bars over 10–20, 20–30, 30–40, 40–50, 50–60 with heights 6, 10, 15, 8, 3. Draw the frequency polygon.
Solution:
Step 1: Find mid-points of each bar
- 10–20: mid = 15, height = 6
- 20–30: mid = 25, height = 10
- 30–40: mid = 35, height = 15
- 40–50: mid = 45, height = 8
- 50–60: mid = 55, height = 3
Step 2: Add imaginary points
- (5, 0) and (65, 0)
Step 3: Plot all 7 points and connect with straight lines.
The polygon closes by touching the x-axis at (5, 0) and (65, 0).
Example 3: Example 3: Finding class marks
Problem: Find the class marks for the classes 100–120, 120–140, 140–160, 160–180, 180–200.
Solution:
Using class mark = (lower + upper) / 2:
- 100–120: (100 + 120)/2 = 110
- 120–140: (120 + 140)/2 = 130
- 140–160: (140 + 160)/2 = 150
- 160–180: (160 + 180)/2 = 170
- 180–200: (180 + 200)/2 = 190
Imaginary class marks: 90 (before), 210 (after) — both with frequency 0.
Example 4: Example 4: Reading a frequency polygon
Problem: A frequency polygon has vertices at (5, 0), (15, 6), (25, 12), (35, 18), (45, 10), (55, 4), (65, 0). Construct the frequency distribution table.
Solution:
Class width = 10 (difference between successive class marks)
| Class Interval | Class Mark | Frequency |
|---|---|---|
| 10–20 | 15 | 6 |
| 20–30 | 25 | 12 |
| 30–40 | 35 | 18 |
| 40–50 | 45 | 10 |
| 50–60 | 55 | 4 |
Total frequency: 6 + 12 + 18 + 10 + 4 = 50.
Example 5: Example 5: Comparing two distributions
Problem: The marks of Section A and Section B in an exam are:
| Marks | 0–20 | 20–40 | 40–60 | 60–80 | 80–100 |
|---|---|---|---|---|---|
| Section A | 4 | 10 | 16 | 12 | 8 |
| Section B | 6 | 14 | 18 | 8 | 4 |
Draw frequency polygons for both on the same axes.
Solution:
Class marks: 10, 30, 50, 70, 90
Imaginary points: (−10, 0) and (110, 0) for both.
Section A points: (−10, 0), (10, 4), (30, 10), (50, 16), (70, 12), (90, 8), (110, 0)
Section B points: (−10, 0), (10, 6), (30, 14), (50, 18), (70, 8), (90, 4), (110, 0)
Observation: Section B has more students scoring in the 20–60 range, while Section A has more in the 60–100 range.
Example 6: Example 6: Total students from frequency polygon
Problem: A frequency polygon has the following coordinates: (5, 0), (15, 5), (25, 10), (35, 20), (45, 15), (55, 8), (65, 0). Find the total number of data points.
Solution:
- Frequencies (excluding the zero-frequency endpoints): 5, 10, 20, 15, 8
- Total = 5 + 10 + 20 + 15 + 8 = 58
Answer: Total data points = 58.
Example 7: Example 7: Age distribution
Problem: Ages of patients in a hospital: 0–10 (15), 10–20 (25), 20–30 (18), 30–40 (12), 40–50 (10). Draw the frequency polygon. Which age group has the most patients?
Solution:
Class marks and points:
- (−5, 0), (5, 15), (15, 25), (25, 18), (35, 12), (45, 10), (55, 0)
Peak: The highest point is (15, 25), corresponding to class 10–20.
Answer: The age group 10–20 years has the most patients (25).
Example 8: Example 8: Area under polygon equals histogram area
Problem: Verify that the area under a frequency polygon equals the area of the corresponding histogram for the data: 0–10 (4), 10–20 (8), 20–30 (6).
Solution:
Histogram area:
- = (10 × 4) + (10 × 8) + (10 × 6) = 40 + 80 + 60 = 180 sq units
Polygon area:
- The polygon extends from (−5, 0) to (5, 4) to (15, 8) to (25, 6) to (35, 0).
- Using the shoelace formula or geometric decomposition, the area = 180 sq units.
Verified: Both areas are equal.
Real-World Applications
Applications of Frequency Polygons:
- Comparing distributions: Multiple frequency polygons on the same axes allow quick visual comparison of data sets (e.g., performance of two classes, rainfall across years).
- Trend identification: The shape of the polygon shows whether data is symmetric, skewed, or bimodal.
- Demographics: Population pyramids and age-group distributions are often represented as frequency polygons.
- Business: Sales data, customer age profiles, and product price distributions are visualised using frequency polygons.
- Weather analysis: Comparing temperature or rainfall distributions across seasons or regions.
- Education: Comparing exam results across sections, schools, or academic years.
Key Points to Remember
- A frequency polygon is drawn by plotting (class mark, frequency) and joining with straight lines.
- Class mark = (Lower limit + Upper limit) / 2.
- The polygon is closed by extending to the x-axis using imaginary classes at both ends (frequency = 0).
- From a histogram: connect the mid-points of the top edges of the bars.
- Area under the frequency polygon = area under the histogram.
- Frequency polygons are ideal for comparing two or more distributions.
- The peak of the polygon corresponds to the modal class.
- The polygon touches the x-axis at (first class mark − class width) and (last class mark + class width).
- Unlike histograms, frequency polygons are not filled — they are open line graphs.
- Multiple frequency polygons on the same axes should use different colours or line styles for clarity.
Practice Problems
- Draw a frequency polygon for: 0–5 (3), 5–10 (7), 10–15 (12), 15–20 (8), 20–25 (4).
- Find the class marks for classes 50–60, 60–70, 70–80, 80–90, 90–100.
- A frequency polygon has vertices at (10, 0), (20, 8), (30, 15), (40, 10), (50, 5), (60, 0). Find the total frequency and the modal class.
- Draw frequency polygons for two sets of data on the same graph: Set A: 0–10 (5), 10–20 (12), 20–30 (8); Set B: 0–10 (3), 10–20 (10), 20–30 (14).
- Convert the following histogram into a frequency polygon: bars over 20–30, 30–40, 40–50, 50–60 with heights 6, 14, 10, 4.
- The class marks of a frequency polygon are 15, 25, 35, 45, 55. What are the class intervals?
Frequently Asked Questions
Q1. What is a frequency polygon?
A frequency polygon is a line graph drawn by plotting class marks (mid-points of class intervals) against frequencies, and joining the plotted points with straight lines. It is closed by extending to the x-axis at both ends.
Q2. How is a frequency polygon different from a histogram?
A histogram uses rectangular bars with area representing frequency. A frequency polygon uses points connected by lines. The polygon can be drawn on top of a histogram by connecting mid-points of bar tops. Both enclose the same area.
Q3. Why do we add imaginary classes at the ends?
To close the polygon. The imaginary classes (one before the first, one after the last) have frequency 0. This ensures the polygon touches the x-axis and encloses an area equal to the histogram.
Q4. What is a class mark?
Class mark = (Lower limit + Upper limit) / 2. It is the mid-point of a class interval. For example, the class mark of 20–30 is (20 + 30)/2 = 25.
Q5. Can we compare two distributions using a frequency polygon?
Yes. This is one of the main advantages of frequency polygons. Two or more polygons can be plotted on the same axes for easy visual comparison. This is not practical with histograms, which would overlap.
Q6. How do you draw a frequency polygon from a histogram?
Mark the mid-point of the top edge of each bar. Connect these mid-points with straight lines. Extend the line to the x-axis at both ends (at the class marks of imaginary classes with frequency 0).
Q7. Is frequency polygon in the CBSE Class 9 syllabus?
Yes. Frequency polygons are part of CBSE Class 9 Mathematics, Chapter: Statistics, under graphical representation of data.
Q8. What does the shape of a frequency polygon tell us?
The shape reveals the distribution pattern: symmetric (bell-shaped), skewed left (tail on left), skewed right (tail on right), or bimodal (two peaks). The peak corresponds to the modal class.
Related Topics
- Histogram of Grouped Data
- Frequency Distribution Table
- Statistics - Collection and Presentation
- Ogive (Cumulative Frequency Curve)
- Mean of Ungrouped Data
- Median of Ungrouped Data
- Mean of Grouped Data
- Median of Grouped Data
- Mode of Grouped Data
- Cumulative Frequency Distribution
- Empirical Relationship Between Mean, Median, Mode
- Mean by Assumed Mean Method
- Mean by Step-Deviation Method
- Statistics in Real Life
