Mode of grouped data

The mode of a data set is the value that occurs most frequently. For ungrouped data, this is straightforward , just find the value with the highest count. For grouped data, you can't pinpoint a single value (since each class contains many distinct values), so we instead find the modal class , the class with the highest frequency , and use a formula to estimate where in that class the actual mode lies.

The modal class is the class with the highest frequency.

For example, in the table

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
$f$	$4$	$6$	$10$	$5$	$3$

the modal class is $20-30$ (frequency $10$ , the highest).

The mode formula

Within the modal class, the mode is estimated by

$\text{Mode} = L + \frac{f_1 - f_0}{2 f_1 - f_0 - f_2} \times h,$

where:

$L$ = lower limit of the modal class.
$f_1$ = frequency of the modal class.
$f_0$ = frequency of the class before the modal class.
$f_2$ = frequency of the class after the modal class.
$h$ = class size.

For the example: $L = 20, f_1 = 10, f_0 = 6, f_2 = 5, h = 10$ .

$\text{Mode} = 20 + \frac{10 - 6}{20 - 6 - 5} \times 10 = 20 + \frac{4}{9} \times 10 = 20 + 40/9 \approx 24.44$ .

Intuition behind the formula

The mode formula is a weighted shift from the lower limit of the modal class. If the class before has higher frequency than the class after, the actual mode is shifted toward the lower end of the modal class; if the after class is heavier, the mode shifts toward the upper end.

The factor $(f_1 - f_0)/(2 f_1 - f_0 - f_2)$ is always between $0$ and $1$ (assuming $f_1$ is genuinely the largest), giving a "fractional position" within the modal class.

Multiple modes

If two classes have the same (highest) frequency, the data is bimodal. The board exam usually constructs data with a unique modal class to keep things clean.

If the highest frequency is the first class, we cannot apply the formula directly because $f_0$ doesn't exist (use $f_0 = 0$ ). Similarly if the highest is the last class, use $f_2 = 0$ . The formula still works.

Worked examples

Example 1. Find the mode of the data: classes $0-10, 10-20, 20-30, 30-40, 40-50$ with frequencies $7, 8, 12, 13, 10$ .

Modal class is $30-40$ (frequency $13$ ). $L = 30, f_1 = 13, f_0 = 12, f_2 = 10, h = 10$ .

$\text{Mode} = 30 + \frac{13 - 12}{26 - 12 - 10} \times 10 = 30 + \frac{1}{4} \times 10 = 32.5$ .

Example 2. Find the mode: classes $10-15, 15-20, 20-25, 25-30, 30-35$ with frequencies $4, 6, 5, 8, 3$ .

Wait , frequency $8$ is higher than $6$ . Modal class: $25-30$ ( $f_1 = 8$ ). $L = 25, f_0 = 5, f_2 = 3, h = 5$ .

$\text{Mode} = 25 + \frac{8 - 5}{16 - 5 - 3} \times 5 = 25 + \frac{3}{8} \times 5 = 25 + 1.875 = 26.875$ .

Example 3. Find the mode: classes $0-20, 20-40, 40-60, 60-80, 80-100$ with frequencies $10, 35, 52, 61, 38$ .

Modal class: $60-80$ ( $f_1 = 61$ ). $L = 60, f_0 = 52, f_2 = 38, h = 20$ .

$\text{Mode} = 60 + \frac{61 - 52}{122 - 52 - 38} \times 20 = 60 + \frac{9}{32} \times 20 = 60 + 5.625 = 65.625$ .

Example 4. A class of $40$ students has heights as below:

Height (cm)	$140-150$	$150-160$	$160-170$	$170-180$
$f$	$8$	$14$	$12$	$6$

Find the mode.

Modal class: $150-160$ ( $f_1 = 14$ ). $L = 150, f_0 = 8, f_2 = 12, h = 10$ .

$\text{Mode} = 150 + \frac{14 - 8}{28 - 8 - 12} \times 10 = 150 + \frac{6}{8} \times 10 = 157.5$ cm.

Example 5. If the mean of a data is $50$ and the median is $45$ , find the mode using the empirical relation.

Empirical relation: $\text{Mode} \approx 3 \cdot \text{Median} - 2 \cdot \text{Mean} = 3(45) - 2(50) = 135 - 100 = 35$ .

Empirical relation

For approximately symmetric distributions, the three measures are related by

$\boxed{\text{Mode} \approx 3 \cdot \text{Median} - 2 \cdot \text{Mean}.}$

This relation is not exact (it is an empirical observation about moderately skewed distributions), but it is good enough for board-exam questions that give you any two and ask for the third.

Try it yourself

Find the mode: classes $0-10, 10-20, 20-30, 30-40$ with frequencies $5, 9, 12, 6$ .
Find the mode: classes $5-15, 15-25, 25-35, 35-45, 45-55$ with frequencies $3, 8, 12, 10, 7$ .
Marks of $80$ students:

Marks	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Students	$7$	$14$	$20$	$25$	$14$
Find the mode.

Find the mode: classes $0-20, 20-40, 40-60, 60-80$ with frequencies $35, 75, 45, 25$ .
Mean = $80$ , median = $75$ . Find mode (empirical).
Mode = $60$ , mean = $72$ . Find median.
A factory's workers' ages:

Age	$15-25$	$25-35$	$35-45$	$45-55$	$55-65$
Number	$6$	$11$	$21$	$23$	$14$
Find the modal age.

Why is the mode formula not applicable when $f_0 = f_1 = f_2$ ?
Mean and mode of a data are $30$ and $40$ . Find median.
Find the mode of $4, 5, 5, 5, 6, 7, 7, 8, 8, 8, 9$ .
Find the modal class and mode: $0-10: 4, 10-20: 9, 20-30: 9, 30-40: 6$ . (Bimodal , discuss.)
The frequency of the modal class is twice the frequency of either neighbour. The class size is $5$ . Find the mode in terms of $L$ .

Pitfalls / Insight

(1) The mode is computed from the modal class and its two neighbours. If the modal class is the first or last, use $f_0 = 0$ or $f_2 = 0$ .

(2) The formula's denominator is $2 f_1 - f_0 - f_2$ . A common arithmetic slip is to forget the $2$ .

(3) The empirical relation $\text{Mode} \approx 3 \cdot \text{Median} - 2 \cdot \text{Mean}$ holds only approximately. Use it only when the question asks for one of the three based on the other two.

(4) For a symmetric distribution, mean = median = mode. For a skewed distribution, they differ; the mode is most influenced by the bulk of the data, the mean by extreme values, and the median is in between.