Histogram vs Frequency Polygon

A histogram cuts the number line into bins of equal width and draws a bar over each bin whose height is the count of values inside it, or the relative frequency. The bars touch because the axis is a continuous scale rather than a list of labels. Every histogram needs a boundary rule, and the usual one puts a value in the bin whose lower bound it meets or exceeds and whose upper bound it falls below.

Thirty exam scores binned by tens give 9 scores in 70 to 80 and 12 in 80 to 90, and three of those students scored exactly 80. Switch to the other convention, where a value belongs to the bin whose upper bound it equals, and those three move down: the counts become 12 and 9, the tallest bar changes, and so does the shape you would report. Nothing in the drawn picture tells you which rule was used.

The misreading is "twelve students scored 85." The bar over 80 to 90 says twelve students scored somewhere in that interval and nothing more. The individual values have been thrown away, which is also why you cannot read a median off a histogram, only the bin the median lands in.

Equal width is doing real work in the definition. When bins differ in width, height stops encoding frequency fairly: a bin of width 20 holding 12 values and a bin of width 10 holding 9 draw bars of height 12 and 9, so the wide one looks bigger, while the densities are $12 / 20 = 0.6$ and $9 / 10 = 0.9$ values per unit and the narrow bin is the more crowded one. With unequal bins the vertical axis has to carry density and you read area, not height.

A histogram also destroys the order the values arrived in, which is what a time plot keeps, and its shape depends on the bin width you choose. Histograms are topic 1.5.

A frequency polygon plots one point per bin, at the bin's midpoint horizontally and the bin's frequency vertically, then joins consecutive points with straight segments. The bins have to be equal in width, because a height stands for a count inside a fixed-width interval and counts from intervals of different sizes are not comparable. The line is usually anchored down to zero at one empty bin below the first and one above the last, which closes the polygon onto the axis and makes the area it encloses equal to the total area of the histogram bars.

Take 40 exam scores binned by tens: 3 in the 50s, 8 in the 60s, 14 in the 70s, 10 in the 80s, 5 in the 90s. The midpoints are 55, 65, 75, 85 and 95, so the plotted points are $(55, 3)$, $(65, 8)$, $(75, 14)$, $(85, 10)$ and $(95, 5)$, anchored at $(45, 0)$ and $(105, 0)$.

Here is the sentence to avoid: "the line passes through 11 at a score of 70, so 11 students scored 70." Nothing was measured at 70. The segment from $(65, 8)$ to $(75, 14)$ is drawn, not observed, and halfway along it the height is simply $(8 + 14) / 2 = 11$. Even the plotted points claim less than they look like they claim: $(75, 14)$ says 14 scores fell somewhere between 70 and 80, and says nothing about any score equalling 75.

Overlaying two groups on one set of axes is the main reason to draw a polygon instead of a histogram, and it only works when both groups use identical bins. If the groups differ in size, plot relative frequency instead of count, so the 70s point above becomes $14 / 40 = 0.35$.

Everything the polygon shows is inherited from the binning underneath it, so bin width still decides the shape you end up describing.