MathsICSE Class 9 · Statistics

Mean and Median of Ungrouped Data with Solved Problems (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

The mean is the sum of the observations divided by their number. The median is the middle value of the data arranged in order: for n odd it is the ((n + 1)/2)th value; for n even it is the average of the (n/2)th and (n/2 + 1)th values. Use sum = mean × n to solve for missing or replaced observations.

After sorting, with six values the median is the average of the 3rd and 4th values: (11 + 12) ÷ 2 = 11.5.

Illustration drawn for QuestionBuddy.

Understand Mean and Median of Ungrouped Data

Mean. For observations $x_{1}, x_{2}, \dots, x_{n}$ :
$ $\overset{x}{ˉ} = \frac{x _{1} + x _{2} + \dots + x _{n}}{n} .$ $
Rearranged: $sum = n \overset{x}{ˉ}$ . This is the key to most problems: *total* is what you can add, subtract, replace.

Median. Arrange the data in ascending (or descending) order.

$n$ odd: median $=$ the $\frac{n + 1}{2}$ -th value.
$n$ even: median $=$ average of the $\frac{n}{2}$ -th and $(\frac{n}{2} + 1)$ -th values.

*Example.* $12, 7, 15, 9, 20, 11$ in order: $7, 9, 11, 12, 15, 20$ ; $n = 6$ ; median $= \frac{11 + 12}{2} = 11.5$ .

Mode and range. The mode is the most frequent value (there can be more than one); range $=$ largest $-$ smallest.

Typical problems.

*Replaced observation:* if the mean of $n$ numbers is $20$ and replacing $15$ by $35$ makes it $21$ , then the total increased by $20$ and the mean by $1$ , so $n = 20$ .
*Excluding a value:* mean of 6 numbers is $8$ (total $48$ ); removing $5$ leaves $43$ for 5 numbers, mean $8.6$ .
*Overlapping groups:* mean of 5 numbers is $12$ (total $60$ ); first three have mean $10$ (total $30$ ); last three have mean $14$ (total $42$ ). The third number is counted twice: $30 + 42 - 60 = 12$ .
*Median with unknown:* for $3, 5, x + 1, 2 x - 1, 11, 13, 15$ in ascending order, the median is the 4th value $2 x - 1$ ; if the median is $9$ then $x = 5$ .
*Effect of adding a constant:* adding $k$ to every observation adds $k$ to the mean and the median; multiplying by $k$ multiplies both by $k$ .

Level. Class IX is a school-exam class. The overlapping-group and replaced-observation questions are at the hardest school-exam level.

Mean and Median of Ungrouped Data: formulas to remember

Mean: $\overset{x}{ˉ} = \frac{\sum x}{n}$; Sum divided by number.
Total: $\sum x = n \overset{x}{ˉ}$; Use for missing values.
Median, n odd: $value at position \frac{n + 1}{2}$; After sorting.
Median, n even: $\frac{x _{n /2} + x _{n /2 + 1}}{2}$; After sorting.
Range: $largest - smallest$; Spread.

How to solve Mean and Median of Ungrouped Data questions

Sort for the median, add for the mean, and use total = mean × n for missing or replaced values.

1Write the data; count $n$ .
2For the mean, add all values and divide by $n$ .
3For the median, sort in order and pick the middle (or average the two middles).
4For a missing or replaced value, work with totals: new total = old total ± change.
5Check that the mean lies between the smallest and largest values.
6Give units and, if needed, decimals.

Mean and Median of Ungrouped 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.

Daily rainfall

Suppose nine days of rainfall (mm) are 12, 7, 15, 9, 20, 11, 15, 8, 11 (hypothetical figures).

Find the mean, median, mode and range.

Try it first, then show the solutionSolution

Sum $= 108$ ; mean $= \frac{108}{9} = 12$ mm.
Sorted: $7, 8, 9, 11, 11, 12, 15, 15, 20$ ; the 5th value is $11$ , the median.
Modes: $11$ and $15$ (each twice). Range $= 20 - 7 = 13$ mm.

Answer: Mean 12 mm; median 11 mm; modes 11 and 15; range 13 mm.

Memory hook: Mean uses the total; median uses the order.

Class average after a correction

Imagine a teacher who finds that one score of 15 was recorded wrongly and should be 35, raising the class average from 20 to 21.

How many students are in the class?

Try it first, then show the solutionSolution

Total increased by $35 - 15 = 20$ .
The mean increased by $1$ , so $\frac{20}{n} = 1$ and $n = 20$ .

Answer: 20 students.

Memory hook: Change in total ÷ n = change in mean.

Tricks to remember Mean and Median of Ungrouped Data

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

Total is king
Mean × n gives the total you can edit.
Use it when: Replaced and missing values.
Sort first
Median needs ordered data.
Use it when: Median questions.
Middle of the middle
For even n, average the two middle values.
Use it when: Even-count medians.

Mean and Median of Ungrouped Data: solved examples

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

Example 1: Mean · 1 mark · MCQBoard level

The mean of $2, 4, 6, 8, 10$ is:
(a) 5
(b) 6
(c) 7
(d) 30

Try it first, then show the solutionSolution

$\frac{30}{5} = 6$ .

Answer: (b) 6

Example 2: Mean and median · 3 marks · Short answerBoard level

Find the mean and median of $8, 12, 15, 9, 11, 15, 10, 12, 9, 19$ .

Try it first, then show the solutionSolution

Sum $= 120$ ; $n = 10$ ; mean $= 12$ .
Sorted: $8, 9, 9, 10, 11, 12, 12, 15, 15, 19$ .
Median $= \frac{11 + 12}{2} = 11.5$ .

Answer: Mean 12; median 11.5

Example 3: A replaced observation · 3 marks · Short answerHarder board level

The mean of $n$ observations is $20$ . When one observation $15$ is replaced by $35$ , the mean becomes $21$ . Find $n$ and the new total.

Try it first, then show the solutionSolution

The total rises by $35 - 15 = 20$ .
The mean rises by $1$ , so $\frac{20}{n} = 1$ and $n = 20$ .
Old total $= 400$ , new total $= 420$ (and $21 \times 20 = 420$ ✓).

Answer: n = 20; new total 420

Example 4: Median with an unknown · 4 marks · Short answerHarder board level

The observations $3, 5, x + 1, 2 x - 1, 11, 13, 15$ are in ascending order and their median is $9$ . Find $x$ and verify the order. Find the range.

Try it first, then show the solutionSolution

$n = 7$ , so the median is the 4th value: $2 x - 1 = 9$ , $x = 5$ .
The data: $3, 5, 6, 9, 11, 13, 15$ — ascending. ✓
Range $= 15 - 3 = 12$ .

Answer: x = 5; range 12

Example 5: Overlapping groups · 4 marks · Short answerHOTS / competency

The mean of five numbers is $12$ . The mean of the first three is $10$ and of the last three is $14$ . Find the third number.

Try it first, then show the solutionSolution

Total of all five $= 60$ .
First three total $= 30$ ; last three total $= 42$ .
The third number is counted in both: $30 + 42 = 72 = 60 + third$ , so the third number is $12$ .

Answer: The third number is 12.

Where students lose marks in Mean and Median of Ungrouped Data

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

Mistake: Not sorting before the median
The median needs ordered data.
Wrong working
Median of 3, 1, 7, 5, 9 is the middle value, 7.
Right working
Sort: 1, 3, 5, 7, 9; median 5.
Do this: Always sort.
Usually costs: Whole answer.
Mistake: Dividing by the wrong count
n changes when values are removed.
Wrong working
$\frac{43}{6}$ .
Right working
$\frac{43}{5}$ .
Do this: Update n.
Usually costs: Whole answer.
Mistake: Averaging the wrong two values
For even n, take the n/2-th and n/2 + 1-th.
Wrong working
Average of 2nd and 3rd for 6 values.
Right working
Average of 3rd and 4th.
Do this: Count positions.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; show total = mean × n explicitly.
Write the sorted list on its own line for the median.
Check that the mean is between the smallest and largest value.
Quote units in context questions.

Practice questions on Mean and Median of Ungrouped 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 $3, 1, 7, 5, 9$ is:
(a) 3
(b) 5
(c) 7
(d) 25
Hint
Sort: 1, 3, 5, 7, 9.
Answer
(b) 5
Q2 · 3 marks · Short answerHarder board level
The mean of 6 numbers is $8$ . One number $5$ is removed. Find the mean of the remaining numbers.
Hint
Total 48.
Answer
$\frac{48 - 5}{5} = 8.6$ .
Q3 · 3 marks · Short answerHarder board level
Find the median of $12, 7, 15, 9, 20, 11$ .
Hint
Average the 3rd and 4th.
Answer
Sorted $7, 9, 11, 12, 15, 20$ ; median $= \frac{11 + 12}{2} = 11.5$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If 5 is added to every observation, the mean increases by 5.
Reason (R): The new total is the old total plus $5 n$ , and dividing by $n$ gives an increase of 5.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Try $1, 2, 3$ and then $6, 7, 8$ .
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Daily rainfall (hypothetical) over 9 days in mm: $12, 7, 15, 9, 20, 11, 15, 8, 11$ .
(a) Find the mean.
(b) Find the median.
(c) Find the range.
Hint
Sort the data for the median.
Answer
(a) 12 mm (b) 11 mm (c) 13 mm

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.Formula for the mean?Show answer
Sum of observations divided by their number.
2.Median for an even number of observations?Show answer
Average of the two middle values.
3.Range?Show answer
Largest minus smallest.
4.Effect of adding 5 to every observation on the median?Show answer
It increases by 5.

Mean and Median of Ungrouped Data: frequently asked questions

Can the mean be larger than every observation?

No, it lies between the smallest and largest.

Is mode in the syllabus?

Mean and median of ungrouped data are the listed measures; the mode can still be read off.

How do I handle weighted means?

Use totals: multiply each group mean by its size and add.