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 \(f\) |
|---|
| 3 | 2 |
| 4 | 5 |
| 5 | 9 |
| 6 | 7 |
| 7 | 4 |
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 |
|---|
| 140–150 | 6 |
| 150–160 | 10 |
| 160–170 | 15 |
| 170–180 | 9 |
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) | Width | Frequency \(f\) | Density \(f/\text{width}\) |
|---|
| 0–5 | 5 | 8 | 1.6 |
| 5–15 | 10 | 18 | 1.8 |
| 15–20 | 5 | 11 | 2.2 |
| 20–30 | 10 | 12 | 1.2 |
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 |
|---|
| 40–50 | 7 |
| 50–60 | 12 |
| 60–70 | 20 |
| 70–80 | 14 |
| 80–90 | 9 |
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
Answer: Modes = 2 and 3 (tie at the highest frequency).
Practice B (grouped, equal width): State the modal class
| Speed (km/h) | Freq |
|---|
| 30–40 | 5 |
| 40–50 | 12 |
| 50–60 | 19 |
| 60–70 | 15 |
Answer: Modal class = 50–60 km/h.
Practice C (grouped, unequal width): Identify the modal class using density
| Interval | Width | Freq | Density |
|---|
| 0–2 | 2 | 6 | 3.0 |
| 2–6 | 4 | 10 | 2.5 |
| 6–7 | 1 | 5 | 5.0 |
| 7–10 | 3 | 7 | 2.33 |
Answer: Modal class = 6–7 (highest density).