Measures of Central Tendency ICSE Class 10: Mean, Median and Mode

What Measures of Central Tendency is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Measures of Central Tendency is in the 6. Statistics (a) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Mean, median and mode of raw data

For $n$ observations:

• Mean $\bar{x}=\frac{\text{sum of observations}}{n}$.
• Median: arrange in order. If $n$ is odd, it is the $\left(\frac{n+1}{2}\right)$th value. If $n$ is even, it is the average of the $\frac{n}{2}$th and $\left(\frac{n}{2}+1\right)$th values.
• Mode: the value that occurs most often (there may be more than one, or none).

Worked example

Find the mean, median and mode of 12, 15, 11, 15, 18, 20, 15, 14, 11.

Try it first, then show the solutionSolution

1. Sum $=131$, $n=9$: mean $=\frac{131}{9}\approx 14.56$.
2. In order: 11, 11, 12, 14, 15, 15, 15, 18, 20. Median is the 5th value: 15.
3. 15 occurs three times: mode $=15$.

Answer: Mean ≈ 14.56, median 15, mode 15

Median with an even number of observations

With $n$ even, there are two middle values; the median is their average, and it need not be one of the data values.

If the data contain an unknown and the median is given, write the median as the average of the two middle terms and solve.

Worked example

The observations 10, 13, 15, $x$, $x+2$, 22, 25, 28 are in ascending order and their median is 18. Find $x$.

Try it first, then show the solutionSolution

1. $n=8$: median $=\frac{4\text{th}+5\text{th}}{2}=\frac{x+x+2}{2}$.
2. $x+1=18$, so $x=17$.

Answer: x = 17

Arrayed data (frequency tables)

For values $x$ with frequencies $f$ (total $N=\Sigma f$):

• Mean $=\frac{\Sigma fx}{\Sigma f}$.
• Median: make a cumulative frequency (c.f.) column. Find the position $\frac{N+1}{2}$ (odd $N$), or the $\frac{N}{2}$th and $\left(\frac{N}{2}+1\right)$th (even $N$). The first value whose c.f. reaches that position is the median.
• Mode: the value with the highest frequency.

Do not group discrete data into classes, and do not average the frequencies.

Worked example

Find the mean, median and mode.

$x$	4	5	6	7	8
$f$	3	5	9	6	2

Try it first, then show the solutionSolution

1. $\Sigma f=25$, $\Sigma fx=12+25+54+42+16=149$: mean $=5.96$.
2. c.f.: 3, 8, 17, 23, 25. The 13th value lies where c.f. first reaches 13, at $x=6$: median $=6$.
3. Highest frequency 9 at $x=6$: mode $=6$.

Answer: Mean 5.96, median 6, mode 6

Grouped mean: direct method

For grouped data, each class is represented by its class mark (mid-value) $x=\frac{\text{lower limit}+\text{upper limit}}{2}$. Then
$\bar{x}=\frac{\Sigma fx}{\Sigma f}.$
This is an estimate, because it assumes every value sits at the centre of its class.

Worked example

Find the mean by the direct method.

Class	0–10	10–20	20–30	30–40	40–50
$f$	5	8	12	9	6

Try it first, then show the solutionSolution

1. Class marks: 5, 15, 25, 35, 45; $\Sigma f=40$.
2. $\Sigma fx=25+120+300+315+270=1030$.
3. Mean $=\frac{1030}{40}=25.75$.

Answer: 25.75

Grouped mean: short-cut (assumed mean) method

Choose an assumed mean $A$ from the class marks (usually a middle one). Let $d=x-A$. Then
$\bar{x}=A+\frac{\Sigma fd}{\Sigma f}.$
The numbers $d$ are smaller than $x$, so the products are easier.

Worked example

Find the mean of the same data (classes 0–10 to 40–50, $f$ = 5, 8, 12, 9, 6) by the short-cut method.

Try it first, then show the solutionSolution

1. Take $A=25$; $d=-20,-10,0,10,20$.
2. $\Sigma fd=-100-80+0+90+120=30$.
3. Mean $=25+\frac{30}{40}=25.75$.

Answer: 25.75 (same as the direct method)

Grouped mean: step-deviation method

When all classes have the same width $i$, divide the deviations by $i$: $t=\frac{x-A}{i}$. Then
$\bar{x}=A+\frac{\Sigma ft}{\Sigma f}\times i.$
The values of $t$ are small whole numbers ($-2,-1,0,1,2,\dots$), so this is the quickest method. If a question names the method, you must use that method.

Worked example

Find the mean of the same data by the step-deviation method.

Try it first, then show the solutionSolution

1. $A=25$, $i=10$; $t=-2,-1,0,1,2$.
2. $\Sigma ft=-10-8+0+9+12=3$.
3. Mean $=25+\frac{3}{40}\times 10=25.75$.

Answer: 25.75

Median class, modal class and discontinuous classes

• Modal class: the class with the highest frequency.
• Median class: the class containing the $\frac{N}{2}$th observation; find it from the cumulative frequencies.
• Discontinuous (inclusive) classes such as 10–19, 20–29: the class mark is still the average of the limits (14.5, 24.5, …). For boundaries, subtract 0.5 from each lower limit and add 0.5 to each upper limit (9.5–19.5, 19.5–29.5, …).

Worked example

For the table with classes 0–10 to 40–50 and $f$ = 5, 8, 12, 9, 6, find the median class and the modal class.

Try it first, then show the solutionSolution

1. $N=40$, $\frac{N}{2}=20$. c.f.: 5, 13, 25, 34, 40. The 20th observation is in 20–30.
2. Highest frequency 12 is also in 20–30.

Answer: Median class 20–30; modal class 20–30

Maths in real life: Measures of Central Tendency

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.

Topics in this chapter

• Mean of Grouped Data
• Median of Raw and Arrayed Data
• Mode and Modal Class

Measures of Central Tendency formula sheet

Mean of raw data

$\bar{x}=\frac{\Sigma x}{n}$

n = number of observations.

Median of raw data

$\text{odd }n:\left(\frac{n+1}{2}\right)\text{th};\text{even }n:\text{average of }\frac{n}{2}\text{th and }\left(\frac{n}{2}+1\right)\text{th}$

Arrange in order first.

Direct method

$\bar{x}=\frac{\Sigma fx}{\Sigma f}$

x = value or class mark.

Short-cut method

$\bar{x}=A+\frac{\Sigma fd}{\Sigma f},d=x-A$

A = assumed mean, a class mark.

Step-deviation method

$\bar{x}=A+\frac{\Sigma ft}{\Sigma f}\times i,t=\frac{x-A}{i}$

i = common class width.

Class mark

$x=\frac{\text{lower limit}+\text{upper limit}}{2}$

Same for inclusive classes like 10–19 (14.5).

How the board asks Measures of Central Tendency

The question patterns to expect, and the method that works for each.

Tricks to remember Measures of Central Tendency

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

• Mean, Median, Mode = Add, Middle, Most

  Add and divide; the middle one; the most frequent.

  Use it when: Keeping the three averages apart.

• A from the middle class mark

  Pick A as a class mark near the middle with a large frequency: the t-values become −2, −1, 0, 1, 2.

  Use it when: Fast step-deviation working.

• Even count, two in the middle

  Even number of observations: average the two middle ones.

  Use it when: Median questions with unknowns.

• c.f. finds the middle

  For any frequency table, the median comes from the cumulative frequency column.

  Use it when: Median of arrayed data; median class.

Measures of Central Tendency: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Measures of Central Tendency

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

• ✗Mistake: Median of an even number of observations

  CISCE examiners' comments, 2024: in an MCQ with ten ordered observations, candidates equated x + 4 to the median instead of averaging the two middle terms.

  Wrong working

  x + 4 = 64

  Right working

  $\frac{x+(x+4)}{2}=64$, so $x=62$

  ✓Do this: Even count: average the two middle values.

  Usually costs: 1 mark.

• ✗Mistake: Not using the method asked

  CISCE examiners' comments, 2024: candidates did not use the step-deviation method as asked, chose an assumed mean that was not a class mark, and took the class width as 5 instead of 10.

  ✓Do this: Read the method in the question; take A from the class marks; i is the class width (upper − lower limit).

  Usually costs: Method marks, even if the mean is right.

• ✗Mistake: Counting 0 as a natural number

  CISCE examiners' comments, 2025: for the mean of the first 9 natural numbers, candidates divided 45 by 10.

  ✓Do this: Natural numbers start at 1; the first 9 are 1 to 9, so the mean is 45 ÷ 9 = 5.

  Usually costs: 1 mark.

• ✗Mistake: Wrong method for a frequency table

  CISCE examiners' comments, 2025: candidates grouped discrete data into classes, took the average of the frequencies as the median, did not show a cumulative frequency column, and did not round the mean to the nearest whole number as asked.

  ✓Do this: For discrete x: mean = Σfx ÷ Σf; median from the c.f. column; round only as the question says.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Arithmetic slips in Σfx

  CISCE examiners' comments, 2024 and 2025 note calculation errors in the products fx.

  ✓Do this: Use the step-deviation method when allowed, or check Σfx by a second method.

  Usually costs: The final answer mark.

Practice questions on Measures of Central Tendency

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

1. Q1 · 1 mark · MCQBoard level

   The median of 7, 3, 9, 5, 11 is:
   (a) 5
   (b) 7
   (c) 9
   (d) 7.5

   Hint

   Order first.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The class mark of the class 25–35 is:
   (a) 10
   (b) 25
   (c) 30
   (d) 35

   Hint

   Average of the limits.

   Answer

   (c)

3. Q3 · 1 mark · MCQHarder board level

   The mean of five numbers is 18. If one number is removed, the mean of the rest is 16. The number removed is:
   (a) 2
   (b) 16
   (c) 18
   (d) 26

   Hint

   90 − 64.

   Answer

   (d)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): In a grouped frequency table, the modal class is the class with the highest frequency.
   Reason (R): The mode is the value that occurs most often.
   (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

   Most frequent value lies in the most frequent class.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   The mean of $x+1$, $x+3$, $x+5$, $x+7$ and $x+9$ is 15. Find $x$, the median, and the new mean if 2 is added to each number.

   Hint

   Sum = 5x + 25.

   Answer

   $5x+25=75$, $x=10$. Median $=x+5=15$. New mean $=17$.

6. Q6 · 4 marks · Short answerHarder board level

   The mean of this distribution is 6. Find $k$, then the median.

   $x$	2	4	6	8	10
   $f$	5	3	8	$k$	4

   Hint

   Σfx = 110 + 8k.

   Answer

   $\frac{110+8k}{20+k}=6$ gives $k=5$. $N=25$; c.f. 5, 8, 16, …; the 13th value is 6: median $=6$.

7. Q7 · 3 marks · Long answerBoard level

   Find the mean by the direct method.

   Class	0–6	6–12	12–18	18–24	24–30
   $f$	4	5	7	3	1

   Hint

   Class marks 3, 9, 15, 21, 27.

   Answer

   $\Sigma fx=12+45+105+63+27=252$; mean $=\frac{252}{20}=12.6$.

8. Q8 · 4 marks · Long answerHarder board level

   Find the mean, the median class and the modal class.

   Class	1–5	6–10	11–15	16–20	21–25
   $f$	4	7	12	5	2

   Hint

   Class marks 3, 8, 13, 18, 23.

   Answer

   $\Sigma fx=360$, $N=30$, mean $=12$. c.f. 4, 11, 23: the 15th observation is in 11–15 (10.5–15.5): median class. Modal class 11–15.

9. Q9 · 4 marks · Long answerHOTS / competency

   The mean of 20 observations was found to be 15. Later it was found that one observation, 23, had been copied as 32. Find the correct mean. If the observation is removed altogether instead, what is the mean of the remaining 19?

   Hint

   Correct the total first.

   Answer

   Wrong total $=300$; correct total $=300-32+23=291$; correct mean $=14.55$. Without it: $\frac{291-23}{19}=\frac{268}{19}\approx 14.11$.

10. Q10 · 4 marks · Long answerHarder board level

    Find the mean by the step-deviation method and check it by the short-cut method.

    Class	20–30	30–40	40–50	50–60	60–70
    $f$	5	8	12	10	5

    Hint

    A = 45, i = 10.

    Answer

    $\Sigma ft=-10-8+0+10+10=2$; mean $=45+\frac{2}{40}\times 10=45.5$. Short-cut: $\Sigma fd=20$, $45+\frac{20}{40}=45.5$ ✓

11. Q11 · 1 mark · MCQHOTS / competency

    The mean of a set of numbers is 10. If each number is multiplied by 3 and then 2 is added, the new mean is:
    (a) 30
    (b) 32
    (c) 12
    (d) 36

    Hint

    The mean follows the same operations.

    Answer

    (b)

12. Q12 · 5 marks · Long answerHOTS / competency

    Class A has 30 students with mean score 62; class B has 20 students with mean score 72. (a) Find the combined mean. (b) Five students from B, whose mean score is 80, move to A. Find the new means of A and B. (c) Explain why both class means changed but the combined mean did not.

    Hint

    Track totals.

    Answer

    (a) $\frac{1860+1440}{50}=66$. (b) A: $\frac{1860+400}{35}\approx 64.57$; B: $\frac{1440-400}{15}\approx 69.33$. (c) The five students are above A's mean (raising it) and above B's mean (so removing them lowers it), but the overall total 3300 and count 50 are unchanged.

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. 1.Median of n observations, n odd?Show answer

   The ((n + 1)/2)th value in order.

2. 2.Median of n observations, n even?Show answer

   Average of the (n/2)th and (n/2 + 1)th values.

3. 3.Step-deviation formula?Show answer

   Mean = A + (Σft ÷ Σf) × i, with t = (x − A)/i.

4. 4.Class mark of 20–30?Show answer

   25

5. 5.Class mark of 10–19?Show answer

   14.5

6. 6.Modal class?Show answer

   The class with the highest frequency.

7. 7.Median class?Show answer

   The class containing the (N/2)th observation.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Do you arrange data before finding the median?
• ☐Do you average the two middle values when n is even?
• ☐Can you build a c.f. column quickly?
• ☐Do you take A from the class marks?
• ☐Do you use the method named in the question?
• ☐Do you round the mean only as asked?

Measures of Central Tendency: frequently asked questions