A frequency polygon plots each class's frequency against the midpoint of its interval — never the upper or lower boundary. This is because the midpoint is the best single representative of all the data hidden inside that interval.

For the interval $20 < w \le 30$: midpoint $= \tfrac{20+30}{2} = 25$.

Plot each (midpoint, frequency) pair, then join consecutive points with straight line segments drawn with a ruler — never a freehand curve. The line should not close back to the first point; it floats above the axis as an open line graph. The one exception: if a class at either end of the range genuinely has a frequency of 0, the line correctly touches the x-axis at that class's midpoint.

• Plotting at an endpoint instead of the midpoint — this distorts the whole shape of the graph.
• Drawing a smooth curve — that's a cumulative frequency diagram, not a frequency polygon.
• Joining the last point back to the first, forming a closed shape.

The table shows the weights of 100 boxes. Draw a frequency polygon and state the modal class.

1. Midpoints: $2, 6, 10, 14, 18$.
2. Coordinate pairs to plot: $(2,10), (6,17), (10,28), (14,25), (18,20)$.
3. Join the five points in order with straight ruled segments — the line floats above the axis at both ends since no interval has frequency 0.
4. The highest point is $(10, 28)$, so the modal class is $8 < w \le 12$.

As a bonus check: the probability a randomly chosen box weighs $8$ kg or less is $\tfrac{10+17}{100} = \tfrac{27}{100}$ — frequency polygons and probability questions are frequently combined in the same exam item.

Both display continuous grouped data, and a frequency polygon can even be thought of as the line joining the midpoints of a histogram's bar-tops. The key practical difference is that two or more frequency polygons can be drawn on the same axes and compared cleanly, whereas overlaid histograms tend to obscure one another — which is why a question comparing two datasets will usually ask for polygons rather than histograms.