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Pearson Edexcel 4MA1 IGCSE  ·  Topic G7.16  ·  Higher tier

G7.16 Histograms

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01 — Explanation

Why isn't a histogram just a bar chart for continuous data?

A bar chart's bar height shows frequency. A histogram's bar area shows frequency — because class widths can be unequal, and area is the only way to avoid a wide class looking artificially huge. This means the yy-axis is never labelled "Frequency"; it's always Frequency Density:

Frequency Density=FrequencyClass Width\text{Frequency Density}=\frac{\text{Frequency}}{\text{Class Width}}

Bars always touch (continuous data has no gaps between classes), and they can never be reordered, since the xx-axis is a genuine numerical scale.

02 — Explanation

How do I find the class width when the data looks "gapped" (e.g. 10–19, 20–29)?

This is rounded, not genuinely gapped, data — the real boundary between 1010–1919 and 2020–2929 is the midpoint, 19.519.5. The fastest reliable method: subtract the lower limits of two consecutive classes.

CW for 10–19=20−10=10\text{CW for }10\text{–}19 = 20-10=10

Exam Tip: Don't subtract the limits within one class (e.g. 19−10=919-10=9) — that ignores the rounding gap and gives the wrong width.

03 — Explanation

How do I draw a histogram from a frequency table?

For each class: find the class width, then FD=FCWFD=\dfrac{F}{CW}, then draw a bar of that width and height. The bars must touch and the yy-axis must be labelled Frequency Density.

Worked example: Class 40≤d<6040\le d<60 with frequency 1212: CW=20CW=20, so FD=1220=0.6FD=\dfrac{12}{20}=0.6.

04 — Explanation

How do I find a frequency from a completed histogram?

Rearrange the same relationship: F=FD×CWF=FD\times CW — read the bar's height (FD) and width (CW) straight off the axes.

Worked example: a bar spans 6060 to 7070 with height 0.80.8: F=0.8×10=8F=0.8\times10=8.

05 — Explanation

How do I find the hidden frequency density scale when the axis isn't labelled in real units?

Use one bar whose true frequency you're told, convert it to the frequency density it must represent, then divide by its measured height (e.g. in squares) to find what one square is worth. Apply that conversion factor to read every other bar.

06 — Explanation

How do I estimate the median or a quartile from grouped/histogram data?

Use linear interpolation — assume the data is spread evenly across each class.

Estimate=L+(n2−Ff)×w\text{Estimate} = L + \left(\frac{\frac{n}{2}-F}{f}\right)\times w

where LL is the lower boundary of the class containing the median, FF is the cumulative frequency before that class, ff is that class's own frequency, and ww is its width. (For Q1Q_1 use position n4\frac n4; for Q3Q_3 use 3n4\frac{3n}4, with the matching class's L,F,f,wL,F,f,w.)

Worked exam-style question: Weights (kg): 4≤w<84\le w<8 (F=4F{=}4), 8≤w<108\le w<10 (F=16F{=}16), 10≤w<1210\le w<12 (F=19F{=}19), 12≤w<1512\le w<15 (F=12F{=}12), 15≤w<3015\le w<30 (F=9F{=}9). Total n=60n=60. Estimate the median.

  1. Median position: 602=30\dfrac{60}{2}=30th value.
  2. Cumulative frequencies: 4, 20, 39, 51, 604,\ 20,\ 39,\ 51,\ 60. The 30th value falls inside the 10≤w<1210\le w<12 class (since CF passes 30 between 20 and 39).
  3. L=10L=10, F=20F=20 (CF before), f=19f=19, w=2w=2.
Median≈10+(30−2019)×2≈11.05 kg\text{Median} \approx 10+\left(\frac{30-20}{19}\right)\times2\approx11.05\text{ kg}

Exam Tip: Don't confuse FF (cumulative frequency before the class) with ff (frequency of the class) — swapping them is the most common error in this formula.

07 — Common mistakes

Common mistakes to avoid

  • Treating frequency density as if it were frequency, especially when reading bar heights directly.
  • Getting the class width wrong for "gapped" rounded data.
  • Using the class midpoint instead of proper interpolation for the median/quartiles.
  • Comparing two histograms with different class widths or scales as if they were directly comparable.