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Measures of central tendency – The mode and the modal class

📘 Notes

Year 12 (9709) — The Mode and the Modal Class

Mode is the value that occurs most frequently in a dataset. For grouped data (values in classes), the modal class is the class with the largest frequency (or the largest frequency density if class widths are unequal).

  • Ungrouped/discrete data: the mode is an actual data value.
  • Grouped/continuous data: the exact mode is unknown; we identify the modal class and can estimate the modal value using a formula.
  • There can be no mode (all frequencies equal) or multiple modes (bi-/multi-modal).

1) Ungrouped (raw) data — finding the mode

Example 1 (raw list): Find the mode of 7, 9, 7, 10, 11, 7, 9, 8

Frequencies: 7 appears 3 times, 9 appears 2 times, others ≤ 1.

Mode = 7.

Example 2 (bimodal): Find the mode(s) of 5, 6, 6, 7, 7, 8

6 and 7 both occur twice (highest and tied).

Modes = 6 and 7 (bimodal).

2) Discrete frequency table — mode

The mode is the value (row) with the largest frequency.

Score \(x\)Frequency \(f\)
32
45
59
67
74
Solution (show)

Largest frequency is \(9\) at \(x=5\).

Mode = 5.

3) Grouped data (equal class widths) — modal class

Identify the class with the greatest frequency; that is the modal class.

Height (cm)Frequency
140–1506
150–16010
160–17015
170–1809
Solution (show)

Largest frequency = 15 for 160–170.

Modal class = 160–170 cm.

4) Grouped data (unequal widths) — use frequency density

If class widths differ, compare frequency density: \[ \text{freq. density} \;=\; \frac{f}{\text{class width}}. \] The class with the greatest density is the modal class (this matches histogram “tallest bar”).

Time (min)WidthFrequency \(f\)Density \(f/\text{width}\)
0–5581.6
5–1510181.8
15–205112.2
20–3010121.2
Solution (show)

Greatest density = 2.2 for 15–20.

Modal class = 15–20 min.

5) Estimating the modal value (grouped data)

For grouped continuous data with equal (or similar) widths, an approximate modal value can be obtained using the interpolation formula:

\[ \boxed{ \text{Mode} \approx L \;+\; \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \;\times\; h } \] where:

  • \(L\): lower boundary of the modal class,
  • \(h\): class width of the modal class,
  • \(f_1\): frequency of the modal class,
  • \(f_0\): frequency of the class just below,
  • \(f_2\): frequency of the class just above.
Mass (kg)Freq
40–507
50–6012
60–7020
70–8014
80–909
Solution (show)

\(L=60\), \(h=10\), modal class freq \(f_1=20\), below \(f_0=12\), above \(f_2=14\).

\[ \text{Mode} \approx 60 + \frac{20-12}{2\cdot 20 - 12 - 14}\times 10 \;=\; 60 + \frac{8}{14}\times 10 \;=\; 60 + 5.714\;\approx\; \mathbf{65.7\text{ kg}}. \]

6) Interpreting answers & exam tips

  • State clearly: for grouped data you usually give the modal class. Only give a single modal value if asked to estimate (then use the formula and boundaries).
  • If widths are unequal, identify the modal class using frequency density, not just raw frequencies.
  • Mention if the data are bi-modal or no mode, when appropriate.
  • Use class boundaries (e.g. 60–70 means 59.5–69.5 if data are recorded to the nearest unit) when you apply the formula.

7) Practice tables (with answers)

Practice A (discrete): Find the mode
MarkFreq
13
26
36
44

Answer: Modes = 2 and 3 (tie at the highest frequency).

Practice B (grouped, equal width): State the modal class
Speed (km/h)Freq
30–405
40–5012
50–6019
60–7015

Answer: Modal class = 50–60 km/h.

Practice C (grouped, unequal width): Identify the modal class using density
IntervalWidthFreqDensity
0–2263.0
2–64102.5
6–7155.0
7–10372.33

Answer: Modal class = 6–7 (highest density).