Frequency Polygons and Histograms

Frequency polygons and histograms are used to display continuous grouped data. Histograms with unequal class widths are a Higher-tier topic and are a favourite of Edexcel examiners.

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Key Vocabulary

Term	Definition
Frequency polygon	A line graph plotted at the midpoints of class intervals against frequency
Histogram	A bar chart for continuous data where the area of each bar represents the frequency
Frequency density	Frequency ÷ class width; used as the vertical axis in histograms with unequal class widths
Class width	Upper class boundary − lower class boundary

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Histogram vs Bar Chart — Decision Tree

Students often mix up histograms and bar charts. Use this flowchart.

flowchart TD
    A["Is the data continuous?"] -->|No| B["Use a bar chart<br/>(gaps between bars)"]
    A -->|Yes| C["Are all class widths equal?"]
    C -->|Yes| D["Histogram with<br/>frequency on y-axis"]
    C -->|No| E["[H] Histogram with<br/>frequency density<br/>on y-axis"]

    style B fill:#3498db,color:#fff
    style D fill:#27ae60,color:#fff
    style E fill:#e74c3c,color:#fff

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Frequency Polygons

A frequency polygon is drawn by plotting points at the midpoint of each class interval against the frequency, then joining the points with straight lines.

Worked Example 1

Draw a frequency polygon for the time (t seconds) 50 students took to complete a puzzle.

Time (t seconds)	Frequency	Midpoint
$0 < t \leq 10$0<t≤10	5	5
$10 < t \leq 20$10<t≤20	12	15
$20 < t \leq 30$20<t≤30	18	25
$30 < t \leq 40$30<t≤40	10	35
$40 < t \leq 50$40<t≤50	5	45

Solution: Plot the points (5, 5), (15, 12), (25, 18), (35, 10), (45, 5) on a graph with time on the x-axis and frequency on the y-axis. Join them with straight line segments. Modal class is $20 < t \leq 30$20<t≤30 (the peak of the polygon).

Notes:

• You can optionally close the polygon by extending to the midpoints of hypothetical classes before and after (frequency = 0).
• Frequency polygons are useful for comparing two data sets on the same axes — plot two polygons in different colours.

Worked Example 2

A frequency polygon is to be drawn for this data:

Class	Frequency
0–10	8
10–20	15
20–30	22
30–40	11
40–50	4

What points should be plotted?

Solution: Midpoints are 5, 15, 25, 35, 45. Plot: (5, 8), (15, 15), (25, 22), (35, 11), (45, 4), then join with straight lines.

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Histograms with Equal Class Widths

When all class widths are equal, a histogram looks like a standard bar chart but with no gaps between bars (because the data is continuous).

The vertical axis is frequency (since all bars have the same width, the area is proportional to frequency).

Key difference from a bar chart: Bars touch because the data is continuous; there are no gaps between classes.

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Histograms with Unequal Class Widths [H]

When class widths are unequal, using frequency on the vertical axis would be misleading — a wide class would appear to have a higher frequency just because its bar is wider.

Instead, we use frequency density on the vertical axis.

Frequency density = frequency ÷ class width

The area of each bar equals the frequency for that class.

Worked Example 3 [H]

The table shows the ages of 100 people at a concert. Calculate the frequency density for each class and describe the histogram.

Age (a years)	Frequency	Class width	Frequency density
$0 < a \leq 10$0<a≤10	15	10	15 ÷ 10 = 1.5
$10 < a \leq 20$10<a≤20	25	10	25 ÷ 10 = 2.5
$20 < a \leq 30$20<a≤30	30	10	30 ÷ 10 = 3.0
$30 < a \leq 50$30<a≤50	20	20	20 ÷ 20 = 1.0
$50 < a \leq 80$50<a≤80	10	30	10 ÷ 30 = 0.33…

Solution:

1. Class width = upper boundary − lower boundary.
2. Frequency density = frequency ÷ class width.
3. On the x-axis mark class boundaries (0, 10, 20, 30, 50, 80); on the y-axis plot frequency density (up to 3).
4. Draw bars with no gaps. The tallest bar is $20 < a \leq 30$20<a≤30 (FD = 3.0). The bar for $50 < a \leq 80$50<a≤80 is the widest but very short (FD = 0.33…).

Worked Example 4 [H] — Finding frequency from a histogram

A histogram bar spans 30 to 50 on the x-axis and has frequency density 2.5. How many data items does it represent?

Solution:

• Class width = 50 − 30 = 20
• Frequency = frequency density × class width = 2.5 × 20 = 50 items (the area of the bar).

Worked Example 5 [H] — Completing a table from a histogram

A histogram is partially described. The $10 < t \leq 20$10<t≤20 bar has frequency density 3 and represents 30 students. Complete the frequency column.

Time (t minutes)	Frequency density	Class width	Frequency
$0 < t \leq 10$0<t≤10	2	10	?
$10 < t \leq 20$10<t≤20	3	10	30
$20 < t \leq 25$20<t≤25	4	5	?
$25 < t \leq 40$25<t≤40	1	15	?

Solution:

• $0 < t \leq 10$0<t≤10: f = 2 × 10 = 20
• $20 < t \leq 25$20<t≤25: f = 4 × 5 = 20
• $25 < t \leq 40$25<t≤40: f = 1 × 15 = 15
• Total students = 20 + 30 + 20 + 15 = 85

Worked Example 6 [H] — Estimating from part of a bar

Using the histogram in Worked Example 5, estimate the number of students who took more than 22 minutes.

Solution:

• From 22 to 25 (part of the $20 < t \leq 25$20<t≤25 class, which contains 20 students): proportion = (25 − 22) ÷ 5 = 3/5 of the class. Frequency = (3/5) × 20 = 12.
• From 25 to 40 (whole class): 15 students.
• Estimate = 12 + 15 = 27 students.

Key assumption: Data is evenly distributed within each class.

Worked Example 7 [H] — Histogram with varying widths

100 delivery drivers travel the following distances (d km) in one day:

Distance (d km)	Frequency
$0 < d \leq 20$0<d≤20	10
$20 < d \leq 30$20<d≤30	15
$30 < d \leq 40$30<d≤40	30
$40 < d \leq 60$40<d≤60	28
$60 < d \leq 100$60<d≤100	17

(a) Calculate the frequency density for each class. (b) Estimate the number of drivers who travelled between 35 and 50 km.

Solution:

(a)

Class	Width	FD
$0 < d \leq 20$0<d≤20	20	10 ÷ 20 = 0.5
$20 < d \leq 30$20<d≤30	10	15 ÷ 10 = 1.5
$30 < d \leq 40$30<d≤40	10	30 ÷ 10 = 3.0
$40 < d \leq 60$40<d≤60	20	28 ÷ 20 = 1.4
$60 < d \leq 100$60<d≤100	40	17 ÷ 40 = 0.425