POP

E9.7 Histograms

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01 — Explanation

Why can't I just plot frequency on a histogram?

In an ordinary bar chart every bar has equal width, so height alone fairly represents frequency. But once continuous data is grouped into unequal class widths, a wide bar plotted by raw frequency would look artificially large just because it spans more of the scale — not because it contains more data.

02 — Explanation

How do I calculate frequency density?

\text{Frequency density} = \frac{\text{frequency}}{\text{class width}}

so that it's the area of a bar, not its height, that represents frequency: $\text{area} = \text{frequency density}\times\text{class width} = \text{frequency}$ .

Example: ages of $140$ museum visitors.

Age	$f$	Width	Frequency density
$0<a\le10$	$20$	$10$	$2.0$
$10<a\le20$	$35$	$10$	$3.5$
$20<a\le30$	$30$	$10$	$3.0$
$30<a\le50$	$40$	$20$	$2.0$
$50<a\le80$	$15$	$30$	$0.5$

Bars are drawn with these heights, touching with no gaps, since the underlying data is continuous.

Examiner Tip

When class widths are unequal, the vertical axis must be frequency density, never raw frequency — plotting frequency directly is the single most common histogram error.

03 — Explanation

How do I find a frequency from part of a bar?

Multiply that bar's frequency density by the sub-width actually needed, not the full class width, and not a mid-interval value (that technique belongs to estimating a mean, not this one).

Example: using the table above, find the number of visitors aged between $15$ and $40$ .

\underbrace{5\times3.5}_{15\text{–}20} + \underbrace{30}_{20\text{–}30\text{ (whole class)}} + \underbrace{10\times2.0}_{30\text{–}40} = 17.5+30+20 = 67.5 \approx 68 \text{ visitors}

04 — Explanation

How do I work backwards from a drawn histogram to find a missing frequency?

Read the frequency density directly from the height of the bar (using the scale on the vertical axis), then multiply by that bar's class width — this is simply the area calculation from above, run in reverse from a picture rather than a table.

Worked Example

A histogram bar spans the class $30<a\le50$ (width $20$ ) and has a height of $1.8$ on the frequency density axis. Find the frequency this bar represents.

Frequency $=$ frequency density $\times$ width.
Frequency $= 1.8 \times 20 = 36$ .

Examiner Tip: Always double-check which axis you're reading a value from — height gives frequency density, not frequency itself. Multiplying only the height by nothing (forgetting the width entirely) is a common way marks are lost when reading values directly off a drawn histogram.

05 — Explanation

How do I estimate the total number of items from a whole histogram?

Multiply every bar's frequency density by its own class width to recover that bar's frequency, then add all the bars' frequencies together — this reconstructs the total from the areas alone, without ever needing the original raw data table.

Examiner Tip

A quick sanity check after any histogram calculation is to confirm the recovered frequencies sum to the total stated in the question — if the table said $140$ visitors and the reconstructed frequencies only add to $132$ , an earlier frequency density or width has been misread.