Frequency polygon

Give a student a table of numbers — heights of forty children, marks of a class, minutes taken to finish a task — and it says very little at a glance. Draw the same data as a frequency polygon and the shape speaks instantly: where the values cluster, where they thin out, whether the distribution has one clear peak or two. A frequency polygon is a line graph built from a grouped frequency table, obtained by plotting the frequency of each class against its class mark and joining the points with straight lines, closed at both ends onto the horizontal axis. This article is the final piece of the Class 9 ICSE Statistics chapter: constructing a frequency polygon directly from class marks and frequencies, on graph paper, without needing a histogram first. It stays inside that scope — the histogram and the ogive belong to Class X and are not covered here.

The core insight

A frequency polygon plots each class's frequency against its class mark, $\text{class mark} = \frac{l_1+l_2}{2}$ , joins the points with straight lines in order, and adds one zero-frequency class at each end so the figure closes onto the horizontal axis.

The core idea

A frequency polygon turns a grouped frequency table into a single connected shape by fixing one point per class and joining those points in sequence.

The class mark. Every class interval has a lower limit $l_1$ and an upper limit $l_2$ . Its class mark (or mid-value) is the point exactly halfway between them, found by averaging the two limits:

\text{Class mark} = \frac{l_1+l_2}{2}

where: $l_1$ is the lower class limit, $l_2$ is the upper class limit, and the class mark is the single representative x-value used for the whole class when plotting.

The plotting rule. For each class, plot the point (class mark, frequency) on graph paper — class marks along the horizontal axis, frequencies along the vertical axis — choosing a scale that spreads the data without wasting space. Two extra classes of the same width, one just below the lowest class and one just above the highest, are added with frequency zero, purely so the line has somewhere to start and end on the axis. Every plotted point, including the two zero-frequency ones, is then joined to its neighbour with a straight line segment, in order of increasing class mark, giving a closed many-sided figure — the polygon.

Continuous classes only. The class-mark rule only makes sense once the table is in continuous form (no gaps between classes). A discontinuous table must first be converted to continuous class intervals before any class mark is calculated.

Figure

A frequency polygon for the heights table rising from zero at 125 to a peak of 16 at class mark 155 then falling back to zero at 185

Derivation AWhere the class-mark formula comes from, and why n classes need n plus 2 points

The goal is to show algebraically why the midpoint formula for a class mark is forced by the single requirement that it sit exactly halfway between the two class limits, and then to see why a table of $n$ continuous classes always needs exactly $n+2$ plotted points to close into a proper frequency polygon touching the horizontal axis at both ends.

Step 1. Call the class mark $m$ . "Halfway between $l_1$ and $l_2$ " means $m$ is the same distance from $l_1$ as it is from $l_2$ : the gap $m-l_1$ equals the gap $l_2-m$ . Writing that as an equation gives $m-l_1=l_2-m$ .

Step 2. Solve for $m$ . Add $m$ to both sides and add $l_1$ to both sides: $2m=l_1+l_2$ , so $m=\dfrac{l_1+l_2}{2}$ . This confirms the class mark is simply the average of the two limits — no other point on the interval is equidistant from both ends.

Step 3. Count the points needed for the whole polygon. A table with $n$ continuous classes gives exactly $n$ class marks, hence $n$ points of the form (class mark, frequency). Each of these sits strictly above the axis (frequency $\ge 0$ ), so joining only these $n$ points would leave the line "floating," never touching the horizontal axis at either end.

Step 4. Add the two closing points. One extra class of the same width, with frequency $0$ , is placed immediately before the first class and one immediately after the last; each contributes one more point, both sitting exactly on the axis. The total point count is therefore \text{points to plot} = n+2 where: $n$ is the number of classes in the original continuous table. For the heights table below, $n=5$ classes give $5+2=7$ points, matching the list plotted in the worked example. Verified: with $l_1=130,\ l_2=140$ , the equation $m-130=140-m$ gives $2m=270,\ m=135$ , matching $\frac{130+140}{2}=135$ ; and $n=5$ classes give $n+2=7$ points, matching the seven points listed in the worked example below.

Worked example

The following continuous frequency table shows the number of students in various height ranges (in cm):

Height (cm)	130–140	140–150	150–160	160–170	170–180
Frequency	4	10	16	8	2

Step 1. Find the class marks using $\text{class mark}=\frac{l_1+l_2}{2}$ : $\frac{130+140}{2}=135$ , $\frac{140+150}{2}=145$ , $\frac{150+160}{2}=155$ , $\frac{160+170}{2}=165$ , $\frac{170+180}{2}=175$ .

Step 2. Add zero-frequency classes at each end, same width as the rest. Before $130$ – $140$ , add $120$ – $130$ (class mark $125$ , frequency $0$ ). After $170$ – $180$ , add $180$ – $190$ (class mark $185$ , frequency $0$ ).

Step 3. List the seven points to plot, in order of increasing class mark: (125,0),\ (135,4),\ (145,10),\ (155,16),\ (165,8),\ (175,2),\ (185,0) This matches the $n+2=5+2=7$ count from the derivation above.

Step 4. Plot these seven points on graph paper — class mark on the horizontal axis, frequency on the vertical axis — and join them in order with straight line segments. The polygon rises from $0$ at $125$ , peaks at $16$ over class mark $155$ (the class $150$ – $160$ , the highest-frequency class), then falls back to $0$ at $185$ . Answer: the polygon has its peak at class mark 155 cm, showing that most students' heights lie in the 150–160 cm range.

Competitive stretch

Check the point count rule on a second table: time (in minutes) taken by 25 workers, classes $0$ – $5,\ 5$ – $10,\ 10$ – $15,\ 15$ – $20,\ 20$ – $25$ with frequencies $2,6,9,5,3$ . Here $n=5$ , so $n+2=7$ points are needed: $(-2.5,0),\ (2.5,2),\ (7.5,6),\ (12.5,9),\ (17.5,5),\ (22.5,3),\ (27.5,0)$ — again seven points, peaking at class mark $12.5$ (the $10$ – $15$ class, frequency $9$ ). Even when the added lower class runs into negative territory, as here, it is kept purely to close the polygon graphically and is not meant to represent a real negative time.

Common traps

Plotting frequency against the class limits instead of the class mark — the horizontal axis must always use the midpoint of each class, never its lower or upper limit.
Forgetting the two extra zero-frequency classes at either end, which leaves the polygon "open" instead of a closed figure touching the axis.
Trying to compute class marks from a discontinuous table — convert to continuous classes first, or the midpoints and the resulting shape are meaningless.
Joining the plotted points with a smooth curve instead of straight line segments — by definition, a frequency polygon is made of straight lines.

Ask the tutor: want a check on your class-mark arithmetic or your choice of graph scale before you plot — ask the AI tutor to look it over.

Why use the class mark and not the class limits when plotting?

The class mark represents the whole class with a single number sitting exactly at its centre, so the plotted point is a fair stand-in for every value inside that class. Using a limit instead would push the point to one edge of the class, distorting the shape of the polygon and making it inconsistent with how a histogram bar is centred over the same interval.

What happens if I forget to add the two zero-frequency classes?

Without them, your first and last plotted points sit above the axis with nothing to join them down to zero, so the line never touches the horizontal axis and the figure stays open rather than forming a closed polygon. Examiners specifically check for these two extra points, so always add one class of the same width just before the first and just after the last, both with frequency zero.

Can I draw a frequency polygon straight from a table that isn't continuous, like 130-139, 140-149?

No — the class marks calculated from a discontinuous table (with gaps like the one unit between 139 and 140) are slightly wrong, and the classes will not sit edge-to-edge on the horizontal axis. Convert the table to continuous form first, by adjusting each boundary by half the gap, then compute class marks from the corrected continuous limits before plotting anything.

Frequency polygon

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads