MathsICSE Class 10 · Measures of Central Tendency

Mean of Grouped Data by Direct, Short-cut and Step-deviation Methods (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

For grouped data, replace each class by its class mark $x$ (mid-point) and use $\bar{x} = \frac{\Sigma fx}{\Sigma f}$. The short-cut method uses $\bar{x} = A + \frac{\Sigma fd}{\Sigma f}$ with $d = x - A$; the step-deviation method uses $\bar{x} = A + \frac{\Sigma ft}{\Sigma f} \times i$ with $t = \frac{x - A}{i}$. All three agree.

Understand Mean of Grouped Data

The CISCE 2027 syllabus asks for the mean of grouped data, continuous and discontinuous, by all three methods: direct $\frac{Σ f x}{Σ f}$ ; short-cut $A + \frac{Σ f d}{Σ f}$ with $d = x - A$ ; step-deviation $A + \frac{Σ f t}{Σ f} \times i$ with $t = \frac{x - A}{i}$ . If a question names a method, only that method earns full marks (2024 Q4(ii)).

Class marks. Grouped data hide the exact values, so each class is represented by its mid-point, the class mark $x = \frac{lower limit + upper limit}{2}$ .

For continuous classes such as 20–30, $x = 25$ . For discontinuous (inclusive) classes such as 20–29, $x = \frac{20 + 29}{2} = 24.5$ . For the mean you do not need to convert inclusive classes to continuous ones: the class marks are the same either way.

Why the three methods agree. $Σ f (x - A) = Σ f x - A Σ f$ , so $A + \frac{Σ f d}{Σ f} = \frac{Σ f x}{Σ f}$ for any $A$ . Dividing every $d$ by the class width $i$ and multiplying back by $i$ changes nothing either.

Choosing the method.

Direct: small class marks and few classes.
Short-cut: large class marks; choose $A$ as a class mark near the middle.
Step-deviation: equal class widths; then $t = - 2, - 1, 0, 1, 2, \dots$ and the arithmetic is smallest.

**Choosing $A$ and $i$ .** Take $A$ from the class marks (usually the middle one, or the one with the largest frequency) so that one $t$ is 0. Take $i$ as the class width, the difference between consecutive lower limits (for 20–29, 30–39, $i = 10$ , not 9).

Missing frequencies. If the mean is given, write $Σ f$ and $Σ f x$ in terms of the unknown, set $\frac{Σ f x}{Σ f}$ equal to the mean and solve. With two unknowns you also need the total frequency.

Rounding. Keep fractions in the working and round the mean only as the question asks.

Mean of Grouped Data: formulas to remember

Class mark: $x = \frac{lower limit + upper limit}{2}$; 20–29 gives 24.5; 20–30 gives 25.
Direct method: $\overset{x}{ˉ} = \frac{Σ f x}{Σ f}$; Best for small numbers.
Short-cut method: $\overset{x}{ˉ} = A + \frac{Σ f d}{Σ f}, d = x - A$; $A$ is a class mark (assumed mean).
Step-deviation method: $\overset{x}{ˉ} = A + \frac{Σ f t}{Σ f} \times i, t = \frac{x - A}{i}$; $i$ = class width.
Why they agree: $Σ f (x - A) = Σ f x - A Σ f$; Any $A$ gives the same mean.

How to solve Mean of Grouped Data questions

Build the table column by column, then substitute once at the end.

1Write the class marks $x$ in a column; for inclusive classes, average the stated limits.
2For the short-cut or step-deviation method, choose $A$ from the class marks (middle class) and $i$ = class width.
3Add a column for $d = x - A$ or $t = \frac{x - A}{i}$ ; the $t$ values should be small integers.
4Add the product column ( $f x$ , $f d$ or $f t$ ) and find $Σ f$ and the column total.
5Substitute into the formula of the method named in the question.
6Round only the final mean, to the accuracy asked, and give the unit.

Mean of Grouped 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.

Average runs at a cricket academy

Suppose a cricket academy records the season totals of 30 batters: 0–20 runs (4 batters), 20–40 (8), 40–60 (10), 60–80 (5) and 80–100 (3).

What is the mean season total, to 2 decimal places?

Try it first, then show the solutionSolution

Class marks: 10, 30, 50, 70, 90.
$Σ f x = 40 + 240 + 500 + 350 + 270 = 1400$ ; $Σ f = 30$ .
Mean $= \frac{1400}{30} \approx 46.67$ runs.

Answer: About 46.67 runs

Memory hook: No exact scores? Use the middle of each class.

Average commute by step-deviation

Imagine you note your bus journey time on 25 school days: 20–30 minutes on 5 days, 30–40 on 9 days, 40–50 on 7 days and 50–60 on 4 days.

Find the mean journey time by the step-deviation method.

Try it first, then show the solutionSolution

Class marks 25, 35, 45, 55; $A = 35$ , $i = 10$ ; $t = - 1, 0, 1, 2$ .
$Σ f t = - 5 + 0 + 7 + 8 = 10$ .
Mean $= 35 + \frac{10}{25} \times 10 = 39$ minutes.

Answer: 39 minutes

Memory hook: Pick A in the middle, and the t-values stay tiny.

Tricks to remember Mean of Grouped Data

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

A from the marks, i from the width
The assumed mean is always one of the class marks; the step is always the class width.
Use it when: Avoiding the two errors CISCE examiners reported in 2024.
t counts classes
With equal widths, $t$ just counts how many classes you are from $A$ : $- 2, - 1, 0, 1, 2$ .
Use it when: Writing the $t$ column without dividing.
Same answer three ways
If you have time, check the step-deviation mean with $\frac{Σ f x}{Σ f}$ .
Use it when: Catching arithmetic slips.

Mean of Grouped Data: solved examples

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

Example 1: Step-deviation from the totals · 1 mark · MCQBoard level

In the step-deviation method, $A = 45$ , $i = 10$ and $\frac{Σ f t}{Σ f} = - 0.6$ . The mean is:
(a) 39
(b) 44.4
(c) 45.6
(d) 51

Try it first, then show the solutionSolution

Mean $= 45 + (- 0.6) \times 10 = 45 - 6 = 39$ .

Answer: (a)

Example 2: Direct method · 3 marks · Short answerBoard level

Find the mean of the following distribution by the direct method.

Class	0–10	10–20	20–30	30–40	40–50
Frequency	4	9	14	8	5

Try it first, then show the solutionSolution

Class marks $x$ : 5, 15, 25, 35, 45; $Σ f = 40$ .
$f x$ : 20, 135, 350, 280, 225; $Σ f x = 1010$ .
Mean $= \frac{1010}{40} = 25.25$ .

Answer: 25.25

Example 3: Short-cut method · 3 marks · Short answerHarder board level

The monthly electricity use of 40 houses is shown. Using the short-cut method, find the mean number of units.

Units	50–70	70–90	90–110	110–130	130–150	150–170
Houses	4	7	11	9	6	3

Try it first, then show the solutionSolution

Class marks: 60, 80, 100, 120, 140, 160. Take $A = 100$ .
$d = x - 100$ : $- 40, - 20, 0, 20, 40, 60$ ; $f d$ : $- 160, - 140, 0, 180, 240, 180$ .
$Σ f = 40$ , $Σ f d = 300$ .
Mean $= 100 + \frac{300}{40} = 107.5$ units.

Answer: 107.5 units

Example 4: Step-deviation with discontinuous classes · 4 marks · Short answerHarder board level

Using the step-deviation method, find the mean marks.

Marks	10–19	20–29	30–39	40–49	50–59	60–69
Students	4	6	13	15	8	4

Try it first, then show the solutionSolution

Class marks: 14.5, 24.5, 34.5, 44.5, 54.5, 64.5. Take $A = 44.5$ and $i = 10$ (consecutive lower limits differ by 10).
$t = - 3, - 2, - 1, 0, 1, 2$ ; $f t$ : $- 12, - 12, - 13, 0, 8, 8$ .
$Σ f = 50$ , $Σ f t = - 21$ .
Mean $= 44.5 + \frac{- 21}{50} \times 10 = 44.5 - 4.2 = 40.3$ marks.

Answer: 40.3 marks

Example 5: One missing frequency · 4 marks · Short answerHOTS / competency

The mean of the following distribution is 52.4. Find $p$ .

Class	0–20	20–40	40–60	60–80	80–100
Frequency	6	10	$p$	12	8

Try it first, then show the solutionSolution

Class marks: 10, 30, 50, 70, 90. $Σ f = 36 + p$ .
$Σ f x = 60 + 300 + 50 p + 840 + 720 = 1920 + 50 p$ .
$\frac{1920 + 50 p}{36 + p} = 52.4$ , so $1920 + 50 p = 1886.4 + 52.4 p$ .
$33.6 = 2.4 p$ , so $p = 14$ .

Answer: $p = 14$

Example 6: Case study: two missing frequencies · 5 marks · Case-basedHOTS / competency

A survey of 50 families records the distance (km) they travel to a weekly market. The mean distance is 32.8 km.

Distance	0–10	10–20	20–30	30–40	40–50	50–60
Families	4	$a$	10	12	$b$	5

(a) Write two equations in $a$ and $b$ .
(b) Solve them.
(c) Check the mean by the step-deviation method with $A = 35$ .

Try it first, then show the solutionSolution

(a) Total: $31 + a + b = 50$ , so $a + b = 19$ . Class marks 5, 15, …, 55: $Σ f x = 20 + 15 a + 250 + 420 + 45 b + 275 = 965 + 15 a + 45 b$ , and this equals $32.8 \times 50 = 1640$ . So $15 a + 45 b = 675$ , i.e. $a + 3 b = 45$ .
(b) Subtracting, $2 b = 26$ : $b = 13$ and $a = 6$ .
(c) $t = - 3, - 2, - 1, 0, 1, 2$ ; $f t$ : $- 12, - 12, - 10, 0, 13, 10$ ; $Σ f t = - 11$ . Mean $= 35 + \frac{- 11}{50} \times 10 = 32.8$ ✓

Answer: $a = 6$ , $b = 13$

Where students lose marks in Mean of Grouped Data

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

Mistake: Not using the method named in the question
The question tests that method; a correct mean by another method loses the method marks.
Wrong working
Direct method when 'step-deviation' was asked
Right working
Show the $t$ and $f t$ columns and $A + \frac{Σ f t}{Σ f} \times i$
Do this: Underline the method in the question.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q4(ii).
Mistake: Assumed mean not a class mark
Then no $t$ is 0 and the $t$ values are fractions.
Wrong working
$A = 130$ for classes 100–110, 110–120, …
Right working
$A = 135$ (a class mark)
Do this: Pick $A$ from the class-mark column.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q4(ii): longer working and errors.
Mistake: Wrong class width
The width is the difference between consecutive lower limits.
Wrong working
$i = 5$ for classes 100–110, 110–120
Right working
$i = 10$
Do this: Subtract two consecutive lower limits.
Usually costs: Reported in the CISCE examiners' comments on 2024 Q4(ii).
Mistake: Using a class limit instead of the class mark
Each class is represented by its mid-point.
Wrong working
$x = 20$ for 20–30
Right working
$x = 25$
Do this: Write the class-mark column first.
Usually costs: Every product, and so the mean.

Exam tips

Asked in 2024 Q4(ii) (3 marks): step-deviation mean of movie durations; the expected working used class marks 105 to 155, $A = 135$ and $i = 10$ , giving 127.4 minutes.
Draw the full table with headings $x$ , $f$ , $t$ , $f t$ ; the columns themselves earn method marks.
Choose $A$ near the middle of the class marks so that the $t$ values are small and balanced.
For discontinuous classes, the class marks end in .5, but the class width is still the gap between lower limits.

Practice questions on Mean of Grouped 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
For the classes 15–25, 25–35, 35–45, 45–55, a student takes $A = 30$ and $i = 10$ . The value of $t$ for the class 45–55 is:
(a) 1
(b) 2
(c) 15
(d) 20
Hint
$t = \frac{50 - 30}{10}$ .
Answer
(b)
Q2 · 3 marks · Short answerBoard level
Forty students timed how long they took to finish a puzzle. Find the mean time by the direct method, to 2 decimal places.
Minutes 5–10 10–15 15–20 20–25 25–30
Students 3 9 14 10 4
Hint
Class marks 7.5, 12.5, ….
Answer
$Σ f x = 22.5 + 112.5 + 245 + 225 + 110 = 715$ ; mean $= \frac{715}{40} = 17.875 \approx 17.88$ minutes.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The mean found by the short-cut method does not depend on the value chosen for $A$ .
Reason (R): $Σ f (x - A) = Σ f x - A Σ f$ .
(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
Divide (R) by $Σ f$ and add $A$ .
Answer
(c)
Q4 · 4 marks · Short answerHarder board level
Using the step-deviation method, find the mean age of 60 people at a camp, to 2 decimal places.
Age (years) 10–20 20–30 30–40 40–50 50–60 60–70
People 6 11 15 13 9 6
Hint
$A = 35$ , $i = 10$ .
Answer
$t = - 2, - 1, 0, 1, 2, 3$ ; $Σ f t = - 12 - 11 + 0 + 13 + 18 + 18 = 26$ . Mean $= 35 + \frac{26}{60} \times 10 \approx 39.33$ years.
Q5 · 3 marks · Short answerHarder board level
Find the mean of the following distribution.
Class 1–5 6–10 11–15 16–20 21–25
Frequency 7 10 16 11 6
Hint
Class mark of 1–5 is 3.
Answer
Class marks 3, 8, 13, 18, 23; $Σ f x = 21 + 80 + 208 + 198 + 138 = 645$ ; mean $= \frac{645}{50} = 12.9$ .
Q6 · 3 marks · Short answerHarder board level
The mean of the following distribution is 34.25. Find $k$ .
Class 10–20 20–30 30–40 40–50 50–60
Frequency 5 8 $k$ 9 3
Hint
$Σ f x = 845 + 35 k$ , $Σ f = 25 + k$ .
Answer
$845 + 35 k = 34.25 (25 + k) = 856.25 + 34.25 k$ , so $0.75 k = 11.25$ and $k = 15$ .
Q7 · 3 marks · Short answerHOTS / competency
In the step-deviation working for a table, $Σ f = 40$ , $Σ f t = - 12$ , $A = 65$ and $i = 10$ .
(a) Find the mean.
(b) Another student uses the same table with $A = 55$ . Without the table, find his $Σ f t$ and show that he gets the same mean.
Hint
Lowering $A$ by one class raises every $t$ by 1.
Answer
(a) $65 + \frac{- 12}{40} \times 10 = 62$ . (b) Each $t$ rises by 1, so $Σ f t = - 12 + 40 = 28$ ; mean $= 55 + \frac{28}{40} \times 10 = 62$ ✓

Minutes	5–10	10–15	15–20	20–25	25–30
Students	3	9	14	10	4

Age (years)	10–20	20–30	30–40	40–50	50–60	60–70
People	6	11	15	13	9	6

Class	1–5	6–10	11–15	16–20	21–25
Frequency	7	10	16	11	6

Class	10–20	20–30	30–40	40–50	50–60
Frequency	5	8	$k$	9	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.Step-deviation formula?Show answer
$\overset{x}{ˉ} = A + \frac{Σ f t}{Σ f} \times i$ , $t = \frac{x - A}{i}$
2.Short-cut formula?Show answer
$\overset{x}{ˉ} = A + \frac{Σ f d}{Σ f}$ , $d = x - A$
3.Class mark of 40–49?Show answer
44.5
4.Where must $A$ come from?Show answer
The class marks.
5.Does the choice of $A$ change the mean?Show answer
No.

Mean of Grouped Data: frequently asked questions

What is the step-deviation method for the mean?

Take an assumed mean A from the class marks and the class width i, find t = (x − A) ÷ i for each class, and use mean = A + (Σft ÷ Σf) × i.

Which method should I use for the mean of grouped data?

Use the method the question names. If none is named, any of the three is accepted; the step-deviation method is quickest when the classes have equal widths.

How do I find the class mark of a discontinuous class like 20–29?

Average the stated limits: (20 + 29) ÷ 2 = 24.5. You do not need to convert to continuous classes to find the mean.