Edexcel GCSE Maths

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(Histograms)

Understanding Histograms

Understanding Histograms

 

How Do I Interpret a Histogram?

Histograms show grouped continuous data. Unlike a bar chart, the area of each bar represents the frequency — not just the height.

Frequency=Frequency Density×Class Width\text{Frequency} = \text{Frequency Density} \times \text{Class Width}

Watch Out:

  • The y-axis might be labelled frequency density or simply frequency — check carefully!
  • If all class widths are the same, height might be proportional to frequency.

 

Steps for Interpreting a Histogram

1. Work Out Missing Frequencies

Use:

Frequency=Density×Class Width\text{Frequency} = \text{Density} \times \text{Class Width}

2. Estimate Frequencies From Parts of a Bar

Sometimes you’ll be asked for frequency between two points within a class interval.

Treat this like a proportion of the whole bar’s area.

3. Compare Distributions

You can compare shapes, centers, and spreads of two histograms, but only if:

  • They use the same class intervals.
  • They use the same frequency density scale.

 

Example

The histogram shows the time (in minutes) taken by runners to complete a trail race.

 

diagram of frequency density for example question

 

You’re given part of the frequency table:

 

Time (mins)Frequency
30t<4030 \leq t \lt 406
40t<5040 \leq t \lt 5012
50t<5550 \leq t \lt 5510
55t<6055 \leq t \lt 6015
60t<9060 \leq t \lt 90?

 

(a) Use the Histogram to Find the Missing Frequency

The bar for 60t<9060 \leq t \lt 90:

  • Frequency density = 0.4
  • Class width = 90 - 60 = 30

Frequency=0.4×30=12\text{Frequency} = 0.4 \times 30 = 12

So, the missing frequency is 12

 

(b) Use the Table to Complete the Histogram

The bar for 30t<4030 \leq t \lt 40 is missing:

  • Frequency = 6
  • Class width = 10

Frequency Density=610=0.6\text{Frequency Density} = \frac{6}{10} = 0.6

Draw a bar from 30 to 40 with a height of 0.6

 

histogram is missing bar added in

 

 
(c) Estimate the Number of Runners Who Took More Than 58 Minutes

You're looking at:

  • Part of the 55t<6055 \leq t \lt 60 bar: only 2 minutes (from 58 to 60)
  • Plus all of 60t<9060 \leq t \lt 90

Bar for 55 ≤ t < 60:

  • Frequency density = 3
  • Width from 58 to 60 = 2

Frequency=3×2=6\text{Frequency} = 3 \times 2 = 6

From part (a), we know there are 12 runners in the 60–90 interval.

6+12=186 + 12 = 18

Estimated number of runners who took more than 58 minutes is 18

 

diagram of histogram with area shaded above 58 mins

 

 

Tuity Tip

Hover me!

If the histogram looks odd (e.g. a tall narrow bar next to a short wide bar), check frequency densities carefully

Always show your area = frequency working — it’s an easy method mark

Use histograms to compare how spread out or bunched up your data is

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