S3 - Histograms (Higher)

S3 - Histograms (Higher)

Construct and interpret histograms for continuous and grouped discrete data, including unequal class widths. Use frequency density to make area represent frequency, recover a missing scale, and estimate counts within a class while stating the assumption.

Frequency is area (Higher)

One histogram bar is 55 units wide and 44 units high. Another is 1010 units wide and 33 units high. Which represents more observations? The taller bar loses: its area is 2020, while the shorter bar's area is 3030. A histogram asks you to look at area, not height alone.

You already know

A grouped frequency table hides individual observations inside intervals. Midpoints can estimate a mean, but the class width is the upper boundary minus the lower boundary.

Suppose 2020 observations are spread across a width of 55. There are 20÷5=420\div5=4 observations per unit of width. Across a width of 1010, the same frequency would give only 22 per unit. That “per unit” height is frequency density.

Histogram and frequency density

A histogram represents grouped numerical data with touching bars whose widths match the class intervals. Bar area is proportional to frequency. Frequency density is frequency divided by class width. Unlike a categorical bar chart, the horizontal axis is a continuous numerical scale, and a wide class must occupy more space.

The vertical scale makes the rectangle's area give the class frequency directly.

Frequency density

frequency density=frequencyclass widthfrequency=class width×frequency density\begin{aligned} \text{frequency density}&=\frac{\text{frequency}}{\text{class width}}\\ \text{frequency}&=\text{class width}\times\text{frequency density} \end{aligned}

For a class a<x⩽ba<x\leqslant b, its width is b−ab-a, not its midpoint. Use the same horizontal units throughout.

Not on the formulae sheet: learn it.

Label both axes and choose a convenient regular vertical scale before drawing bars.

Worked example 1: equal class widths

Question. Time classes 0<t⩽50<t\leqslant5, 5<t⩽105<t\leqslant10, 10<t⩽1510<t\leqslant15 have frequencies 10,20,1510,20,15. Work out frequency densities and draw the histogram. Non-calculator.

Plan. Divide each frequency by its width 55, then draw touching rectangles.

first density=10÷5=2frequency per unitsecond density=20÷5=4same class widththird density=15÷5=3same class width\begin{aligned} \text{first density}&=10\div5=2 &&\text{frequency per unit}\\ \text{second density}&=20\div5=4 &&\text{same class width}\\ \text{third density}&=15\div5=3 &&\text{same class width} \end{aligned}

Why this step? Equal widths make heights proportional to frequencies, but using density still gives the correct area rule.

Answer. Draw bars over 00 to 55, 55 to 1010, 1010 to 1515, with heights 2,4,32,4,3. Label the vertical axis “Frequency density”.

Check. Areas are 5(2)=105(2)=10, 5(4)=205(4)=20, 5(3)=155(3)=15, matching the table.

Change only the last upper boundary from 1515 to 2020. Its frequency stays 1515, so its height must fall.

Worked example 2: one wider class

Question. Classes 0<t⩽50<t\leqslant5, 5<t⩽105<t\leqslant10, 10<t⩽2010<t\leqslant20 have the same frequencies 10,20,1510,20,15. Draw the new histogram. Non-calculator.

Plan. Keep the first two densities, but divide the last frequency by the new width 1010.

last width=20−10=10subtract boundarieslast density=15÷10=1.5spread over twice the widthbar heights=2,4,1.5first two are unchanged\begin{aligned} \text{last width}&=20-10=10 &&\text{subtract boundaries}\\ \text{last density}&=15\div10=1.5 &&\text{spread over twice the width}\\ \text{bar heights}&=2,4,1.5 &&\text{first two are unchanged} \end{aligned}

Why this step? A bar twice as wide needs half the height to keep its frequency, and therefore its area, unchanged.

Answer. Draw touching bars with heights 2,4,1.52,4,1.5 over the three stated intervals.

Check. The last area is 10×1.5=1510\times1.5=15, just as before.

[DIAGRAM: asset_name: Equal and unequal histogram widths; asset_slug: aqa8300_s06_construct; caption: Doubling the last class width halves its density while keeping frequency 15; description: Two frequency-density histograms show the two worked tables on identical time and density scales. Left widths 5,5,5 and heights 2,4,3. Right widths 5,5,10 and heights 2,4,1.5. The last bars are labelled frequency 15.]
Doubling the last class width halves its density while keeping frequency 15

The bar's horizontal boundaries and vertical height must both be correct.

Watch out

Wrong: draw heights 10,20,1510,20,15 with unequal widths. The last bar then has too much area. Right: use frequency density, not frequency, for its height.

Work out the heights before reaching for a ruler.

Reading frequencies and a missing scale (Higher)

Reading a histogram reverses its construction: multiply width by density. If the density scale is missing, a known bar can establish how drawing area relates to frequency.

For example, a bar with frequency 2020 and width 55 must have density 44, whatever physical height appears on the page. Use that known height to calibrate the vertical scale, then read the other heights consistently.

Worked example 3: a labelled density axis

Question. A bar covers 10<t⩽2010<t\leqslant20 and has frequency density 33. Work out its frequency. Non-calculator.

Plan. Multiply the interval width by its height on the density scale.

width=20−10=10class intervalfrequency=10×3=30rectangle area\begin{aligned} \text{width}&=20-10=10 &&\text{class interval}\\ \text{frequency}&=10\times3=30 &&\text{rectangle area} \end{aligned}

Why this step? Density counts observations per unit of width, so multiply by the number of width units.

Answer. 3030 observations.

Check. 30÷10=330\div10=3, recovering the density.

Change only the available scale: the axis now has no density numbers, but one frequency is known.

Worked example 4: using a known bar

Question. On a histogram grid, a reference bar is 55 horizontal units wide and 44 grid units high. It represents 2020 observations. Another bar is 1010 units wide and 33 grid units high. Work out its frequency. Non-calculator.

Plan. Find observations per unit of drawn area from the reference bar.

reference area=5×4=20grid-area unitsfrequency per area unit=20÷20=1calibrate using known frequencyother area=10×3=30same drawing scalesother frequency=30×1=30apply calibration\begin{aligned} \text{reference area}&=5\times4=20 &&\text{grid-area units}\\ \text{frequency per area unit}&=20\div20=1 &&\text{calibrate using known frequency}\\ \text{other area}&=10\times3=30 &&\text{same drawing scales}\\ \text{other frequency}&=30\times1=30 &&\text{apply calibration} \end{aligned}

Why this step? With an unlabelled vertical scale, a numerical grid height need not already be the frequency density. The known bar tells you the conversion.

Answer. 3030 observations, so the shorter bar represents more than the reference bar.

Check. Its area is 1.51.5 times the reference area; 1.5×20=301.5\times20=30.

[DIAGRAM: asset_name: Calibrating an unlabelled histogram; asset_slug: aqa8300_s06_scale; caption: A known bar sets the area-to-frequency scale before another bar is read; description: Touching rectangles have horizontal widths 5 and 10 and drawn heights 4 and 3. Horizontal grid lines are visible but vertical tick labels are absent. The first bar is labelled frequency 20 and the second frequency unknown. Brackets state the widths and the grid heights.]
A known bar sets the area-to-frequency scale before another bar is read

Use areas in consistent drawing units. Mixing centimetres for one bar with squares for another destroys the comparison.

Watch out

Wrong: compare only the heights 44 and 33, then conclude the second frequency is smaller. Right: compare areas 2020 and 3030, because the widths differ.

The scale factor in the next diagram description is not 11.

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