Statistics

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Subtopic 3 of 5·← Median and mode of raw data·Range, variance and standard deviation →

Median and mode of grouped data

For class-interval data:

Median (continuous classes, width $h$ ): locate the median class where cumulative frequency first reaches $n/2$ , then $\text{median}=L+\frac{\frac{n}{2}-cf}{f}\,h$ ( $L$ = lower limit of median class, $cf$ = cumulative frequency before it, $f$ = its frequency).

Mode: the modal class has the highest frequency $f_1$ (neighbours $f_0$ below, $f_2$ above): $\text{mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h$

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The grouped median

For class-interval data with class width $h$ , build the cumulative frequency (cf) column. The median class is where cf first reaches $n/2$ . Then: $\text{median} = L + \frac{\frac{n}{2} - cf_{\text{prev}}}{f}\,h$ $L$ = lower limit of the median class, $f$ = its frequency, $cf_{\text{prev}}$ = cumulative frequency before that class. The two classic slips: using $n/2$ 's cf inside the numerator (must be the previous cf), and forgetting grouped data always uses $\frac{n}{2}$ , never $\frac{n+1}{2}$ .

The grouped mode

The modal class has the highest frequency $f_1$ ; its neighbours are $f_0$ (below) and $f_2$ (above): $\text{mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}\,h$ If the modal class sits at an edge, the missing neighbour counts 0. The denominator $2f_1 - f_0 - f_2$ is positive whenever $f_1$ is a strict maximum.

Missing frequency questions

A frequency is unknown, the median (or mode) given. Reverse the median formula: the known cf column pins the median class, giving one linear equation in the unknown frequency. Solve for it; verify it keeps the median class where you assumed.

Grouped mean via midpoints

$\bar{x} = \frac{\sum f_i m_i}{\sum f_i}$ with $m_i$ the class midpoints. CGL prefers the shortcut/assumed-mean method internally but the direct formula answers every exam-sized table. Convert inclusive classes (10–19) to exclusive (9.5–19.5) before computing if the formula demands continuity.

Worked example: grouped median

Classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 12, 9, 6 ( $n = 40$ ). Cumulative frequencies: 5, 13, 25, 34, 40. Since $\frac{n}{2} = 20$ , the median class is 20–30 (the first cf that reaches 20). With $L = 20$ , $cf_{\text{prev}} = 13$ , $f = 12$ , $h = 10$ : $\text{median} = 20 + \frac{20 - 13}{12} \times 10 \approx 25.83$

Worked example: grouped mode and mean

Same table: the modal class is 20–30 ( $f_1 = 12$ , $f_0 = 8$ , $f_2 = 9$ ). $\text{mode} = 20 + \frac{12 - 8}{24 - 8 - 9} \times 10 = 20 + \frac{40}{7} \approx 25.71$ Mean via midpoints 5, 15, 25, 35, 45: $\frac{25 + 120 + 300 + 315 + 270}{40} = 25.75$ . All three measures sit close together, so the table is nearly symmetric — a free sanity check on your arithmetic.

Quick revision

Median: cf column → median class → $L + \frac{n/2 - cf_{\text{prev}}}{f}h$ .
Mode: tallest class → $L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}h$ .
Grouped always uses $n/2$ .
Missing frequency: reverse the median formula — one linear equation.
Mean: midpoints × frequencies ÷ total frequency.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Grouped medianvery common3 practice Q

How to spot it:

A frequency table with class intervals; the median asked — occasionally with a missing frequency.

cf column → first cf $\ge n/2$ → $L + \frac{n/2 - cf_{\text{prev}}}{f}h$.

Build the cf column; locate the median class by $n/2$ .
Read $L$ , $f$ , $h$ , $cf_{\text{prev}}$ from the table.
Substitute; keep the fraction exact.

Why: the formula is mechanical once the median class is right — all errors live in that step.

Example: Classes 0-10, 10-20, 20-30, 30-40, 40-50 have frequencies 4, 6, 10, 8, 2. The median is:

$n = 30$ , $n/2 = 15$ : cf reaches 20 at 20-30. Median $= 20 + \frac{15-10}{10} \times 10 = 25$ .

Type 2: Grouped modecommon2 practice Q

How to spot it:

A frequency table; the mode asked — the modal class is the tallest one.

$\text{mode} = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2}h$.

Identify the modal class (highest frequency).
Note $f_0$ , $f_2$ from the neighbours (0 if at an edge).
Substitute; the answer sits inside the modal class.

Why: the formula shifts L toward the heavier neighbour — that is all it does.

Example: Classes 10-20, 20-30, 30-40, 40-50, 50-60 have frequencies 5, 8, 12, 6, 3. The mode is:

Modal class 30-40: $30 + \frac{12-8}{24-8-6} \times 10 = 30 + 4 = 34$ .

Type 3: Missing frequency from the mediancommon2 practice Q

How to spot it:

One frequency unknown; the median of the grouped data given — find it.

Assume the median class from the known structure, reverse the formula: $n/2 - cf_{\text{prev}} = \frac{f(\text{median} - L)}{h}$, solve for the unknown.

Write the cf column in terms of the unknown.
Force the median to fall in the plausible class; substitute.
Solve the linear equation; verify the class assumption holds.

Why: the median formula inverted is one linear equation — nothing deeper.

Example: The median of grouped data with classes 0-20 (f=8), 20-40 (f=x), 40-60 (f=10), 60-80 (f=6) is 38. Find x ( $n = 24+x$ ).

$n/2 = 12 + \frac{x}{2}$ ; median class 20-40: $20 + \frac{12 + \frac{x}{2} - 8}{x} \times 20 = 38 \Rightarrow \frac{4 + \frac{x}{2}}{x} = \frac{18}{20} \Rightarrow 80 + 10x = 18x \Rightarrow x = 10$ .

Type 4: Grouped mean via midpointsoccasional3 practice Q

How to spot it:

A small frequency table; the mean asked (or a missing frequency via the mean).

$\bar{x} = \frac{\sum f_i m_i}{\sum f_i}$; midpoint $m_i = \frac{L_i + U_i}{2}$.

Write midpoints and multiply by frequencies.
Sum the products; divide by total frequency.
For inclusive classes, convert to exclusive first if required.

Why: the grouped mean treats each class as concentrated at its midpoint — the exam tables are small enough for direct work.

Example: Classes 0-10, 10-20, 20-30 have frequencies 5, 8, 7. The mean (via midpoints) is:

$\frac{5 \times 5 + 8 \times 15 + 7 \times 25}{20} = \frac{330}{20} = 16.5$ .

Formulas

Grouped median

\text{median}=L+\frac{\frac{n}{2}-cf}{f}\,h

Grouped mode

\text{mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\,h

Cumulative frequency

cf_i=f_1+f_2+\cdots+f_i

Grouped mean (midpoints)

\bar{x}=\frac{\sum f_i m_i}{\sum f_i},\quad m_i=\frac{L_i+U_i}{2}

Shortcut tricks

⚡ Find the median class by cf, not by eye

Build the cf column, stop at the first cf $\ge n/2$ . That row's lower limit is $L$ . Everything else is substitution.

Example: Classes 0-10, 10-20, 20-30, 30-40, 40-50 have frequencies 4, 6, 10, 8, 2. Find the median.

$n=30$ , $n/2=15$ ; cf reaches 20 at 20-30. $L=20$ , $cf=10$ , $f=10$ , $h=10$ : median $=20+\frac{5}{10}\times10=25$ .

⚡ Modal class first, formula second

The modal class is just the tallest bar. If it sits at an edge, the missing neighbour counts 0 in the formula.

Example: Classes 10-20, 20-30, 30-40, 40-50, 50-60 have frequencies 5, 8, 12, 6, 3. Find the mode.

Modal class 30-40: $30+\frac{12-8}{24-8-6}\times10=30+\frac{4}{10}\times10=34$ .

Where students lose marks

Using $n/2$ instead of $(n+1)/2$ in the grouped median — grouped data always uses $n/2$ .
Taking $cf$ as the cumulative frequency including the median class (it must exclude it).
Reading class limits as inclusive when the data is exclusive (10-20 vs 10-19.5).

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

Suggested time 11 min · wrong answers go to your mistake notebook automatically.