Find mean, median, mode of given data set — grouped frequency distribution

Question

The following table shows marks obtained by 50 students in a test:

Marks	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	12	10

Find the mean, median, and mode of this data.

(CBSE 2023, similar pattern — 5 marks)

Solution — Step by Step

Finding the Mean (Assumed Mean Method)

Take assumed mean $a = 25$ (midpoint of the modal class). Class width $h = 10$ .

Marks	$f_i$	$x_i$ (midpoint)	$d_i = x_i - 25$	$u_i = d_i/10$	$f_i u_i$
0-10	5	5	-20	-2	-10
10-20	8	15	-10	-1	-8
20-30	15	25	0	0	0
30-40	12	35	10	1	12
40-50	10	45	20	2	20
Total	50				14

\bar{x} = a + h \times \frac{\sum f_i u_i}{\sum f_i} = 25 + 10 \times \frac{14}{50} = 25 + 2.8 = \mathbf{27.8}

Finding the Median

$N/2 = 50/2 = 25$ . Cumulative frequencies: 5, 13, 28, 40, 50. The cumulative frequency first crosses 25 at the class 20-30. So the median class is 20-30.

\text{Median} = l + \frac{\frac{N}{2} - cf}{f} \times h

Where $l = 20$ , $cf = 13$ (CF before median class), $f = 15$ (frequency of median class), $h = 10$ .

\text{Median} = 20 + \frac{25 - 13}{15} \times 10 = 20 + \frac{12}{15} \times 10 = 20 + 8 = \mathbf{28}

Finding the Mode

Modal class = 20-30 (highest frequency = 15).

\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h

Where $l = 20$ , $f_1 = 15$ , $f_0 = 8$ (preceding class), $f_2 = 12$ (succeeding class), $h = 10$ .

\text{Mode} = 20 + \frac{15 - 8}{2(15) - 8 - 12} \times 10 = 20 + \frac{7}{10} \times 10 = 20 + 7 = \mathbf{27}

Why This Works

For grouped data, we cannot find exact values since individual observations are unknown. We assume all values in a class are concentrated at the midpoint ( $x_i$ ). The assumed mean method reduces arithmetic by shifting the origin.

The median formula uses linear interpolation within the median class — it assumes observations are evenly spread within that class interval. Similarly, the mode formula interpolates within the modal class using the frequencies of neighbouring classes.

Notice: Mean (27.8) $\approx$ Median (28) $\approx$ Mode (27). For roughly symmetric distributions, these three measures cluster together. The empirical relationship $\text{Mode} \approx 3 \times \text{Median} - 2 \times \text{Mean}$ can be used as a quick cross-check.

Alternative Method

For the mean, you can use the direct method: calculate $\sum f_i x_i$ directly. But with larger numbers, step-deviation saves time and reduces errors. CBSE examiners accept both methods for full marks.

In CBSE board exams, the step-deviation method is faster and less error-prone. Examiners also appreciate it because it shows you know the more sophisticated approach. Use $a$ = midpoint of the class with highest frequency.

Common Mistake

In the median formula, students often use the cumulative frequency OF the median class instead of the cumulative frequency BEFORE the median class. The $cf$ in the formula is the CF of the class just before the median class. Getting this wrong shifts your answer significantly.