Ninth Grade Mathematics: Frequency Polygons: Construction & Analysis
While histograms represent grouped distributions using columnar rectangular areas, a frequency polygon represents the distribution as a continuous, dynamic line graph. Frequency polygons are especially powerful because they allow multiple datasets to be superimposed onto the exact same set of axes, providing an immediate visual comparison of two populations.
Definition and Concept of a Frequency Polygon
A frequency polygon is a closed geometric polygon formed by joining the midpoints of the tops of the rectangles in a histogram (or by plotting class marks directly against frequencies) using straight line segments, and anchoring the ends to the horizontal axis.
Histograms and the Frequency Polygon:
Frequency
^ * (25, 14)
15 | +
| * | * (35, 10)
10 | * | | |
| * | | | | * (45, 4)
5 | * | | | | | |
+----*--+--+--+---+---+--+--*---->
(5) 10 20 30 40 50 60 (70)
Frequency Polygon anchors to axis at midpoints (5) and (70)!
The points plotted have coordinates (Class Mark, Frequency). Connecting these points creates an undulating polygonal curve that traces the contour of the frequency distribution.
Method 1: Construction with a Histogram
When a histogram has already been constructed:
Steps with a Histogram:
1. Draw the standard histogram for the given continuous frequency table.
2. Mark the midpoint of the horizontal top of each rectangular bar.
3. Connect these midpoints sequentially using straight line segments.
4. To complete the figure into a closed polygon, extend the line segments
at both ends to meet the horizontal axis:
- Connect the first midpoint to the class mark of an imaginary preceding
interval (with frequency 0).
- Connect the last midpoint to the class mark of an imaginary succeeding
interval (with frequency 0).
The Invariance of Area Principle
An extraordinary mathematical property of this construction is that the area enclosed by the frequency polygon is strictly equal to the total area of the histogram!
Area of Frequency Polygon == Total Area of Histogram
When you connect the midpoints, the line segments cut off small triangular corners from the rectangles. However, each cut-off triangle is exactly congruent to the small triangular space added outside the rectangle below the line. By the Congruent Area Axiom, the area lost equals the area gained, keeping the total area invariant.
Method 2: Direct Construction without a Histogram
A frequency polygon can be constructed directly from a frequency table without drawing any rectangular bars by calculating the class mark of each interval:
Class Mark Formula:
Lower Limit + Upper Limit
Class Mark = ---------------------------
2
Class Interval Class Mark (x) Frequency (f) Plotted Coordinate (x, f)
--------------------------------------------------------------------------------------
[ Preceding ] 5 0 (5, 0) <--- Anchor
10 - 20 15 4 (15, 4)
20 - 30 25 12 (25, 12)
30 - 40 35 16 (35, 16)
40 - 50 45 8 (45, 8)
50 - 60 55 2 (55, 2)
[ Succeeding ] 65 0 (65, 0) <--- Anchor
Plot the points (x, f) on a coordinate grid, connect them with a ruler, and anchor the ends to (5, 0) and (65, 0).
Comparing Two Distributions
The greatest analytical advantage of frequency polygons is comparative analysis:
Comparing Test Scores: Section A (Solid line) vs Section B (Dashed line)
Frequency
^ Section A Section B
20 | * *
| / \ / \
15 | / \ / \
| * / \ * / \
10 | / \ / \ / \ / \
| / \ / \ / \ / \
5 | / * \ / * \
+---*-------------------*---*-------------------*--->
0 50 0 50 (Marks)
Plotting two histograms on the same axis results in a confusing jumble of overlapping bars. But two frequency polygons can be superimposed cleanly, instantly revealing differences in central clustering, spread, and skewness between the two sections.
Chapter Exercises
Exercise 1: Calculate the class marks for the following frequency distribution: 140 - 150, 150 - 160, 160 - 170, 170 - 180, and 180 - 190. What are the coordinates of the two anchoring points on the horizontal axis required to close the polygon?
Exercise 2: Explain why the area of a frequency polygon constructed from a histogram is mathematically equal to the total area of the histogram rectangles. Draw an ASCII diagram showing the congruent triangles involved.
Exercise 3: Two sections of ninth-grade students have their scores summarized in frequency tables with identical class intervals. Describe how a frequency polygon enables an analyst to compare the overall academic performance of Section A versus Section B.
Solutions and Explanations
Solution 1: Compute class marks using (Lower Limit + Upper Limit) / 2: for 140 - 150, (140 + 150)/2 = 145; for 150 - 160, (150 + 160)/2 = 155; for 160 - 170, (160 + 170)/2 = 165; for 170 - 180, (170 + 180)/2 = 175; for 180 - 190, (180 + 190)/2 = 185. The uniform class size is 10. The imaginary preceding interval is 130 - 140, having class mark 135 and frequency 0, which gives the lower anchor coordinate (135, 0). The imaginary succeeding interval is 190 - 200, having class mark 195 and frequency 0, which gives the upper anchor coordinate (195, 0).
Solution 2: When the top midpoint of one rectangle is connected to the top midpoint of the next rectangle, the straight line segment slices through the boundary between the two rectangles:
+-------*-------+
| /| |
| / | |
| A / | B |
| / | |
+--*----+-------+
The line segment cuts off a right triangle A from inside the taller rectangle, while enclosing an empty right triangle B above the lower rectangle. Because the line segment passes through the midpoint between the two rectangles, triangle A and triangle B are congruent by the ASA Congruence Criterion (sharing equal base segments, vertical opposite angles, and right angles). By the Congruent Area Axiom, Area(Triangle A) = Area(Triangle B). The area lost inside the rectangle is exactly balanced by the area gained outside it. Summing these cancellations across all class intervals proves that the total area enclosed under the frequency polygon is strictly equal to the total area of the histogram.
Solution 3: A frequency polygon enables direct comparative analysis because it represents each dataset as a single continuous profile line rather than a solid columnar mass. By plotting Section A as a solid line and Section B as a dashed line on the same coordinate axes, an analyst can immediately compare their performance: if Section B's polygon peak is shifted significantly to the right of Section A's peak, Section B scored systematically higher on average; if Section A's polygon has a wide, flattened profile while Section B's is narrow and tall, Section A exhibits much greater variability in student ability. These visual patterns can be detected simultaneously without any confusing visual obstruction.