Download Chapter 13 Class 10 Maths Statistics Notes PDF

These Class 10 Maths Statistics Notes PDF resources explain how to calculate the mean, median and mode of grouped data using the methods required by CBSE. The chapter notes below include formulas, symbol definitions, three methods of finding mean, solved examples, missing-frequency questions, common mistakes, competency-based practice and a downloadable revision section.

The current NCERT Mathematics textbook labels Statistics as Chapter 13, while older books and older learning resources may call it Chapter 14. The current CBSE syllabus focuses on the algebraic calculation of mean, median and mode for grouped data.

Class 10 Statistics chapter overview

Class 10 Statistics teaches students to summarize grouped data using mean, median and mode and to interpret the results in real-life contexts.

The CBSE Class 10 Mathematics syllabus for 2026–27 places Statistics and Probability together in Unit VII, worth 11 marks out of the 80-mark theory paper; this is a combined unit allocation, not a guaranteed chapter-wise mark allocation for Statistics alone. 

The official 2026–27 CBSE Class 10 syllabus specifically names direct, assumed-mean and step-deviation methods for mean and algebraic methods for median and mode. It also says bimodal situations should be avoided. 

Download Class 10 Maths Statistics Notes PDF

The downloadable Class 10 Maths Statistics Notes PDF should contain the same formulas, examples and syllabus notes as the searchable page rather than hiding essential material inside images.

Is Statistics Chapter 13 or Chapter 14?

Statistics is Chapter 13 in the current NCERT Class 10 Mathematics reprint, but it appeared as Chapter 14 in an older textbook sequence.

The current NCERT chapter PDF is headed “Statistics 13” and marked as a 2026–27 reprint. NCERT’s older rationalisation booklet still refers to “Chapter 14: Statistics,” which explains why many older notes and search results use the previous number. 

Use the chapter title rather than relying only on its number:

• Current wording: Class 10 Maths Chapter 13 Statistics
• Older wording: Class 10 Maths Chapter 14 Statistics

NCERT’s rationalisation booklet lists the graphical representation of cumulative-frequency distributions as dropped content that should not be assessed. The current NCERT reprint still discusses ogives in the chapter introduction, so schools may use the material for conceptual support; students should follow their teacher’s instructions. 

Important Statistics terms and definitions

A grouped-data question becomes easier when every class, frequency and formula symbol is identified before calculation begins.

Core definitions

Grouped and ungrouped data

Suppose the commute times of eight students are:

12, 18, 21, 24, 26, 32, 35, 38

This is ungrouped data because every observation is written separately.

It can be grouped as follows:

Grouping makes a large dataset easier to summarize, but it also means the exact individual observations are no longer visible.

Class mark

The class mark is the midpoint of a class interval.

Class mark xi = (lower class limit + upper class limit) ÷ 2

For the class 20–30:

xi = (20 + 30) ÷ 2

xi = 25

Class width

For continuous classes such as 10–20, 20–30 and 30–40:

h = 20 − 10 = 10

The same result is obtained using class boundaries.

Inclusive and exclusive intervals

Exclusive intervals do not include the upper limit in that class, while inclusive intervals include both displayed limits.

• Exclusive form: 10–20, 20–30, 30–40
• Inclusive form: 10–19, 20–29, 30–39

To convert inclusive intervals 10–19 and 20–29 into continuous intervals:

1. Find the gap: 20 − 19 = 1.
2. Divide the gap by two: 1 ÷ 2 = 0.5.
3. Subtract 0.5 from each lower limit.
4. Add 0.5 to each upper limit.

The converted intervals are:

• 9.5–19.5
• 19.5–29.5
• 29.5–39.5

Use class boundaries in the median and mode formulas when discontinuous inclusive classes must first be made continuous.

Cumulative frequency

Cumulative frequency is the running total of frequencies up to a class.

To recover ordinary frequencies from cumulative frequencies, subtract consecutive cumulative values:

12 − 4 = 8

24 − 12 = 12

Statistics symbol glossary

Class 10 Statistics formula sheet

The Class 10 Statistics formulas calculate the mean, median and mode of a grouped frequency distribution.

These formulas and methods appear in the current NCERT Statistics chapter and the current CBSE syllabus. 

Mean by direct method

Mean = Σ(fi × xi) ÷ Σfi

Mean by assumed-mean method

Mean = A + [Σ(fi × di) ÷ Σfi]

where:

di = xi − A

Mean by step-deviation method

Mean = A + h × [Σ(fi × ui) ÷ Σfi]

where:

ui = (xi − A) ÷ h

Median of grouped data

Median = l + [((N ÷ 2) − cf) ÷ f] × h

Mode of grouped data

Mode = l + [(f1 − f0) ÷ (2f1 − f0 − f2)] × h

Empirical relationship

Mode ≈ 3 × Median − 2 × Mean

Equivalent forms are:

3 × Median ≈ Mode + 2 × Mean

Median ≈ (Mode + 2 × Mean) ÷ 3

Mean ≈ (3 × Median − Mode) ÷ 2

The relationship is empirical and approximate; it is not an exact identity for every possible distribution.

How to choose the correct mean method

Choose the mean method that produces the shortest and least error-prone calculation while preserving the same final answer.

Method-selection process

1. Calculate the class marks.
2. Check whether fi × xi products are easy.
   • If yes, use the direct method.
3. Check whether the class marks are large.
   • If yes, choose a central class mark as A and use assumed mean.
4. Check whether the class widths are equal and deviations are multiples of h.
   • If yes, step deviation usually gives the smallest numbers.
5. Use the method requested in the question, even when another method looks faster.

Can the assumed mean be any class mark?

The assumed mean can generally be any convenient class mark, but a value near the centre of the distribution keeps deviations small.

A poor choice is not mathematically invalid; it simply produces larger calculations.

Can step deviation be used with unequal class widths?

The standard Class 10 step-deviation setup is most convenient when class widths are equal.

If class widths differ, one common value of h may not simplify every deviation into easy integers. Use the direct or assumed-mean method unless the question provides a suitable common factor.

Mean of grouped data

The mean of grouped data is a weighted average obtained by multiplying each class mark by its frequency.

The following original dataset records the one-way commute times of 40 students.

Direct-method example

First calculate each class mark:

Mean = Σ(fi × xi) ÷ Σfi

Mean = 1060 ÷ 40

Mean = 26.5

Answer: The estimated mean commute time is 26.5 minutes.

The answer is an estimate because every observation in a class is represented by its class mark.

Assumed-mean example

Use the same data and choose A = 25.

Mean = A + [Σ(fi × di) ÷ Σfi]

Mean = 25 + (60 ÷ 40)

Mean = 25 + 1.5

Mean = 26.5

Answer: The estimated mean commute time is again 26.5 minutes.

Step-deviation example

Here A = 25 and h = 10.

ui = (xi − A) ÷ h

Mean = A + h × [Σ(fi × ui) ÷ Σfi]

Mean = 25 + 10 × (6 ÷ 40)

Mean = 25 + 1.5

Mean = 26.5

Answer: The estimated mean commute time is 26.5 minutes.

Why all three methods give the same answer

The three mean methods rearrange the same weighted calculation, so a correctly completed table must produce the same mean.

A different result signals an arithmetic, sign or table error.

Missing frequency when the mean is given

Find x if the mean of the following distribution is 24.

The class marks are 5, 15, 25, 35 and 45.

Σfi = 3 + 5 + x + 4 + 2

Σfi = 14 + x

Σ(fi × xi) = (3 × 5) + (5 × 15) + (x × 25) + (4 × 35) + (2 × 45)

Σ(fi × xi) = 15 + 75 + 25x + 140 + 90

Σ(fi × xi) = 320 + 25x

Use the mean formula:

24 = (320 + 25x) ÷ (14 + x)

336 + 24x = 320 + 25x

x = 16

Answer: The missing frequency is 16.

Common mean mistakes

1. Using class limits instead of class marks.
2. Adding frequencies incorrectly.
3. Forgetting negative signs in di or ui.
4. Using class width as the assumed mean.
5. Multiplying A by h in the wrong place.
6. Rounding intermediate values too early.
7. Giving a number without a unit or interpretation.

Median of grouped data

The median of grouped data is calculated from the class whose cumulative frequency first reaches or exceeds N ÷ 2.

How to identify the median class

Use this process:

1. Add the frequencies to find N.
2. Calculate N ÷ 2.
3. Create a less-than cumulative-frequency column.
4. Find the first cumulative frequency that is greater than or equal to N ÷ 2.
5. The corresponding class is the median class.

Using the commute-time data:

N = 40

N ÷ 2 = 20

The first cumulative frequency greater than or equal to 20 is 24, so the median class is 20–30.

Why N ÷ 2 is used

The median divides an ordered dataset into two equal halves, so its position is based on half of the total frequency.

For N = 40, the middle position lies around the 20th observation. Cumulative frequency identifies the class containing that position.

What cf means

In the median formula, cf is the cumulative frequency before the median class—not the frequency of the median class.

For the median class 20–30:

• l = 20
• N = 40
• cf = 12
• f = 12
• h = 10

Fully solved median example

Median = l + [((N ÷ 2) − cf) ÷ f] × h

Median = 20 + [((40 ÷ 2) − 12) ÷ 12] × 10

Median = 20 + [(20 − 12) ÷ 12] × 10

Median = 20 + (8 ÷ 12) × 10

Median = 20 + 6.67

Median ≈ 26.67

Answer: The estimated median commute time is approximately 26.67 minutes.

Reasonableness check: 26.67 lies within the median class 20–30, so the result is plausible.

Median from a less-than cumulative-frequency table

Suppose the data is given as:

Recover the frequencies by subtraction:

• First frequency: 4
• Second frequency: 12 − 4 = 8
• Third frequency: 24 − 12 = 12
• Fourth frequency: 34 − 24 = 10
• Fifth frequency: 40 − 34 = 6

Then apply the ordinary median procedure.

Missing frequency when the median is given

Find x if the median is 24.

The median class is 20–30.

N = 5 + x + 15 + 7 + 3

N = 30 + x

l = 20, cf = 5 + x, f = 15 and h = 10

Substitute into the formula:

24 = 20 + {[((30 + x) ÷ 2) − (5 + x)] ÷ 15} × 10

4 = [(10 − x ÷ 2) ÷ 15] × 10

60 = 100 − 5x

5x = 40

x = 8

Answer: The missing frequency is 8.

Common median mistakes

• Selecting the class whose cumulative frequency is nearest to N ÷ 2, instead of the first one greater than or equal to it
• Using the frequency of the median class as cf
• Using the wrong lower class boundary
• Forgetting to convert inclusive classes into continuous intervals
• Using the number of classes as N
• Failing to check whether the result lies inside the median class

Mode of grouped data

The mode of grouped data is estimated from the class with the highest frequency and the frequencies immediately before and after it.

Identifying the modal class

In the commute-time dataset, the highest frequency is 12.

Therefore, the modal class is 20–30.

Understanding f0, f1 and f2

Fully solved mode example

Mode = l + [(f1 − f0) ÷ (2f1 − f0 − f2)] × h

Here:

l = 20, h = 10, f0 = 8, f1 = 12 and f2 = 10

Mode = 20 + [(12 − 8) ÷ (24 − 8 − 10)] × 10

Mode = 20 + (4 ÷ 6) × 10

Mode = 20 + 6.67

Mode ≈ 26.67

Answer: The estimated modal commute time is approximately 26.67 minutes.

The answer lies inside the modal class 20–30, so it passes the reasonableness check.

Missing frequency when the mode is given

Find x if the mode is 25.

The modal class is 20–30, so:

l = 20, h = 10, f0 = x, f1 = 12 and f2 = 8

25 = 20 + [(12 − x) ÷ (24 − x − 8)] × 10

5 = [(12 − x) ÷ (16 − x)] × 10

1 ÷ 2 = (12 − x) ÷ (16 − x)

16 − x = 24 − 2x

x = 8

Answer: The missing frequency is 8.

What if two classes have the same highest frequency?

A distribution with two prominent modes is called bimodal, but the current CBSE Class 10 syllabus says bimodal situations should be avoided.

Do not force the grouped-data mode formula onto an ambiguous modal class unless the question provides additional instructions.

Common mode mistakes

• Taking f0 as the lowest frequency
• Taking f1 as the total frequency
• Selecting non-adjacent frequencies as f0 and f2
• Using the lower limit when a class boundary is required
• Forgetting the denominator 2f1 − f0 − f2
• Accepting a result outside the modal class without checking the calculation

Relationship between mean, median and mode

For a moderately skewed distribution, mean, median and mode may be connected approximately by Mode = 3 × Median − 2 × Mean.

Finding the median

If mean = 24 and mode = 27:

3 × Median = Mode + 2 × Mean

3 × Median = 27 + 48

3 × Median = 75

Median = 25

Finding the mean

If median = 28 and mode = 30:

Mean = (3 × Median − Mode) ÷ 2

Mean = (84 − 30) ÷ 2

Mean = 27

Is the empirical relationship always exact?

No; it is an approximate relationship and should not replace a direct calculation when the full frequency distribution is available.

For the commute-time example:

• Mean = 26.5
• Median ≈ 26.67
• Mode from the grouped-mode formula ≈ 26.67

Using the empirical relationship:

Mode ≈ 3 × 26.67 − 2 × 26.5

Mode ≈ 27.01

The small difference shows why the relationship is approximate.

Mean vs median vs mode

Real-life and competency-based Statistics questions

Competency-based Statistics questions require students to calculate accurately and explain what the result means in context.

The 2026–27 CBSE curriculum emphasizes problem-solving, modelling, mathematical communication, data analytics and real-life applications rather than procedural recall alone. 

How an extreme value affects the mean

Consider five daily reading times:

20, 22, 24, 25, 29

Mean = 120 ÷ 5

Mean = 24

Median = 24

Now replace 29 with an extreme value of 80:

20, 22, 24, 25, 80

Mean = 171 ÷ 5

Mean = 34.2

Median = 24

Interpretation: One extreme value increases the mean from 24 to 34.2 minutes, while the median remains 24 minutes. The median better represents the typical reading time in the second dataset.

Case study: school commute times

Use the earlier grouped distribution.

Question 1: Which interval contains the greatest number of students?

Answer: 20–30 minutes.

Question 2: What is the estimated mean commute time?

Answer: 26.5 minutes.

Question 3: What percentage of students commute for less than 30 minutes?

Percentage = [(4 + 8 + 12) ÷ 40] × 100

Percentage = 60%

Answer: 60%.

Question 4: A school claims that “most students travel for less than 20 minutes.” Is the claim supported?

Students travelling for less than 20 minutes:

4 + 8 = 12

Percentage = (12 ÷ 40) × 100

Percentage = 30%

Answer: No. Only 30% of the students travel for less than 20 minutes.

Mathematics Basic practice path

Focus on:

1. Calculating class marks
2. Completing frequency and cumulative-frequency tables
3. Identifying median and modal classes
4. Substituting into formulas
5. Interpreting one result in a sentence

Mathematics Standard practice path

Add:

1. Missing-frequency equations
2. Comparing mean methods
3. Identifying an incorrect solution
4. Selecting the most suitable measure
5. Defending a conclusion using data
6. Solving unfamiliar case studies

Cumulative frequency and ogives

Cumulative frequency is still useful for identifying the median class, while ogive drawing should be treated as enrichment unless it is required by your school.

The current CBSE syllabus lists algebraic calculation of median and mode but does not explicitly list graphical median calculation. NCERT’s rationalisation booklet identifies graphical cumulative-frequency representation as dropped, non-assessed content in the older sequence. 

Less-than cumulative frequency

For less-than cumulative frequency, add frequencies from the first class onwards.

More-than cumulative frequency

For more-than cumulative frequency, begin with the total and subtract frequencies progressively.

Drawing a less-than ogive

1. Put upper class boundaries on the horizontal axis.
2. Put less-than cumulative frequencies on the vertical axis.
3. Choose and write a suitable scale.
4. Plot every coordinate accurately.
5. Join the plotted points with a smooth increasing curve or as directed by your teacher.
6. Label both axes and include units.

Drawing a more-than ogive

1. Put lower class boundaries on the horizontal axis.
2. Put more-than cumulative frequencies on the vertical axis.
3. Plot the coordinates.
4. Join them to form a decreasing curve.
5. Label axes, scale and units.

Common ogive questions

Should an ogive start at zero?
A less-than curve commonly begins with cumulative frequency zero at the lower boundary before the first class; follow the table and your teacher’s required convention.

Must the scale be written?
Yes. A graph should show the scale and label both axes clearly.

Which values go on the horizontal axis?
Use class boundaries: upper boundaries for a less-than ogive and lower boundaries for a more-than ogive.

How is the median found using two ogives?
The horizontal coordinate of the curves’ intersection estimates the median.

Class 10 Statistics solved questions

Class 10 Statistics questions test formula recall, table construction, calculation, algebra and interpretation at different mark levels.

One-mark questions

1. Find the class mark of 30–40.
   
   Class mark = (30 + 40) ÷ 2
   
   Class mark = 35
2. If N = 60, find N ÷ 2.
   
   N ÷ 2 = 30
3. Which class is the modal class?

1. Answer: 20–30.
2. State the empirical relationship.
   
   Mode ≈ 3 × Median − 2 × Mean

Two-mark question

Find the mean of the values 10, 20 and 30 with frequencies 2, 3 and 5.

Mean = [(2 × 10) + (3 × 20) + (5 × 30)] ÷ (2 + 3 + 5)

Mean = (20 + 60 + 150) ÷ 10

Mean = 23

Answer: 23.

Three-mark question

Find the median class.

N = 40

N ÷ 2 = 20

Cumulative frequencies are:

6, 15, 29, 36, 40

The first cumulative frequency greater than 20 is 29.

Answer: The median class is 20–30.

Five-mark question

Find the mean using step deviation.

Choose A = 35 and h = 10.

Mean = 35 + 10 × (0 ÷ 40)

Mean = 35

Answer: The mean is 35.

Answers-first self-test

1. Class mark of 50–60: 55
2. Total frequency when fi = 4, 7, 9, 5: 25
3. Median position for N = 50: 25th observation
4. Modal class: the class with the highest frequency
5. Best method when equal intervals create simple ui values: step deviation

Also Check: Statistics Class 10 Important Questions

Common mistakes in Class 10 Statistics

Most Statistics errors come from misidentifying symbols, constructing the table incorrectly or substituting a value from the wrong class.

Final reasonableness checks

Before finishing, verify that:

1. Σfi equals the stated total.
2. Every class mark lies inside its interval.
3. The median lies inside the median class.
4. The mode lies inside the modal class.
5. The mean lies within the overall data range.
6. The unit matches the question.
7. The result is sensible in context.

How to revise Class 10 Statistics

A strong Statistics revision session should combine formula recall, one example of each method and an error-checking exercise.

30-minute revision plan

1. Minutes 0–5: Review the symbol glossary.
2. Minutes 5–10: Write every formula from memory.
3. Minutes 10–18: Solve one mean question.
4. Minutes 18–24: Identify a median class and calculate the median.
5. Minutes 24–28: Identify a modal class and calculate the mode.
6. Minutes 28–30: Check answers using the common-mistakes list.

Related Class 10 Maths resources

Related chapter notes and practice papers help connect Statistics formulas with the wider Class 10 Mathematics course.

• Class 10 Probability notes
• Class 10 Maths formula sheet
• Chapter 13 NCERT solutions
• Class 10 Maths sample papers
• Class 10 previous-year papers