Frequency Polygon Calculator

Enter your class intervals and their frequencies to build a frequency polygon: class midpoints, plot points, total frequency, and the estimated grouped mean.

Quick Facts

Method
Class midpoint = (lower + upper) / 2; plot midpoint vs. frequency
Grouped mean estimated as Σ(f·x) / Σf using class midpoints x.

Your Results

Calculated
Total frequency (Σf)
-
Sum of all frequencies
Number of classes
-
Points on the polygon
Estimated mean
-
Σ(f·x) / Σf
Modal class
-
Highest-frequency class

Ready

Enter your class intervals and frequencies, then calculate.

What a frequency polygon is

A frequency polygon is a line graph that displays a grouped frequency distribution. Instead of drawing a bar for each class the way a histogram does, you plot a single point per class — its class midpoint on the horizontal axis against its frequency on the vertical axis — and join consecutive points with straight line segments. The result is a jagged line whose shape reveals the distribution: where data clusters, whether it is symmetric or skewed, and whether it has one peak or several.

Because each class is reduced to one point, frequency polygons are ideal for comparing two or more distributions on the same axes — something that overlapping histograms make visually cluttered.

How the calculation works

The calculator turns your grouped table into plot points in three steps:

  • Class midpoint (class mark): for each interval, midpoint x = (lower boundary + upper boundary) / 2. A class of 20–30 has a midpoint of 25. These midpoints are the x-coordinates of the polygon.
  • Frequency: the count f you recorded for that class becomes the y-coordinate.
  • Total frequency: Σf is the sum of all the class counts — the size of your dataset.

The calculator also estimates the mean of the grouped data using the standard formula mean ≈ Σ(f·x) / Σf, where each class is represented by its midpoint. This is an estimate: the individual values inside a class are unknown, so every observation in a class is treated as if it sat exactly at the midpoint.

Worked example

Take the classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 3, 7, 12, 8, 5. The midpoints are 5, 15, 25, 35, 45. Total frequency Σf = 3 + 7 + 12 + 8 + 5 = 35. The weighted sum Σ(f·x) = 3·5 + 7·15 + 12·25 + 8·35 + 5·45 = 15 + 105 + 300 + 280 + 225 = 925, so the estimated mean = 925 / 35 ≈ 26.43. The plot points are (5, 3), (15, 7), (25, 12), (35, 8) and (45, 5), with the modal class 20–30.

Closing the polygon

By convention a frequency polygon is anchored to the horizontal axis at both ends. You add an imaginary class with zero frequency one interval below the first midpoint and one interval above the last, then run the line down to those points. In the example above that means extra points at (−5, 0) and (55, 0). This closes the figure and makes the area under the polygon equal to the area of the corresponding histogram.

Frequency polygon vs. histogram vs. ogive

A histogram uses bars whose heights are the frequencies; a frequency polygon connects the midpoints of those bar tops with lines. An ogive (cumulative frequency polygon) plots cumulative frequency against the upper class boundary instead of raw frequency against the midpoint, and always rises. Use a frequency polygon when you want to see the shape of a distribution or compare several distributions; use an ogive when you want percentiles, medians, or "how many fall below this value" answers.

Frequently Asked Questions

What is a frequency polygon?
It is a line graph of a grouped frequency distribution. For each class you plot the class midpoint (x-axis) against its frequency (y-axis) and connect the points with straight lines. It is typically anchored to the x-axis at an empty class on each end so the line closes.
How do I find the class midpoint?
Average the lower and upper boundaries of the class: midpoint = (lower + upper) / 2. For a 10–20 class the midpoint is 15. These midpoints become the x-coordinates of the polygon.
Do all classes need equal width?
Frequency polygons are almost always drawn from equal-width classes, and the midpoint spacing then matches the class width. If your classes have unequal widths the plotted line can be misleading, because equal horizontal spacing no longer represents equal ranges of data. In that case a histogram using frequency density is usually the clearer choice.
Why is the mean only an estimate?
Grouped data hides the individual values inside each class. The formula Σ(f·x)/Σf assumes every value in a class equals the class midpoint, which is rarely exactly true. If you still have the raw data, computing the mean directly from it will be more accurate than this grouped estimate.