Class 10 · Chapter 13 · Statistics and Probability unit (11 of 80 marks)

Statistics Class 10: notes and important questions

Revision notes, 10 free important questions with step-marked answers, and a four-step route from the basics to 95+ through 41 board-style questions.

Practise this chapter Join freeFree account, no card: +5 practice questions on every chapter, with instant feedback.

41 questions
11 multiple choice, 4 assertion–reason, 9 very short answer, 9 short answer, 5 long answer, 3 case study
About 12 hours to master

Statistics & Probability unit: 11 of 80 theory marks (Statistics and Probability).

NCERT textbook: Chapter 13 on ncert.nic.in (PDF)

Revision notes

Mean of grouped data

Use the class mark $x_{i} = \frac{lower limit + upper limit}{2}$ as the representative value of each class.

Direct method: $\bar{x} = \frac{\sum f_{i} x_{i}}{\sum f_{i}}$
Assumed mean method: choose $a$ (a central $x_{i}$ ), let $d_{i} = x_{i} - a$ ; $\bar{x} = a + \frac{\sum f_{i} d_{i}}{\sum f_{i}}$
Step-deviation method (equal class width $h$ ): $u_{i} = \frac{x_{i} - a}{h}$ ; $\bar{x} = a + h \cdot \frac{\sum f_{i} u_{i}}{\sum f_{i}}$

All three give the same answer. The last two just reduce the arithmetic.

Mode of grouped data

The modal class is the class with the highest frequency. Then

$Mode = l + (\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}}) \times h$

where $l$ is the lower limit of the modal class, $f_{1}$ its frequency, $f_{0}$ and $f_{2}$ the frequencies of the classes before and after it, and $h$ the class width.

Median of grouped data

Find the cumulative frequencies (cf) and $\frac{n}{2}$ , where $n = \sum f_{i}$ . The median class is the first class whose cf is greater than (or equal to) $\frac{n}{2}$ . Then

$Median = l + (\frac{\frac{n}{2} - c f}{f}) \times h$

where $l$ is the lower limit of the median class, $c f$ the cumulative frequency of the class before it, $f$ its frequency, and $h$ the class width.

Empirical relationship

$3 Median = Mode + 2 Mean$ (approximately true for moderately skewed data).

Worked example 1 (step deviation)

Classes 10–20, 20–30, 30–40, 40–50 with $f = 3, 5, 8, 4$ . With $a = 35$ and $h = 10$ : $u = - 2, - 1, 0, 1$ and $\sum f u = - 6 - 5 + 0 + 4 = - 7$ , $n = 20$ . Mean $= 35 + 10 (\frac{- 7}{20}) = 31.5$ .

Worked example 2 (mode)

For the same data, the modal class is 30–40: $l = 30, f_{1} = 8, f_{0} = 5, f_{2} = 4$ . Mode $= 30 + \frac{3}{16 - 9} \times 10 = 30 + \frac{30}{7} \approx 34.29$ .

Worked example 3 (median)

cf: 3, 8, 16, 20; $\frac{n}{2} = 10$ , so the median class is 30–40 with $c f = 8$ , $f = 8$ . Median $= 30 + \frac{10 - 8}{8} \times 10 = 32.5$ .

Common errors

Using the cf of the median class itself instead of the class before it.
Taking $f_{0}, f_{2}$ from the wrong classes, or using the class with the largest cf as the modal class.
Applying the formulas to non-continuous classes (1–10, 11–20, …). First convert them to 0.5–10.5, 10.5–20.5, …
Forgetting to multiply by $h$ in the step-deviation method.

Board-exam tips

Always show a full table with columns $x_{i}$ , $f_{i} x_{i}$ (or $u_{i}$ , $f_{i} u_{i}$ ) and cf. The table carries method marks.
For missing frequencies, set up two equations: one from $\sum f_{i} = n$ and one from the given mean or median.
Bimodal data is not asked. Graphical representation (ogives) is deleted from the current syllabus.

Topics in this chapter: Mean of grouped data · Empirical relationship · Mode of grouped data · Median of grouped data.

Formulas for this chapter: Chapter 13 on the Class 10 formula sheet (free PDF and A3 poster).

Statistics Class 10: important questions

10 original questions in the board paper's question types, from 1-mark MCQs to a case study, each with a step-marked worked answer. Try each one first. 5 worked answers open here; the other 5 open with a free account (no card).

Download these questions (PDF) 10 questions, 5 worked answers · 5 pages · 209 KB

Section A style · Multiple choice and assertion–reason (1 mark each)

Q1·1 mark·Multiple choiceMedian of grouped data

Using $3 Median = Mode + 2 Mean$ , find the median of a distribution whose mode is $24$ and whose mean is $30$ .

Show worked answer

$\begin{aligned} 3 Median = Mode + 2 Mean = 24 + 60 = 84 \\ \Rightarrow & Median = 28 \end{aligned}$ A1
Correct option: (a) $28$

Q2·1 mark·Multiple choiceMode of grouped data

The classes $0$ – $10$ , $10$ – $20$ , $20$ – $30$ , $30$ – $40$ have frequencies $4$ , $8$ , $12$ , $6$ . The modal class is

Q3·1 mark·Assertion–reasonMean of grouped data

In the following question, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Assertion (A): For any grouped data, the assumed-mean method gives the same mean as the direct method.

Reason (R): The assumed-mean method subtracts the same number $a$ from every class mark and adds $a$ back at the end.

Section B style · Short answer (2 marks each)

Q4·2 marks·Short answerMean of grouped data

Find the mean of the data: values $x$ : $2, 4, 6, 8, 10$ with frequencies $f$ : $3, 5, 6, 4, 2$ .

Show worked answer

Answer:

5.7

$\sum f = 20$ and $\sum f x = 6 + 20 + 36 + 32 + 20 = 114$ M1
Mean $= \frac{114}{20} = 5.7$ A1

Q5·2 marks·Short answerMean of grouped data

The mean of the data below is $13$ . Find $p$ . Class marks $x$ : $5, 10, 15, 20$ ; frequencies $f$ : $2, p, 5, 3$ .

Section C style · Short answer (3 marks each)

Q6·3 marks·Short answerMedian of grouped data

Find the median of the distribution: $0$ – $10$ : $5$ , $10$ – $20$ : $8$ , $20$ – $30$ : $12$ , $30$ – $40$ : $10$ , $40$ – $50$ : $5$ .

Show worked answer

Answer:

\frac{155}{6} \approx 25.83

$n = 40$ , $\frac{n}{2} = 20$ ; cumulative frequencies $5, 13, 25, 35, 40$ , so the median class is $20$ – $30$ M1
Median $= l + \frac{\frac{n}{2} - c f}{f} \times h = 20 + \frac{20 - 13}{12} \times 10$ A1
$= 20 + \frac{35}{6} = \frac{155}{6} \approx 25.83$ A1

Q7·3 marks·Short answerMode of grouped data

Find the mode of the distribution: $0$ – $10$ : $6$ , $10$ – $20$ : $10$ , $20$ – $30$ : $16$ , $30$ – $40$ : $12$ , $40$ – $50$ : $6$ .

Section D style · Long answer (5 marks each)

Q8·5 marks·Long answerMean of grouped data

The daily wages of $50$ workers are: ₹ $100$ – $120$ : $6$ , ₹ $120$ – $140$ : $9$ , ₹ $140$ – $160$ : $15$ , ₹ $160$ – $180$ : $12$ , ₹ $180$ – $200$ : $8$ . Find the mean wage by the step-deviation method, and the modal wage.

Show worked answer

Answer: Mean ₹

152.80

; mode ₹

\frac{460}{3} \approx 153.33

Class marks $110, 130, 150, 170, 190$ ; with $a = 150$ , $h = 20$ : $u = - 2, - 1, 0, 1, 2$ M1
$\sum f u = - 12 - 9 + 0 + 12 + 16 = 7$ A1
Mean $= 150 + 20 \times \frac{7}{50} =$ ₹ $152.80$ A1
Modal class $140$ – $160$ : Mode $= 140 + \frac{15 - 9}{30 - 9 - 12} \times 20$ M1
$= 140 + \frac{40}{3} \approx$ ₹ $153.33$ A1

Q9·5 marks·Long answerMedian of grouped data

The distribution below has $50$ observations and median $36$ . Find $x$ and $y$ . Classes: $0$ – $10$ : $4$ , $10$ – $20$ : $6$ , $20$ – $30$ : $x$ , $30$ – $40$ : $15$ , $40$ – $50$ : $y$ , $50$ – $60$ : $5$ .

Section E style · Case study (4 marks)

Q10·4 marks·Case studyMean of grouped data

Screen-time survey. A class of $40$ students recorded their daily screen time (hours): $0$ – $2$ : $8$ students, $2$ – $4$ : $14$ , $4$ – $6$ : $10$ , $6$ – $8$ : $6$ , $8$ – $10$ : $2$ .

Write the modal class. [1 mark]
Write the median class. [1 mark]
Find the mean daily screen time. [2 marks]

Show worked answer

$2$ – $4$ hours (highest frequency, $14$ ) A1
$\frac{n}{2} = 20$ ; cumulative frequencies $8, 22, \dots$ , so the median class is $2$ – $4$ A1
Class marks $1, 3, 5, 7, 9$ : $\sum f x = 8 + 42 + 50 + 42 + 18 = 160$ M1
Mean $= \frac{160}{40} = 4$ hours A1

Next: practise by Route to 95 step · the revision notes · free MCQs and case studies · all Class 10 assertion–reason questions · the Class 10 syllabus checklist.

Route to 95: four steps

Work through the steps in order. Take each checkpoint closed book, about 1.5 minutes per mark; pass at 80% to move on. Two misses in a row means going back one step.

Step 1

Secure the basics

You can find the mean of grouped data and pick out the modal class correctly.

Read first: Mean of grouped data; Mode of grouped data 9 practice questions · checkpoint: 3 questions, 5 marks, pass 80%

Practise step 1

Step 2

Board standard

You can use the step-deviation method, the mode formula and the median formula with a full table.

Read first: Mean of grouped data; Mode of grouped data; Median of grouped data; Worked examples 1-3 17 practice questions · checkpoint: 4 questions, 9 marks, pass 80%

Practise step 2

Step 3

Full marks on long answers

You can write 5-mark answers with the complete table, find missing frequencies and use the empirical relationship.

Read first: Worked examples 1-3; Empirical relationship; Board-exam tips 6 practice questions · checkpoint: 3 questions, 15 marks, pass 80%

Practise step 3

Step 4

95+ stretch (HOTS)

You can handle inclusive classes, two unknown frequencies and questions that combine mean, median and mode.

Read first: Median of grouped data; Common errors 5 practice questions · checkpoint: 3 questions, 13 marks, pass 80%

Practise step 4

Practice questions

Original questions in the board's styles. Multiple-choice answers are checked as you go; for written answers, compare your working with the step mark scheme and record your marks. 14 of the 41 are competency-based (case studies and questions set in a real-life situation): the "Competency-based" button shows just those.

Q1·1 mark·Multiple choiceMean of grouped data

The class mark of the class 25–35 is

(a)10
(b)25
(c)30
(d)35

Q2·2 marks·Very short answerMean of grouped data

Find the mean of the following data.

Class	10–20	20–30	30–40
Frequency	4	6	10

Q3·2 marks·Very short answerMean of grouped data

For a grouped distribution of the heights (in cm) of students, $a = 150$ , $h = 5$ , $\sum f_{i} u_{i} = 12$ and $\sum f_{i} = 40$ . Find the mean height.

Worked solution·2 marks·Very short answerEmpirical relationship

A teacher finds that the median mark in a class test is $52.5$ and the modal mark is $54$ . Use the empirical relationship to estimate the mean mark.

Full worked solution Free, no sign-in

$2 Mean = 3 Median - Mode$ $= 157.5 - 54$ M1
Mean $= 51.75$ A1

Every question here has a full worked solution. See them all →

Stretch yourself: original probability problems at Intermediate level (ages 13 to 16), with hints and full solutions: Problem I128 · Problem I135 · Problem I142 · Problem I33. All 26 →

Themed lesson: Big tournament statistics maths lesson · Estimation station and class survey maths lesson · Spot the mistake relay and maths taboo maths lesson (starter, activity and worked answers). All themed maths

Knowledge organiser: Statistics on one page (key facts, formulas and mistakes to avoid; printable PDF)

Where marks are lost in Statistics

No table, or a table missing the xᵢ, fᵢxᵢ (or uᵢ, fᵢuᵢ) or cf column. Fix: always draw the full table; its columns carry method marks.
Median: using the cf of the median class instead of the class before it. Fix: write 'cf = cumulative frequency of the class preceding the median class' and circle that number.
Mode: taking f₀ and f₂ from the wrong classes, or choosing the class with the largest cf. Fix: the modal class has the largest f; f₀ is the class just before it, f₂ just after.
Inclusive classes (1-10, 11-20) used as they are. Fix: convert to 0.5-10.5, 10.5-20.5 before using l in the formula.
Step-deviation: forgetting to multiply by h. Fix: write mean = a + h(Σfᵢuᵢ/Σfᵢ) in full before substituting.
Missing frequency questions with one equation only. Fix: use Σfᵢ = n as one equation and the given mean or median as the other.

Exam strategy: how answers are step-marked · competency-based questions

Statistics in our sample papers

Once step 3 is passed, test the chapter inside a full timed paper on the 2026-27 pattern (original papers by us, not official CBSE papers).

Sample paper 1 (Board standard): questions 15, 16, 35
Sample paper 2 (Board standard): questions 17, 35
Sample paper 3 (Above board): questions 17, 31, 36
Sample paper 4 (Above board): questions 17, 35
Sample paper 5 (95+ challenge): questions 31, 35

Want it against the clock? Take a timed 30-mark chapter test on Statistics, new questions each time, or climb the Difficulty ladder, levels 1 to 10, three questions a level (CBSE Pro or the free trial).

Statistics: common questions

What is in CBSE Class 10 Maths Chapter 13, Statistics?

The topics are: Mean of grouped data, Empirical relationship, Mode of grouped data, Median of grouped data. This page has revision notes for each, and 41 board-style practice questions with step-marked solutions.

How many marks is Statistics worth?

It is in the Statistics and Probability unit, which carries 11 of the 80 theory marks. CBSE gives marks by unit, not by chapter.

Where do students lose marks in Statistics?

No table, or a table missing the xᵢ, fᵢxᵢ (or uᵢ, fᵢuᵢ) or cf column. Fix: always draw the full table; its columns carry method marks. Median: using the cf of the median class instead of the class before it. Fix: write 'cf = cumulative frequency of the class preceding the median class' and circle that number.

Are the Statistics important questions free?

Yes. The 10 important questions on this page are free to try, from 1-mark MCQs to a case study. 5 worked answers are open on the page; the others open with a free account (no card).

Are these NCERT questions?

No. Every practice question here is original, written in the board's question styles with step marks. For the textbook itself, use the NCERT link on this page.

Work through the Route to 95 with your progress saved

Save your answers and track your progressA free account remembers the questions you have done and brings you back to this page.

Join free

Already have an account? Sign in

Try it first

Revision notes

Mean of grouped data

Mode of grouped data

Median of grouped data

Empirical relationship

Worked example 1 (step deviation)

Worked example 2 (mode)

Worked example 3 (median)

Common errors

Board-exam tips

Statistics Class 10: important questions

Section A style · Multiple choice and assertion–reason (1 mark each)

Section B style · Short answer (2 marks each)

Section C style · Short answer (3 marks each)

Section D style · Long answer (5 marks each)

Section E style · Case study (4 marks)

Route to 95: four steps

Secure the basics

Board standard

Full marks on long answers

95+ stretch (HOTS)

Practice questions

Full worked solution Free, no sign-in

Where marks are lost in Statistics

Statistics in our sample papers

Statistics: common questions

Work through the Route to 95 with your progress saved