Statistics

🔒 Log in to track

high importance~3 Q in Tier 128 formulas⚡ 10 shortcuts5 subtopics

Overview Learn (5)Formulas (28)Examples (5)Practice (68)

All subtopics·Subtopic 3 of 5

Median and mode of grouped data

🔒 Log in to track

⏱ 5 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Grouped data comes as class intervals with frequencies: how many values fall in each band.

The single values are lost, so we estimate averages from the bands. Every method below uses a cumulative frequency column and the midpoint of each class.

Reading a frequency table

A grouped table only says how many values lie in each band, like 20-30 with frequency 10.

Two columns do all the work:

Midpoint $x$ : the middle of the band, $20$ - $30 \rightarrow 25$ .
Cumulative frequency (cf): a running total of frequencies down the table.

Tip: Build the cf column once. Median questions, quartile questions and distribution questions all read from it.

Mean by midpoints

Treat every value in a class as if it sat at the midpoint.

\bar{x} = \frac{\sum f x}{\sum f}

Classes 0-10 and 10-20 with frequencies 4 and 6: $\sum fx = 4 \times 5 + 6 \times 15 = 110$ , $\sum f = 10$ , mean $= 11$ .

Watch: Midpoints, not class limits. Using 10 for the class 10-20 (instead of 15) is the classic slip.

Median from the cumulative column

Half the values lie below the median, so find the class whose cumulative frequency first reaches $\dfrac{n}{2}$ .

\text{Median} = L + \frac{\frac{n}{2} - c}{f} \times h

$L$ is the lower limit of that class, $c$ the cumulative frequency before it, $f$ its frequency, $h$ its width.

Table: classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 4, 6, 10, 8, 2. Then $n = 30$ , $\dfrac{n}{2} = 15$ . The cf column runs 4, 10, 20, 28, 30; the first cf to reach 15 is 20, so the median class is 20-30.

\text{Median} = 20 + \frac{15 - 10}{10} \times 10 = 25

Rule: The median class is where cf crosses $n/2$ , never simply the biggest frequency.

Mode from the tallest bar

The modal class has the highest frequency. Then interpolate inside it:

\text{Mode} = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h

$f_m$ is the modal class frequency, $f_1$ and $f_2$ the frequencies just before and after.

Same classes with frequencies 5, 8, 12, 6, 3: modal class 20-30, so $L = 20$ , $f_m = 12$ , $f_1 = 8$ , $f_2 = 6$ , $h = 10$ :

\text{Mode} = 20 + \frac{12 - 8}{2 \times 12 - 8 - 6} \times 10 = 20 + 4 = 24

Tip: Equal neighbours ( $f_1 = f_2$ ) put the mode exactly at $L + h/2$ .

A missing frequency from the median

Give the unknown a letter, extend the cf column through it, and apply the median rule.

Classes 10-20, 20-30, 30-40, 40-50 with frequencies 7, 8, $x$ , 12 and median 32. Then $n = 27 + x$ , $\dfrac{n}{2}$ must land in 30-40, where $c = 15$ :

30 + \frac{\frac{27+x}{2} - 15}{x} \times 10 = 32 \Rightarrow x = 5

A missing frequency from the mean

The mean fixes $\sum fx$ . Let the unknown be $f$ and expand both sides.

Classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 5, $f$ , 11, 9, 5 and mean 25:

\frac{840 + 15f}{30 + f} = 25 \Rightarrow 840 + 15f = 750 + 25f \Rightarrow f = 9

Watch: Keep the midpoint of the missing class in the numerator; students often drop it along with the frequency.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Median of grouped data

How to spot it:

A frequency table with equal class widths; the median is asked.

\text{Med} = L + \frac{\frac{n}{2} - c}{f} \times h

Method

Total the frequencies to get n.
Build the cf column and find where it crosses $\frac{n}{2}$ .
Read $L$ , $c$ , $f$ , $h$ from that class and substitute.

Why it works:

Values spread evenly inside the median class, so we walk in proportion.

Try this

Find the median:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	4	6	10	8	2

Show solution

$n = 30$ , so $n/2 = 15$ .
cf: 4, 10, 20, 28, 30. Median class 20-30: $L=20$ , $c=10$ , $f=10$ , $h=10$ .
Median $= 20 + \dfrac{15-10}{10} \times 10 = 25$ .

Answer

Type 2common2 practice Q

Mode of grouped data

How to spot it:

A frequency table; the most frequent class is obvious and the exact mode is asked.

\text{Mode} = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h

Method

Spot the class with the highest frequency.
Note its neighbours' frequencies.
Substitute into the mode formula.

Why it works:

The mode sits inside the tallest bar, pulled toward the taller neighbour.

Try this

Find the mode:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	12	6	3

Show solution

Modal class 20-30: $L = 20$ , $f_m = 12$ , $f_1 = 8$ , $f_2 = 6$ .
Denominator $= 2 \times 12 - 8 - 6 = 10$ .
Mode $= 20 + \dfrac{4}{10} \times 10 = 24$ .

Answer

Type 3occasional2 practice Q

Missing frequency from the median

How to spot it:

One frequency in the table is x or unknown; the median is given.

L + \frac{\frac{n}{2} - c}{f} \times h = \text{Med}

Method

Write n including the unknown.
Extend the cf column through the unknown class.
Apply the median formula and solve.

Why it works:

The stated median names the class, and the formula turns it into an equation.

Try this

The median of the following data is 32. Find x.

Class	10-20	20-30	30-40	40-50
Frequency	7	8	x	12

Show solution

$n = 27 + x$ . Median 32 lies in 30-40: $L = 30$ , $c = 15$ , $f = x$ .
$30 + \dfrac{(27+x)/2 - 15}{x} \times 10 = 32$ .
Solving gives $x = 5$ .

Answer

x = 5

Type 4very common3 practice Q

Mean of grouped data

How to spot it:

A frequency table (or a midpoints row) is given; the mean is asked.

\bar{x} = \frac{\sum f x}{\sum f}

Method

Write the midpoint of each class.
Multiply midpoint by frequency and total them.
Divide by the total frequency.

Why it works:

Each class acts like its whole mass sitting at its midpoint.

Try this

Find the mean:

Midpoint	5	15	25	35
Frequency	4	6	7	3

Show solution

$\sum fx = 20 + 90 + 175 + 105 = 390$ .
$\sum f = 20$ .
Mean $= 390 \div 20 = 19.5$ .

Answer

19.5

Type 5occasional

Missing frequency from the mean

How to spot it:

One frequency is unknown; the mean of the distribution is given.

\frac{\sum f x}{\sum f} = \bar{x}

Method

Compute the known part of $\sum fx$ and of $n$ .
Add the unknown times its midpoint (and once to $n$ ).
Set the ratio equal to the mean and solve.

Why it works:

The mean fixes the total, and the unknown appears in it linearly.

Try this

The mean of the following distribution is 25. Find f.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	f	11	9	5

Show solution

Known $\sum fx = 25 + 275 + 315 + 225 = 840$ ; plus $15f$ .
$n = 30 + f$ .
$\dfrac{840 + 15f}{30 + f} = 25 \Rightarrow 840 + 15f = 750 + 25f \Rightarrow f = 9$ .

Answer

f = 9

Formula sheet

Grouped mean

\bar{x} = \frac{\sum f x}{\sum f}

x is the midpoint of each class.

Grouped median

\text{Med} = L + \frac{\frac{n}{2} - c}{f} \times h

L: lower limit of median class; c: cf before it; f: its frequency; h: width.

Grouped mode

\text{Mode} = L + \frac{f_m - f_1}{2f_m - f_1 - f_2} \times h

f_m: modal class frequency; f_1, f_2: neighbouring frequencies.

Shortcuts that save time

⚡ Build the cf column once, use it thrice

Median, quartiles and 'how many below a value' questions all read the same cumulative column. Write it before touching any formula.

Example

Frequencies 4, 6, 10, 8, 2 for classes 0-10 to 40-50. How many values are below 30?

Show solution

cf column: 4, 10, 20, 28, 30.
Values below 30 sit in the first two classes.
$4 + 6 = 10$ .

Answer

⚡ Cross-check the median class fast

Half of n must fall inside the median class. Glance: the cf before it is below n/2, the cf through it is at or above n/2.

Example

Frequencies 5, 8, 12, 6, 3 for classes 0-10 to 40-50. Which class holds the median?

Show solution

$n = 34$ , so $n/2 = 17$ .
cf runs 5, 13, 25, 31, 34.
17 first crossed in 20-30.

Answer

20-30

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using class limits instead of midpoints for the mean.

Midpoint of 10-20 is 15.

Mistake 02

Picking the class with the highest frequency as the median class.

The median class is where the cumulative frequency crosses n/2.

Mistake 03

Taking c as the cumulative frequency of the median class.

c is the cumulative frequency of the class before it.

Mistake 04

Mixing widths: classes of 10 used with h = 5.

Read h from the actual class width, including any gap.

Mistake 05

Solving a missing-frequency mean without the missing midpoint.

The unknown frequency multiplies its class midpoint in the numerator.

Quick revision

Read this the night before the exam.

Grouped mean $= \dfrac{\sum fx}{\sum f}$ , $x$ = midpoint.
Median $= L + \dfrac{n/2 - c}{f} \times h$ ; median class where cf crosses $n/2$ .
Mode $= L + \dfrac{f_m - f_1}{2f_m - f_1 - f_2} \times h$ .
cf column first; it answers three question types.
Missing frequency: letter in, solve with the given average.
Check the median class: cf below n/2 before, at or above after.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

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

Continue learning