Mean, median, mode for grouped data — when to use each formula

Question

The following data shows marks of 50 students:

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

Find the mean, median, and mode. When should we use each measure?

(CBSE Class 10 pattern)

Solution — Step by Step

flowchart TD
    A["Central Tendency\nfor Grouped Data"] --> B{"What do you need?"}
    B -->|"Average value\n(uses all data)"| C["Mean\nDirect/Assumed Mean/\nStep Deviation method"]
    B -->|"Middle value\n(unaffected by extremes)"| D["Median\nUsing cf and\nmedian class"]
    B -->|"Most frequent value"| E["Mode\nUsing modal class\nformula"]
    C --> F["Best for symmetric\ndata without outliers"]
    D --> G["Best for skewed\ndata or open-ended classes"]
    E --> H["Best for categorical\nor peaked data"]

Marks	$f_i$	$x_i$ (mid-point)	$f_i x_i$
0-10	5	5	25
10-20	8	15	120
20-30	15	25	375
30-40	12	35	420
40-50	10	45	450
Total	50		1390

\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1390}{50} = \mathbf{27.8}

$N/2 = 50/2 = 25$ . We need the class where the 25th observation falls.

Cumulative frequencies: 5, 13, 28, 40, 50.

The 25th observation falls in the class 20-30 (cf just exceeds 25). This is the median class.

\text{Median} = l + \frac{N/2 - cf}{f} \times h = 20 + \frac{25 - 13}{15} \times 10 = 20 + \frac{12}{15} \times 10 = 20 + 8 = \mathbf{28}

where $l = 20$ (lower limit), $cf = 13$ (cumulative frequency before median class), $f = 15$ (frequency of median class), $h = 10$ (class width).

The modal class has the highest frequency: 20-30 ( $f_1 = 15$ ).

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

where $f_0 = 8$ (previous class), $f_2 = 12$ (next class).

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

Why This Works

Mean uses all values and gives the arithmetic average — best for symmetric data. Median finds the middle value and is robust against extreme scores. Mode identifies the peak — where data clusters most.

For this data: mean = 27.8, median = 28, mode = 27. They are close because the data is roughly symmetric. For heavily skewed data, these three would differ significantly, and median would be the most reliable representative.

Alternative Method — Step Deviation for Mean

When numbers are large, use the assumed mean method: pick $a = 25$ (mid-value), then calculate $d_i = x_i - a$ and $u_i = d_i/h$ :

\bar{x} = a + h \times \frac{\sum f_i u_i}{\sum f_i}

This reduces computation significantly in board exams.

In CBSE Class 10, the step deviation method saves time in the exam. Pick the assumed mean as the mid-point of the middle class. Also remember the empirical relationship: $\text{Mode} \approx 3\text{Median} - 2\text{Mean}$ . If you have calculated any two, you can quickly estimate the third as a check.

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 cumulative frequency of all classes before the median class, not including it. Using the wrong $cf$ shifts your answer significantly.

Question

Solution — Step by Step

Why This Works

Alternative Method — Step Deviation for Mean

Common Mistake

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