What is the formula for the Mode of grouped data and what do the variables represent?

\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h - l: Lower limit of modal class (class with highest frequency f_1) - f_1: Frequency of modal class - f_0: Frequency of class preceding modal class - f_2: Frequency of class succeeding modal class - h: Class size

What is the formula for the Median of grouped data?

\text{Median} = l + \left(\frac{\frac{N}{2} - cf}{f}\right) \times h - l: Lower limit of median class (class whose cumulative frequency exceeds \frac{N}{2}) - N = \sum f_i: Total frequency - cf: Cumulative frequency of class PRECEDING median class - f: Frequency of median class - h: Class size

What is the Empirical Relationship between Mean, Median, and Mode for moderately skewed data?

\text{Mode} = 3\text{ Median} - 2\text{ Mean} \iff 3\text{ Median} = \text{Mode} + 2\text{ Mean}. This relationship is frequently tested in 1-mark board MCQs.

How do you convert inclusive (discontinuous) classes like 11-15, 16-20 into continuous classes?

Find half the gap: \frac{16 - 15}{2} = 0.5. Subtract 0.5 from each lower limit and add 0.5 to each upper limit, converting 11-15 to 10.5-15.5 and 16-20 to 15.5-20.5. Modal and median formulas require continuous class boundaries.

Are Ogives (cumulative frequency curves) included in the CBSE 2026–27 exam?

No, Less-than and More-than Ogives and graphical determination of median have been removed from the current CBSE syllabus. Students are tested on calculating Mean, Median, and Mode algebraically.

What is the most common mistake in finding the missing frequencies $x$ and $y$ when Median is given?

Students often use cf of the median class instead of the class PRECEDING the median class. Always remember: in \frac{\frac{N}{2} - cf}{f}, the cf must come from the row strictly above the median class.

All Revision Notes

MathematicsCh-13 15 min comprehensive revision

NCERT Class 10 Mathematics — Chapter 13

Statistics

Mean of grouped data (Direct, Assumed Mean, Step-Deviation methods), Mode of grouped data, Median of grouped data, and the empirical relation between central tendencies.

Quick Key Takeaways:

Direct Method for Mean:

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Assumed Mean Method:

\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad \text{where } d_i = x_i - a

Step-Deviation Method:

\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h, \quad \text{where } u_i = \frac{x_i - a}{h}

Mode of Grouped Data:

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

Median of Grouped Data:

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

Empirical Relationship:

3\text{Median} = \text{Mode} + 2\text{Mean}

Statistics CalculatorCompute mean (direct, assumed, step-dev), median, and mode

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Jump to Concept:1. 1. Measures of Central Tendency & The 3 Mean Algorithms 2. 2. Mode, Median & The Empirical Relationship 3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Health Clinic Blood Pressure Survey 5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

1. Measures of Central Tendency & The 3 Mean Algorithms

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Statistics.

• Mean Computation Methods for Grouped Frequency Distributions

•

Class Mark ( $x_i$ ):

x_i = \frac{\text{Upper Class Limit} + \text{Lower Class Limit}}{2}

•

When to use which method:

•

Direct Method: When

x_i

and

f_i

are small numerical integers.

•

Assumed Mean Method: When

x_i

and

f_i

are moderately large numbers.

•

Step-Deviation Method: When class intervals have equal width

h

and values of

d_i = x_i - a

have a common factor

h

📊 Statistics: Central Tendency & Ogive FormulasVisual Model

Visual schematic mapping the 3 mean algorithms, grouped mode formulation, grouped median class lookup, and the empirical relation.

2. Mode, Median & The Empirical Relationship

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Statistics.

• Mode and Median Formulations for Continuous Grouped Data

•

Modal Class: The class interval with the highest frequency (

f_1

•

l =

Lower limit of modal class

•

f_1 =

Frequency of modal class

•

f_0 =

Frequency of class preceding modal class

•

f_2 =

Frequency of class succeeding modal class

•

h =

Class size

•

Median Class: The class interval whose cumulative frequency is greater than or equal to

\frac{n}{2}

(where

n = \sum f_i

•

l =

Lower limit of median class

•

cf =

Cumulative frequency of class preceding median class

•

f =

Frequency of median class

•

h =

Class size

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure by the Step-Deviation method.
Expenditure (Rs): 100-150 (4), 150-200 (5), 200-250 (12), 250-300 (2), 300-350 (2).

•

Step 1 (Tabular Computation with $a = 225, h = 50$ ):
- 100–150:

x_i = 125, f_i = 4, u_i = \frac{125-225}{50} = -2, f_i u_i = -8

- 150–200:

x_i = 175, f_i = 5, u_i = -1, f_i u_i = -5

- 200–250:

x_i = 225, f_i = 12, u_i = 0, f_i u_i = 0

- 250–300:

x_i = 275, f_i = 2, u_i = +1, f_i u_i = +2

- 300–350:

x_i = 325, f_i = 2, u_i = +2, f_i u_i = +4

•

Step 2 (Summations):

\sum f_i = 25

\sum f_i u_i = (-8 - 5) + (2 + 4) = -13 + 6 = -7

•

Step 3 (Step-Deviation Formula):

\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 225 + \left(\frac{-7}{25}\right) \times 50

\bar{x} = 225 + (-7 \times 2) = 225 - 14 = \mathbf{211}

•

Final Boxed Answer:

\mathbf{\text{Mean Daily Expenditure} = \text{Rs } 211}

• 5-Mark Heavyweight Board Problem / Rider: The median of the following data is 525. Find the values of $x$ and $y$ , if the total frequency is 100.
0-100: 2, 100-200: 5, 200-300: x, 300-400: 12, 400-500: 17, 500-600: 20, 600-700: y, 700-800: 9, 800-900: 7, 900-1000: 4.

•

Step 1 (Cumulative Frequency Table):
- 0–100:

f=2, cf=2

- 100–200:

f=5, cf=7

- 200–300:

f=x, cf=7+x

- 300–400:

f=12, cf=19+x

- 400–500:

f=17, cf=36+x

- 500–600:

f=20, cf=56+x

- 600–700:

f=y, cf=56+x+y

- 700–800:

f=9, cf=65+x+y

- 800–900:

f=7, cf=72+x+y

- 900–1000:

f=4, cf=76+x+y

•

Step 2 (Frequency Sum Equation):

76 + x + y = 100 \implies x + y = 24 \quad \text{--- (1)}

•

Step 3 (Median Class Identification):
Since Median

= 525

, the median class is 500–600.

l = 500, f = 20, cf = 36 + x, h = 100, n/2 = 50

•

Step 4 (Applying Median Formula):

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

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

25 = (14 - x) \times 5

5 = 14 - x \implies \mathbf{x = 9}

•

Step 5 (Finding $y$ ): From (1):

9 + y = 24 \implies \mathbf{y = 15}

•

Final Boxed Answer:

\mathbf{x = 9, \quad y = 15}

4. CBSE Case-Study Modeling: Health Clinic Blood Pressure Survey

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Health Clinic Blood Pressure Survey

A health survey recorded the systolic blood pressure (in mm Hg) of 50 patients: 110-120 (6), 120-130 (10), 130-140 (18), 140-150 (10), 150-160 (6).

•

Q1: Identify the modal class of the distribution.

\rightarrow

Highest frequency is 18

\implies

Modal Class is 130–140.

•

Q2: Calculate the mode of the blood pressure data.

\rightarrow

l = 130, f_1 = 18, f_0 = 10, f_2 = 10, h = 10

\text{Mode} = 130 + \left(\frac{18 - 10}{2(18) - 10 - 10}\right) \times 10 = 130 + \left(\frac{8}{36 - 20}\right) \times 10 = 130 + \left(\frac{8}{16}\right) \times 10 = 130 + 5 = \mathbf{135\text{ mm Hg}}

•

Q3: If Mean = 134 mm Hg and Mode = 135 mm Hg, estimate Median using empirical relation.

\rightarrow

3\text{Median} = \text{Mode} + 2\text{Mean} = 135 + 2(134) = 135 + 268 = 403 \implies \text{Median} = \frac{403}{3} \approx \mathbf{134.33\text{ mm Hg}}

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines

•

1 Mark: Clean frequency distribution table with class marks (

x_i

) and

cf

•

1 Mark: Correct formula citation for Mean/Median/Mode.

•

2 Marks: Step-by-step arithmetic substitution without calculation errors.

•

1 Mark: Final boxed value with appropriate statistical units.

• Common Calculation Traps & Verification Checklist

•

Trap 1: In median calculation, taking

cf

of the median class instead of the preceding class.

•

Trap 2: Forgetting brackets around

(36+x)

50 - (36+x) = 14 - x

(sign error gives

50 - 36 + x = 14 + x

•

Trap 3: Misapplying empirical relation (it is

3\text{Median} = \text{Mode} + 2\text{Mean}

, not

3\text{Mean} = \text{Mode} + 2\text{Median}

Authentic Board Question (3 Marks)Topic: Statistics Mathematical Proof & Computations

State the governing mathematical theorem/formula, show complete geometric or computational steps, and find the exact result for Statistics.

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Given / To Prove / Formula Setup: State the governing theorem (BPT, Tangents, Pythagoras), mensuration formula, or trigonometric ratios with given values.

1 Mark

Step 2: Step-by-Step Derivation / Arithmetic Working: Show complete deductive steps or numerical calculations with

\pi = 22/7

or trigonometric standard values.

1 Mark

Step 3: Final Boxed Answer with Units / Q.E.D.: State the final numerical result with proper square/cubic units (

\text{cm}^2, \text{m}^3

) or formal proof conclusion.

1 Mark

Model Student Answer (Target: Full 3/3 Marks):

To score full 3 marks on Statistics in CBSE Mathematics:

1. Theorem / Formula: Write the standard governing formula or state the geometric theorem clearly.
2. Calculation / Proof: Substitute dimensions systematically or provide deductive step justifications.
3. Final Result: Box the final answer with required units (

\text{cm}, \text{cm}^2, \text{m}^3

) or conclude with "Hence Proved".

Examiner Mark Deduction Traps:

•Always write intermediate calculation lines to earn partial credit under CBSE step-marking.

•State reasons in parentheses (e.g., "[Tangents from an external point are equal]") during geometric proofs.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

1. Direct Method:

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

2. Assumed Mean Method:

\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}

, where

d_i = x_i - a

3. Step-Deviation Method:

\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) h

, where

u_i = \frac{x_i - a}{h}

Statistics

1. Measures of Central Tendency & The 3 Mean Algorithms

2. Mode, Median & The Empirical Relationship

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

4. CBSE Case-Study Modeling: Health Clinic Blood Pressure Survey

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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