MathsICSE Class 10 · Measures of Central Tendency

Median of Raw and Arrayed Data, and the Median Class (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The median is the middle value of data arranged in order. For $n$ observations it is the $\left(\frac{n + 1}{2}\right)$th value when $n$ is odd, and the average of the $\frac{n}{2}$th and $\left(\frac{n}{2} + 1\right)$th values when $n$ is even. For a frequency table, locate these positions with cumulative frequencies.

Order the data first: odd count gives the middle value, even count gives the mean of the two middle values, which may not be in the data.

Illustration drawn for QuestionBuddy.

Understand Median of Raw and Arrayed Data

The CISCE 2027 syllabus asks for the median of raw and arrayed data, and the median class of grouped data. (The median of grouped data itself is read from an ogive, in the next chapter.)

Raw data. Arrange the observations in ascending (or descending) order first. With $n$ observations:

$n$ odd: median $=$ the $(\frac{n + 1}{2})$ th observation.
$n$ even: median $= \frac{1}{2} [\frac{n}{2} th + (\frac{n}{2} + 1) th]$ . The median need not be one of the data.

Unknowns in ordered data. If the data are already in order and the median is given, write the median in terms of the unknown(s) and solve. With 10 observations the median is the average of the 5th and 6th, so if these are $x$ and $x + 4$ , then $\frac{2 x + 4}{2} =$ median (2024 Q1(v)). Check that your answer keeps the data in order.

Arrayed data (a frequency table of single values). Here $N = Σ f$ observations are already in order of $x$ .

1. Add a cumulative frequency (c.f.) column.
2. Find the median position(s): $\frac{N + 1}{2}$ if $N$ is odd; $\frac{N}{2}$ and $\frac{N}{2} + 1$ if $N$ is even.
3. The median is the $x$ whose c.f. first reaches or passes each position. If the two middle positions fall on different values, average them.

Do not take the middle frequency, or the average of the frequencies: the median is an $x$ value, never an $f$ value (2025 Q8(ii)).

Median class (grouped data). With total frequency $N$ , the median class is the class containing the $\frac{N}{2}$ th observation: the first class whose c.f. is at least $\frac{N}{2}$ .

Why the median is useful. It depends only on the middle position, so one very large or very small value does not move it. That makes it a better 'typical value' than the mean for data like house prices or incomes.

Median of Raw and Arrayed Data: formulas to remember

Odd number of observations: $median = (\frac{n + 1}{2}) th value$; Data in order.
Even number of observations: $median = \frac{1}{2} [\frac{n}{2} th + (\frac{n}{2} + 1) th]$; Average of the two middle values.
Arrayed data: $N = Σ f; locate the middle position(s) in the c.f. column$; The answer is an $x$ value.
Median class: $first class with c.f. \geq \frac{N}{2}$; For grouped data.

How to solve Median of Raw and Arrayed Data questions

Order, count, locate: the median is about positions, so get the position right before looking for the value.

1Raw data: arrange in ascending order and count $n$ .
2Decide odd or even, and write the middle position(s): $\frac{n + 1}{2}$ , or $\frac{n}{2}$ and $\frac{n}{2} + 1$ .
3Arrayed data: write the c.f. column; $N$ is the last c.f.
4Read the $x$ value at each middle position (the first c.f. that reaches it); average two values if needed.
5For an unknown, write the median expression in the unknown, solve, and check the order still holds.
6For grouped data, give the median class (the first class with c.f. $\geq \frac{N}{2}$ ).

Median of Raw and Arrayed Data in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Median versus mean for house prices

Suppose the flats on your lane sold for ₹42, 45, 47, 50, 52 and 55 lakh, and one penthouse for ₹210 lakh.

Find the mean and the median price. Which better describes a typical flat?

Try it first, then show the solutionSolution

Mean $= \frac{501}{7} \approx 71.57$ , that is, about ₹71.57 lakh.
Seven values in order: the median is the 4th, ₹50 lakh.
Six of the seven flats cost less than the mean; the median is more typical.

Answer: Mean ≈ ₹71.57 lakh, median ₹50 lakh; the median

Memory hook: One giant value drags the mean, never the median.

Median of arrayed data

Imagine a clinic recording household sizes for 40 visiting families: 2 people (8 families), 3 (14), 4 (11), 5 (5), 6 (2).

What is the median household size?

Try it first, then show the solutionSolution

$N = 40$ : average of the 20th and 21st values.
c.f.: 8, 22, 33, 38, 40. Both the 20th and 21st lie in the size-3 group (c.f. 22).
Median $= 3$ .

Answer: 3 people

Memory hook: c.f. finds the middle family.

Tricks to remember Median of Raw and Arrayed Data

Short memory aids. Say each one aloud once, then use it on a question the same day.

Line them up first
No median without order: sort before you count.
Use it when: Raw data questions.
Even means two
An even count has two middle values: add them and halve.
Use it when: Avoiding the 2024 error of equating one middle term to the median.
c.f. points at x, not f
The cumulative frequency tells you where the middle is; the answer is always read from the $x$ column.
Use it when: Arrayed data and the median class.

Median of Raw and Arrayed Data: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Even number of raw observations · 1 mark · MCQBoard level

The median of 14, 9, 21, 6, 17, 11 is:
(a) 11
(b) 12.5
(c) 14
(d) 15.5

Try it first, then show the solutionSolution

In order: 6, 9, 11, 14, 17, 21.
Six values: median $= \frac{11 + 14}{2} = 12.5$ .

Answer: (b)

Example 2: Unknown in ordered data · 3 marks · Short answerBoard level

The observations 18, 23, 27, $x$ , $x + 6$ , 41, 45, 52 are in ascending order, and their median is 33. Find $x$ and the mean of the observations.

Try it first, then show the solutionSolution

Eight observations: median $= \frac{x + ( x + 6 )}{2} = x + 3$ .
$x + 3 = 33$ , so $x = 30$ . Check order: $27 \leq 30$ and $36 \leq 41$ ✓
Sum $= 18 + 23 + 27 + 30 + 36 + 41 + 45 + 52 = 272$ ; mean $= \frac{272}{8} = 34$ .

Answer: $x = 30$ ; mean 34

Example 3: Arrayed data with an odd total · 3 marks · Short answerHarder board level

Find the median and the mean (to 2 decimal places) of the following distribution.

$x$	4	5	6	7	8	9
$f$	2	6	7	10	4	2

Try it first, then show the solutionSolution

c.f.: 2, 8, 15, 25, 29, 31. $N = 31$ (odd), so the median is the 16th value.
The c.f. first reaches 16 at $x = 7$ (c.f. 25): median $= 7$ .
$Σ f x = 8 + 30 + 42 + 70 + 32 + 18 = 200$ ; mean $= \frac{200}{31} \approx 6.45$ .

Answer: Median 7; mean ≈ 6.45

Example 4: Middle positions on different values · 3 marks · Short answerHarder board level

The goals scored by a team in 40 matches are shown. Find the median number of goals.

Goals	0	1	2	3	4	5
Matches	5	9	6	8	7	5

Try it first, then show the solutionSolution

c.f.: 5, 14, 20, 28, 35, 40. $N = 40$ (even): average of the 20th and 21st values.
The 20th value is 2 (c.f. reaches 20 at 2 goals); the 21st is 3 (c.f. 28).
Median $= \frac{2 + 3}{2} = 2.5$ goals.

Answer: 2.5 goals

Example 5: Median class of grouped data · 2 marks · Short answerHarder board level

The masses of 50 parcels are shown. Find the median class.

Mass (g)	100–120	120–140	140–160	160–180	180–200
Parcels	7	12	18	9	4

Try it first, then show the solutionSolution

$N = 50$ , so $\frac{N}{2} = 25$ .
c.f.: 7, 19, 37, 46, 50. The first c.f. $\geq 25$ is 37.
Median class: 140–160 g.

Answer: 140–160 g

Example 6: Case study: an unknown frequency · 4 marks · Case-basedHOTS / competency

A teacher records the marks (out of 50) of a class in a quiz.

Marks	10	20	30	40	50
Students	4	6	$k$	5	3

(a) If $k = 2$ , find the median.
(b) Find the least value of $k$ for which the median is 30.
(c) Explain why every larger value of $k$ also gives a median of 30.

Try it first, then show the solutionSolution

(a) $k = 2$ : $N = 20$ , c.f. 4, 10, 12, 17, 20. The 10th value is 20 and the 11th is 30, so the median is $\frac{20 + 30}{2} = 25$ .
(b) Ten students scored below 30, so the middle position(s) must be at least 11. $N = 18 + k$ . If $k = 3$ , $N = 21$ and the median is the 11th value, which is 30 ✓. $k = 2$ fails by (a), and $k = 0$ or 1 puts the middle position at or below 10. Least $k = 3$ .
(c) As $k$ grows, the middle position $\frac{N + 1}{2}$ grows only half as fast as the c.f. at 30 ( $10 + k$ ), so it stays inside the 30 group (and above 10).

Answer: (a) 25 (b) $k = 3$

Where students lose marks in Median of Raw and Arrayed Data

The slips examiners see most often, and what to write instead.

Mistake: Equating one middle term to the median
With an even count, the median is the average of two middle terms.
Wrong working
Median 64 with middle terms $x$ and $x + 4$ : $x + 4 = 64$
Right working
$\frac{x + x + 4}{2} = 64$ , so $x = 62$
Do this: Count the observations and write both middle positions.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q1(v): 1 mark.
Mistake: Averaging the frequencies
The median is a value of $x$ , found by position.
Wrong working
Median $= \frac{Σ f}{number of classes}$
Right working
c.f. column, then read $x$ at the middle position
Do this: Always write the c.f. column for arrayed data.
Usually costs: Reported in the CISCE examiners' comments on 2025 Q8(ii).
Mistake: No cumulative frequency shown
Without the c.f. column the examiner cannot see how the position was found.
Wrong working
Median = 13 with no working
Right working
c.f. 3, 5, 7, 13, … ; 12th and 13th values are 13
Do this: Show the c.f. column and the position(s).
Usually costs: Reported in the CISCE examiners' comments on 2025 Q8(ii): method marks.
Mistake: Forgetting to order raw data
The middle of an unordered list is not the median.
Wrong working
Median of 14, 9, 21, 6, 17, 11 as $\frac{21 + 6}{2}$
Right working
Order first: $\frac{11 + 14}{2} = 12.5$
Do this: Rewrite the data in order every time.
Usually costs: The whole question.

Exam tips

Asked in 2024 Q1(v) (1 mark): ten ordered observations with middle terms $x$ and $x + 4$ and median 64; the answer was $x = 62$ .
Asked in 2025 Q8(ii) (3 marks): mean to the nearest whole number and median of a discrete table with $N = 24$ ; the expected answers were mean $13.2 \approx 13$ and median 13.
When the mean is asked 'to the nearest whole number', write the decimal first, then the rounded value.
The CISCE examiners suggest practising the median for raw, discrete, grouped and continuous data.

Practice questions on Median of Raw and Arrayed Data

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The median of the first 10 natural numbers is:
(a) 5
(b) 5.5
(c) 6
(d) 10
Hint
1 to 10; average the 5th and 6th.
Answer
(b)
Q2 · 2 marks · Short answerBoard level
Find the median of 31, 26, 40, 35, 28, 44, 37. If 50 is added to the data, find the new median.
Hint
Order first; the new set has 8 values.
Answer
In order 26, 28, 31, 35, 37, 40, 44: median 35. With 50: average of 35 and 37 $= 36$ .
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The median of 3, 8, 8, 10, 15 does not change if 15 is replaced by 150.
Reason (R): The median depends only on the value(s) in the middle position(s) of the ordered data.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
The 3rd value is still 8.
Answer
(c)
Q4 · 3 marks · Short answerHarder board level
For the distribution below, find (a) the mean, to the nearest whole number, (b) the median.
$x$ 11 12 13 14 15 16
$f$ 4 7 10 9 6 4
Hint
$N = 40$ : 20th and 21st values.
Answer
(a) $Σ f x = 538$ , mean $= 13.45 \approx 13$ . (b) c.f. 4, 11, 21, …; the 20th and 21st values are both 13, so median $= 13$ .
Q5 · 3 marks · Short answerHOTS / competency
The ten observations 5, 8, 12, 15, $x$ , $x + 3$ , 24, 27, 30, 33 are in ascending order and their median is 19.5. Find $x$ . If 33 is then replaced by 13, find the new median.
Hint
Re-order after the replacement.
Answer
$\frac{2 x + 3}{2} = 19.5$ , so $x = 18$ . New ordered data 5, 8, 12, 13, 15, 18, 21, 24, 27, 30: median $= \frac{15 + 18}{2} = 16.5$ .
Q6 · 2 marks · Short answerHarder board level
Find the median class.
Class 0–10 10–20 20–30 30–40 40–50 50–60
Frequency 5 9 14 12 7 3
Hint
$\frac{N}{2} = 25$ .
Answer
c.f. 5, 14, 28, …; first c.f. $\geq 25$ is 28, so the median class is 20–30.
Q7 · 2 marks · Short answerHOTS / competency
The median of five observations is 20. Each observation is multiplied by 3 and then 4 is subtracted. Find the new median, and explain why the order of the observations does not change.
Hint
$y = 3 x - 4$ is increasing.
Answer
$y = 3 x - 4$ keeps larger values larger, so the middle observation stays in the middle: new median $= 3 \times 20 - 4 = 56$ .

$x$	11	12	13	14	15	16
$f$	4	7	10	9	6	4

Class	0–10	10–20	20–30	30–40	40–50	50–60
Frequency	5	9	14	12	7	3

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Median position for $n = 15$ ?Show answer
8th value
2.Median positions for $n = 24$ ?Show answer
12th and 13th; average them
3.In arrayed data, which column gives the answer?Show answer
The $x$ column, located through the c.f.
4.Median class?Show answer
The first class whose c.f. is at least $\frac{N}{2}$ .
5.Does an extreme value change the median?Show answer
No, only the middle position(s) matter.

Median of Raw and Arrayed Data: frequently asked questions

How do I find the median when the number of observations is even?

Arrange the data in order and take the average of the two middle values, the (n/2)th and (n/2 + 1)th observations. The answer need not be one of the data.

How do I find the median of a frequency table?

Write a cumulative frequency column, find the middle position(s) from N = Σf, and read the x value where the cumulative frequency first reaches each position.

What is the median class?

For grouped data, it is the class containing the (N/2)th observation, that is, the first class whose cumulative frequency is at least N/2.