Class 10 Maths Chapter 13: Statistics Notes & Formulas

Exam Weightage & Blueprint

Total: ~7-8 Marks

This chapter falls under Unit VII: Statistics & Probability (11 marks total). As per the latest syllabus, focus is on: Mean, median and mode of grouped data (bimodal situation to be avoided). Note: Ogives are excluded from the current syllabus.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Formulas, Empirical Relation, Mean of simple data
Short Answer	2 or 3	Medium	Finding Mean (Direct method), Mode
Long Answer	4 or 5	Very High	Missing Frequency (Mean/Median), Step-Deviation Method

CBSE 2026–27 Syllabus Status: Statistics

Unit VII: Statistics & Probability • 6–7 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

1. Mean of Grouped Data ($\bar{x}$)

For grouped frequency distributions, the CBSE 2026–27 curriculum includes the Direct Method, Assumed Mean Method and Step-Deviation Method for calculating the mean.

Animated video notes: Play a slide-by-slide lesson with Hinglish audio for all three mean methods and board-style missing-frequency questions. Play Mean video notes

A. Direct Method

$$ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} $$

Use when: Values of class mark ($x_i$) and frequency ($f_i$) are small.

B. Assumed Mean Method

$$ \bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} $$

Where $a$ is assumed mean (middle of $x_i$), and $d_i = x_i - a$.

C. Step-Deviation Method

Board-exam method: The CBSE 2026–27 curriculum includes step-deviation for the mean of grouped frequency distributions. It is especially convenient when all class intervals have a common width $h$.

$$ \bar{x} = a + (\frac{\Sigma f_i u_i}{\Sigma f_i}) \times h $$

Where $u_i = \frac{x_i - a}{h}$, and $h$ is the class size.

Tip: Always double-check your $\Sigma f_i u_i$ sum. One negative sign error can ruin the whole calculation!

2. Mode of Grouped Data

The mode is the value inside the modal class (the class interval with the maximum frequency).

$$ Mode = l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) \times h $$

Formula Breakdown:

$l$: Lower limit of modal class
$h$: Class size
$f_1$: Frequency of modal class (Highest frequency)
$f_0$: Frequency of class preceding modal class
$f_2$: Frequency of class succeeding modal class

3. Median of Grouped Data

The median divides the distribution into two equal halves. It is found using Cumulative Frequency (cf).

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

Critical Steps to find Median Class: 1. Calculate Cumulative Frequency ($cf$) for all classes. 2. Find $n = \Sigma f_i$. Calculate $n/2$. 3. Locate the class whose cumulative frequency is just greater than $n/2$. This is the Median Class.
Warning: In the formula, use the $cf$ of the class preceding the median class, but use the $f$ of the median class itself!

💡 Empirical Relationship (Important for 1 Mark) ?
$$ 3 Median = Mode + 2 Mean $$

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Discontinuous to Continuous Intervals Conversion:
If intervals are given as $118-126, 127-135, \dots$: $$\text{Adjustment factor } = \frac{127 - 126}{2} = 0.5$$ Subtract 0.5 from lower limits, add 0.5 to upper limits $\implies 117.5-126.5, 126.5-135.5$.
Warning: You MUST convert to continuous intervals before determining lower limit $l$ for Median or Mode!
Class Mark ($x_i$) & Class Size ($h$):
• Class Mark $x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$
• Class Size $h = \text{Upper Boundary} - \text{Lower Boundary}$
Comparative Properties of Central Tendencies:
• Mean: Affected by extreme values (outliers); cannot be determined graphically.
• Median: Unaffected by extreme values; best measure for skewed data; can be determined graphically.
• Mode: Represents the most frequent value; can be determined graphically from histogram.

Important PYQs & Solved Examples

📝 PYQ Type 1: Missing Frequencies (Median Given)

Problem: The median of the following data is 525. Find values of $x$ and $y$, if total frequency is 100.

Class	Freq ($f$)	Cum. Freq ($cf$)
0-100	2	2
100-200	5	7
200-300	x	7+x
300-400	12	19+x
400-500	17	36+x
500-600	20	56+x
600-700	y	56+x+y
700-800	9	65+x+y
800-900	7	72+x+y
900-1000	4	76+x+y

Step 1: Form Equation 1 1 Mark

Total frequency $n = 100$. From table, last $cf = 76 + x + y$.

$\Rightarrow 76 + x + y = 100 \Rightarrow x + y = 24$ ... (i)

Step 2: Identify Median Class 0.5 Mark

Given Median = 525. This lies in class 500-600.

$l = 500, f = 20, cf = 36 + x, h = 100$.

Step 3: Apply Formula 1.5 Marks

$525 = 500 + (\frac{50 - (36+x)}{20}) \times 100$

$25 = (14 - x) \times 5$

$5 = 14 - x \Rightarrow x = 9$.

Step 4: Solve for y 1 Mark

From (i): $9 + y = 24 \Rightarrow y = 15$.

Answer: $x = 9, y = 15$.

📝 PYQ Type 2: Missing Frequency in Mean (3 Marks)

Question: The mean pocket allowance is ?18. Find the missing frequency $f$ in the distribution.

Strategy: Use the Direct Method. Construct a table with $x_i$ (class mark) and $f_i x_i$. Then solve equation: $\frac{\Sigma f_i x_i}{\Sigma f_i} = 18$.

📝 PYQ Type 3: Discontinuous Class Intervals (4 Marks)

Scenario: Classes given as 118-126, 127-135... (Example 13.3 Q4).

Solution Step: Convert to continuous intervals by subtracting 0.5 from lower limit and adding 0.5 to upper limit. New classes: 117.5-126.5, 126.5-135.5...

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Continuity Check: Always check if class intervals are continuous (e.g., 0-10, 10-20). If not (e.g., 1-10, 11-20), fix them first!

Step-Deviation Error: Students often forget to multiply by $h$ at the very end of the Step-Deviation calculation.

Wrong CF Selection: In the Median formula, $cf$ is the cumulative frequency of the preceding class, NOT the median class itself. This is the most common error.

💡 Scoring Tips

Table Clarity: Draw neat tables with a ruler. Misaligned columns lead to calculation errors.

Sanity Check: Your calculated Mean, Median, or Mode MUST lie within the boundaries of their respective classes. If you get a Mode of 65 for a modal class 40-50, recheck your math immediately!

Empirical Formula: If a question asks for all three measures, calculate Mean and Median, then use $3 Median = Mode + 2 Mean$ to find Mode quickly (unless specified otherwise).

📝 More Solved Board Questions

Q4. In a distribution, the Mean and Mode are 26.4 and 27.2 respectively. Find the Median. 2 Marks

Sol. Using Empirical Relation: $3 Median = Mode + 2 Mean$

$3 Median = 27.2 + 2(26.4)$

$3 Median = 27.2 + 52.8 = 80.0$

Median = $80/3 = 26.67$ (approx).

Answer: 26.67

🎯 Board Pattern (2018–2025): 1-mark or 2-mark questions often test the empirical relation. Make sure you don't confuse the coefficients (it's 3-Median, not 3-Mean).

📋 Board Revision Checklist

✅ Class Mark ($x_i$) = (Upper Limit + Lower Limit) / 2
✅ Mean (Direct): $\Sigma f_ix_i / \Sigma f_i$
✅ Mean (Assumed): $a + \Sigma f_id_i / \Sigma f_i$
✅ Mean (Step Dev): $a + h \times (\Sigma f_iu_i / \Sigma f_i)$
✅ Modal Class = Class with highest frequency
✅ Discontinuous intervals: subtract 0.5 from lower limit, add 0.5 to upper limit before using $l$
✅ Mean is sensitive to outliers; Median is resistant to outliers
✅ Median and Mode can be found graphically; Mean cannot
✅ Median Class = Class where $cf \geq n/2$
✅ Empirical Relation: $3 Median = Mode + 2 Mean$
✅ Syllabus Check: Ogive/Cumulative Frequency graphs removed

💡 Exam Tip:
Always write the formula before starting the table or calculation. It ensures you get "step marks" even if you make a calculation error later.

CBSE Section E • 4 Marks

Case Study: Hospital ICU Patient Age Distribution

Scenario: During an epidemic outbreak, a district civil hospital records the ages of 80 patients admitted to the respiratory ICU over a week. The data is grouped as follows: Age 15–25 (6 patients), 25–35 (11 patients), 35–45 (21 patients), 45–55 (23 patients), 55–65 (14 patients), and 65–75 (5 patients).

Q1: Identify the Modal Class of the given patient distribution. 1 Mark

Q2: Calculate the Mode of the ages of the admitted patients. 1 Mark

Q3: Using the empirical formula $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$, find the Mean if the Median age is 44.5 years. (OR) In which age bracket does the Median class lie? 2 Marks

📝 10-Mark Self-Assessment Test

Time: 25 Mins

Solve these high-frequency CBSE Board questions on paper, then reveal step-by-step solutions to check your marks.

Q1 (1 Mark - MCQ): If the mode of a frequency distribution is 55 and the mean is 50, then its median using empirical relation is:
(a) 51.67 (b) 52.5 (c) 53.33 (d) 50

Q2 (2 Marks - VSA): Find the mode of the following data:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 5, 8, 12, 6, 4

Q3 (3 Marks - SA): The mean of the following distribution is 18. Find the missing frequency $f$:
Class: 11-13 (3), 13-15 (6), 15-17 (9), 17-19 (13), 19-21 ($f$), 21-23 (5), 23-25 (4)

Q4 (4 Marks - LA): The median of the following data is 28.5. If the total frequency is 60, find the values of $x$ and $y$:
Class: 0-10 (5), 10-20 ($x$), 20-30 (20), 30-40 (15), 40-50 ($y$), 50-60 (5)

Sol 1: $3 Median = Mode + 2 Mean = 55 + 2(50) = 155$.
$$Median = \frac{155}{3} ≈ 51.67$$ Correct Option: (a) 51.67 [1 Mark]

Sol 2: Highest frequency $= 12 \implies$ Modal class is $20-30$.
$l = 20, f_1 = 12, f_0 = 8, f_2 = 6, h = 10$ [1 Mark].
$$Mode = l + (\frac{f_1 - f_0}{2f_1 - f_0 - f_2}) \times h = 20 + (\frac{12 - 8}{24 - 8 - 6}) \times 10$$ $$= 20 + (\frac{4}{10}) \times 10 = 20 + 4 = 24 [1 Mark]$$ Answer: $Mode = 24$ [2 Marks]

Sol 3: Compute class marks $x_i$ and products $f_i x_i$:
$12(3)=36, 14(6)=84, 16(9)=144, 18(13)=234, 20(f)=20f, 22(5)=110, 24(4)=96$.
$\Sigma f_i = 40 + f$ [1 Mark].
$\Sigma f_i x_i = 704 + 20f$ [1 Mark].
$$Mean = \frac{\Sigma f_i x_i}{\Sigma f_i} \implies 18 = \frac{704 + 20f}{40 + f}$$ $$18(40 + f) = 704 + 20f \implies 720 + 18f = 704 + 20f \implies 2f = 16 \implies f = 8 [1 Mark]$$ Answer: Missing frequency $f = 8$ [3 Marks]

Sol 4: Cumulative frequency table:
$cf$: $5, 5+x, 25+x, 40+x, 40+x+y, 45+x+y$.
Total frequency $= 60 \implies 45 + x + y = 60 \implies x + y = 15$ ...(1) [1 Mark].
Median $= 28.5 \implies$ Median class is $20-30$.
$l = 20, n/2 = 30, cf = 5+x, f = 20, h = 10$ [1 Mark].
$$Median = l + (\frac{n/2 - cf}{f}) \times h \implies 28.5 = 20 + (\frac{30 - (5+x)}{20}) \times 10$$ $$8.5 = \frac{25 - x}{2} \implies 17 = 25 - x \implies x = 8 [1 Mark]$$ From (1): $8 + y = 15 \implies y = 7$ [1 Mark].
Answer: $x = 8, y = 7$ [4 Marks]

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. Which of the following cannot be determined graphically?

2. If the mean of the data is 27 and median is 33, then the mode is:

3. The class mark of the class 100-150 is:

4. In the formula for Mean by Assumed Mean Method, $d_i$ is:

5. The sum of the frequencies is denoted by:

Frequently Asked Questions: Class 10 Maths Chapter 13

What is the marks weightage of Statistics in CBSE Class 10 Maths?

Statistics carries around 6 to 7 marks (in Unit 7: Statistics & Probability, total 11 marks), almost always featuring 1 MCQ and a 3-mark or 5-mark question calculating mean, median, or mode.

Are Ogives (cumulative frequency graphs) in the syllabus?

No, graphical representation of cumulative frequency distribution (Less than and More than ogives) has been removed from the rationalized CBSE Class 10 syllabus.

What is the empirical relationship between mean, median, and mode?

The empirical formula is Mode = 3*Median - 2*Mean. This formula frequently appears in 1-mark board exam MCQs.

What is the step-deviation formula for calculating the mean of grouped data?

x_bar = a + h * (sum(fi * ui) / sum(fi)), where a is the assumed mean, h is class width, and ui = (xi - a) / h.

Chapter 13 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 13: Statistics

Overview