Mathematics 9709 · Probability & Statistics 1

Mode and the modal class

An average is one number that stands in for the whole set.
The mode is the simplest average: the most popular value.
On a bar chart it is the tallest bar; on a histogram it is the tallest density bar.
The only real trap is grouped data with unequal class widths.

Mode = tallest bar = highest density

For raw or discrete data the mode is the value with the highest frequency (there can be none, or two). It is the only average that works for qualitative data (favourite colour).

Worked example

Start simple: the mode of a raw list

$7,3,7,9,3,7$ → 7 appears most → mode $7$ . $4,4,7,7$ → bimodal (4 and 7). $2,4,6,8$ → every value once → no mode.

Answer

\text{mode can be one, several, or none}

For grouped data you can only name the modal class.
In a histogram, “tallest bar” means highest frequency density.
Frequency density = frequency ÷ class width.
Equal widths: biggest frequency and biggest density agree.
Unequal widths: only density tells the truth.

Get the true class boundaries (rounded labels hide half a unit each side).
width = upper boundary − lower boundary.
density = frequency ÷ width.
Modal class = the row with the largest density (not the largest frequency).
Quote it as an interval, e.g. $10.5\le x<12.5$ .

270 pencils. The 4–7 class has the MOST pencils (100) but the lowest bar; the narrow 11–12 class is tallest, so it is the modal class.

Worked example

Find the modal class (unequal widths)

Pencil lengths (nearest cm): 4–7 has 100, 8–10 has 90, 11–12 has 80. True boundaries give widths $4,\ 3,\ 2$ , so densities are $100/4=25,\ 90/3=30,\ 80/2=40$ .

Answer

\text{modal class } 11\text{–}12\ (10.5\le x<12.5)

Modal class = tallest bar = biggest frequency ÷ width. Equal widths? frequency alone is fine. Unequal? only density tells the truth.

Situation	Modal class is…
Equal class widths	the biggest frequency (density agrees)
Unequal widths	the biggest density — must divide by width
Grouped, asked for 'the mode'	illegal — give a modal class/interval
Qualitative data	mode is the only average available

Examiner note

State the answer as an interval with boundaries (not “11–12”). For a no-modal-class question, write the line “equal densities ⇒ frequencies in the ratio of the widths” to earn the method mark.

Common mistake

With unequal widths, picking the class with the biggest frequency instead of the biggest density. And for grouped data, never quote a single “mode” — only a modal class.

Why density, not frequency

A histogram shows frequency by area, so a wide class can hold many items yet sit low. “Most concentrated” means tallest — and height is density, not raw count.

Your turn— tap to reveal the worked answer (no modal class)

Two classes have width ratio $3:2$ and there is no modal class. The first has frequency 48. Find the second.

No modal class means equal densities, so frequencies follow the widths: $f_2=48\times\tfrac23$ .

Answer

f_2=32

PreviousCumulative frequency graphs NextThe mean

Mode and the modal class

An average is one number that stands in for the whole set.
The mode is the simplest average: the most popular value.
On a bar chart it is the tallest bar; on a histogram it is the tallest density bar.
The only real trap is grouped data with unequal class widths.

Mode = tallest bar = highest density

For raw or discrete data the mode is the value with the highest frequency (there can be none, or two). It is the only average that works for qualitative data (favourite colour).

Worked example

Start simple: the mode of a raw list

$7,3,7,9,3,7$ → 7 appears most → mode $7$ . $4,4,7,7$ → bimodal (4 and 7). $2,4,6,8$ → every value once → no mode.

Answer

\text{mode can be one, several, or none}

For grouped data you can only name the modal class.
In a histogram, “tallest bar” means highest frequency density.
Frequency density = frequency ÷ class width.
Equal widths: biggest frequency and biggest density agree.
Unequal widths: only density tells the truth.

Get the true class boundaries (rounded labels hide half a unit each side).
width = upper boundary − lower boundary.
density = frequency ÷ width.
Modal class = the row with the largest density (not the largest frequency).
Quote it as an interval, e.g. $10.5\le x<12.5$ .

270 pencils. The 4–7 class has the MOST pencils (100) but the lowest bar; the narrow 11–12 class is tallest, so it is the modal class.

Worked example

Find the modal class (unequal widths)

Pencil lengths (nearest cm): 4–7 has 100, 8–10 has 90, 11–12 has 80. True boundaries give widths $4,\ 3,\ 2$ , so densities are $100/4=25,\ 90/3=30,\ 80/2=40$ .

Answer

\text{modal class } 11\text{–}12\ (10.5\le x<12.5)

Modal class = tallest bar = biggest frequency ÷ width. Equal widths? frequency alone is fine. Unequal? only density tells the truth.

Situation	Modal class is…
Equal class widths	the biggest frequency (density agrees)
Unequal widths	the biggest density — must divide by width
Grouped, asked for 'the mode'	illegal — give a modal class/interval
Qualitative data	mode is the only average available

Examiner note

Common mistake

With unequal widths, picking the class with the biggest frequency instead of the biggest density. And for grouped data, never quote a single “mode” — only a modal class.

Why density, not frequency

A histogram shows frequency by area, so a wide class can hold many items yet sit low. “Most concentrated” means tallest — and height is density, not raw count.

Your turn— tap to reveal the worked answer (no modal class)

Two classes have width ratio $3:2$ and there is no modal class. The first has frequency 48. Find the second.

No modal class means equal densities, so frequencies follow the widths: $f_2=48\times\tfrac23$ .

Answer

f_2=32