Histograms

A histogram plots class boundaries on the horizontal axis and frequency on the vertical axis; each class interval gets a rectangle whose height equals its frequency. Because the horizontal scale is continuous — every value between the boundaries is a possible data value — adjacent bars touch, with no gaps. This is the one structural difference from an ordinary bar graph, which represents discrete, unrelated categories and is drawn with gaps between bars.

Reading the modal class needs no construction at all: it is simply the class under the tallest bar, spotted by inspection.

Locating the mode needs one geometric step. Identify the modal-class bar. Draw a line from its top-left corner to the top-left corner of the bar immediately to its right, and a second line from its top-right corner to the top-right corner of the bar immediately to its left. These two diagonals cross at a point inside the modal bar; drop a perpendicular from that crossing point to the horizontal axis, and the foot of the perpendicular is the mode. This construction is not a coincidence — it is a compass-and-ruler version of the algebraic modal-class formula used elsewhere in the same chapter for grouped data:

where: $l$ is the lower boundary of the modal class, $f_1$ is the frequency of the modal class, $f_0$ and $f_2$ are the frequencies of the classes just before and just after it, and $i$ is the class width. The two routes — drawing diagonals, or plugging into this formula — are two ways of finding the same interpolated point inside the modal bar.

The mode construction locates the crossing point inside the modal bar using nothing beyond similar triangles: this section shows why the perpendicular foot lands exactly at $l + \dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$, so the ruler-and-pencil picture and the plug-in formula are guaranteed to agree, letting you use either one and trust the other as a check.

Step 1. Set up coordinates with the modal class starting at $x = l$ and ending at $x = l+i$. Its bar has height $f_1$. The bar to its left has height $f_0$, and the bar to its right has height $f_2$. Line A joins $(l, f_1)$ — top-left of the modal bar — to $(l+i, f_2)$ — top-right of the next bar, matched in height by that bar's actual top edge at $x=l+i$; Line B joins $(l+i, f_1)$ to $(l, f_0)$ in the same way.

Step 2. Write Line A's height as a function of horizontal distance $x-l$ from the left edge: it falls from $f_1$ to $f_2$ over width $i$, so its height is $f_1 - \dfrac{(f_1-f_2)}{i}(x-l)$. Line B rises from $f_0$ to $f_1$ over the same width, so its height is $f_0 + \dfrac{(f_1-f_0)}{i}(x-l)$.

Step 3. At the crossing point the two heights are equal. Set them equal and let $x - l = m$ be the unknown horizontal offset of the mode from $l$: $f_1 - \dfrac{(f_1-f_2)}{i}m = f_0 + \dfrac{(f_1-f_0)}{i}m$.

Step 4. Collect the $m$ terms on one side: $f_1 - f_0 = \dfrac{m}{i}\big[(f_1-f_2) + (f_1-f_0)\big] = \dfrac{m}{i}\big[2f_1 - f_0 - f_2\big]$. Solving, $m = \dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$, so the mode is $l + m = l + \dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$ — exactly the modal-class formula, so the diagonal-lines construction and the algebraic formula are provably the same calculation, one done with a ruler and one done with arithmetic.

Draw a histogram for the distribution below, state the modal class, and estimate the mode both graphically and algebraically.

Step 1. Set up axes with the class boundaries $0, 10, 20, 30, 40, 50$ on the horizontal axis and frequency on the vertical axis, and draw five touching bars of heights $5, 12, 20, 9, 4$. The tallest bar has height $20$ over the class $20$–$30$, so the modal class is 20–30.

Step 2. For the graphical mode, join the top-left corner of the $20$–$30$ bar, point $(20,20)$, to the top-left corner of the $30$–$40$ bar, point $(30,9)$; join the top-right corner $(30,20)$ to the top-right corner of the $10$–$20$ bar, $(20,12)$. Where these two lines cross, drop a perpendicular to the horizontal axis; a carefully drawn diagram reads off close to $24$.

Step 3. Confirm algebraically using the modal-class formula with $l = 20$, $f_1 = 20$, $f_0 = 12$, $f_2 = 9$, $i = 10$: $\text{Mode} = 20 + \dfrac{20-12}{2(20)-12-9}\times 10 = 20 + \dfrac{8}{19}\times 10 \approx 20 + 4.21$.

The mode is approximately 24.2, matching the graphical reading, and confirming the two methods are the same interpolation seen from different angles.

• Leaving gaps between bars. The data is continuous, so bars must touch; gaps are a bar-graph feature and cost marks in a histogram.
• Plotting class marks instead of class boundaries. The horizontal axis needs the actual interval boundaries $(0, 10, 20, \ldots)$, not midpoints, or bar widths misalign.
• Joining the wrong corners for the mode construction. Always connect the modal bar's top corners diagonally to the adjacent bars' top corners on the matching side — left corner to the right neighbour, right corner to the left neighbour.
• Forgetting unequal class widths need proportional bar widths. ICSE problems mostly use equal-width classes, but if widths differ, the bar width on the axis must scale with the class width too.

Ask the tutor: if you want to practise the corner-joining construction on your own frequency table, or see why the similar-triangles proof works with different numbers, ask the AI tutor to walk through it with you.