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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: xˉ=∑fixi∑fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}
Assumed Mean Method: xˉ=a+∑fidi∑fi,where di=xi−a\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, \quad \text{where } d_i = x_i - a
Step-Deviation Method: xˉ=a+(∑fiui∑fi)×h,where ui=xi−ah\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: Mode=l+(f1−f02f1−f0−f2)×h\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h
Median of Grouped Data: Median=l+(n2−cff)×h\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h
Empirical Relationship: 3Median=Mode+2Mean3\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 Algorithms2. 2. Mode, Median & The Empirical Relationship3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Health Clinic Blood Pressure Survey5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric
1

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
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Class Mark (xix_i): xi=Upper Class Limit+Lower Class Limit2x_i = \frac{\text{Upper Class Limit} + \text{Lower Class Limit}}{2}
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When to use which method:
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Direct Method: When xix_i and fif_i are small numerical integers.
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Assumed Mean Method: When xix_i and fif_i are moderately large numbers.
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Step-Deviation Method: When class intervals have equal width hh and values of di=xi−ad_i = x_i - a have a common factor hh.
📊 Statistics: Central Tendency & Ogive FormulasVisual Model
EMPIRICAL RELATIONSHIP: 3 Median = Mode + 2 Mean1. MEAN (x̄)Direct: ∑fᵢxᵢ / ∑fᵢAssumed: a + (∑fᵢdᵢ/∑fᵢ)2. MEDIAN (M)l + [(N/2 - cf)/f] × hN = ∑fᵢ, cf = prev cum freq3. MODE (Z)l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × hf₁ = modal freq, f₀=prev

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

2

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
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Modal Class: The class interval with the highest frequency (f1f_1).
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l=l = Lower limit of modal class
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f1=f_1 = Frequency of modal class
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f0=f_0 = Frequency of class preceding modal class
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f2=f_2 = Frequency of class succeeding modal class
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h=h = Class size
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Median Class: The class interval whose cumulative frequency is greater than or equal to n2\frac{n}{2} (where n=∑fin = \sum f_i).
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l=l = Lower limit of median class
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cf=cf = Cumulative frequency of class preceding median class
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f=f = Frequency of median class
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h=h = Class size
3

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).
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Step 1 (Tabular Computation with a=225,h=50a = 225, h = 50):
- 100–150: xi=125,fi=4,ui=125−22550=−2,fiui=−8x_i = 125, f_i = 4, u_i = \frac{125-225}{50} = -2, f_i u_i = -8
- 150–200: xi=175,fi=5,ui=−1,fiui=−5x_i = 175, f_i = 5, u_i = -1, f_i u_i = -5
- 200–250: xi=225,fi=12,ui=0,fiui=0x_i = 225, f_i = 12, u_i = 0, f_i u_i = 0
- 250–300: xi=275,fi=2,ui=+1,fiui=+2x_i = 275, f_i = 2, u_i = +1, f_i u_i = +2
- 300–350: xi=325,fi=2,ui=+2,fiui=+4x_i = 325, f_i = 2, u_i = +2, f_i u_i = +4
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Step 2 (Summations):
∑fi=25\sum f_i = 25
∑fiui=(−8−5)+(2+4)=−13+6=−7\sum f_i u_i = (-8 - 5) + (2 + 4) = -13 + 6 = -7
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Step 3 (Step-Deviation Formula):
xˉ=a+(∑fiui∑fi)×h=225+(−725)×50\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 225 + \left(\frac{-7}{25}\right) \times 50
xˉ=225+(−7×2)=225−14=211\bar{x} = 225 + (-7 \times 2) = 225 - 14 = \mathbf{211}
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Final Boxed Answer: Mean Daily Expenditure=Rs 211\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 xx and yy, 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.
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Step 1 (Cumulative Frequency Table):
- 0–100: f=2,cf=2f=2, cf=2
- 100–200: f=5,cf=7f=5, cf=7
- 200–300: f=x,cf=7+xf=x, cf=7+x
- 300–400: f=12,cf=19+xf=12, cf=19+x
- 400–500: f=17,cf=36+xf=17, cf=36+x
- 500–600: f=20,cf=56+xf=20, cf=56+x
- 600–700: f=y,cf=56+x+yf=y, cf=56+x+y
- 700–800: f=9,cf=65+x+yf=9, cf=65+x+y
- 800–900: f=7,cf=72+x+yf=7, cf=72+x+y
- 900–1000: f=4,cf=76+x+yf=4, cf=76+x+y
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Step 2 (Frequency Sum Equation):
76+x+y=100  ⟹  x+y=24— (1)76 + x + y = 100 \implies x + y = 24 \quad \text{--- (1)}
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Step 3 (Median Class Identification):
Since Median =525= 525, the median class is 500–600.
l=500,f=20,cf=36+x,h=100,n/2=50l = 500, f = 20, cf = 36 + x, h = 100, n/2 = 50.
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Step 4 (Applying Median Formula):
Median=l+(n2−cff)×h\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h
525=500+(50−(36+x)20)×100525 = 500 + \left(\frac{50 - (36 + x)}{20}\right) \times 100
25=(14−x)×525 = (14 - x) \times 5
5=14−x  ⟹  x=95 = 14 - x \implies \mathbf{x = 9}
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Step 5 (Finding yy): From (1): 9+y=24  ⟹  y=159 + y = 24 \implies \mathbf{y = 15}.
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Final Boxed Answer: x=9,y=15\mathbf{x = 9, \quad y = 15}
4

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).
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Q1: Identify the modal class of the distribution. →\rightarrow Highest frequency is 18   ⟹  \implies Modal Class is 130–140.
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Q2: Calculate the mode of the blood pressure data. →\rightarrow l=130,f1=18,f0=10,f2=10,h=10l = 130, f_1 = 18, f_0 = 10, f_2 = 10, h = 10. Mode=130+(18−102(18)−10−10)×10=130+(836−20)×10=130+(816)×10=130+5=135 mm Hg\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}}.
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Q3: If Mean = 134 mm Hg and Mode = 135 mm Hg, estimate Median using empirical relation. →\rightarrow 3Median=Mode+2Mean=135+2(134)=135+268=403  ⟹  Median=4033≈134.33 mm Hg3\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

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
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1 Mark: Clean frequency distribution table with class marks (xix_i) and cfcf.
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1 Mark: Correct formula citation for Mean/Median/Mode.
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2 Marks: Step-by-step arithmetic substitution without calculation errors.
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1 Mark: Final boxed value with appropriate statistical units.
• Common Calculation Traps & Verification Checklist
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Trap 1: In median calculation, taking cfcf of the median class instead of the preceding class.
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Trap 2: Forgetting brackets around (36+x)(36+x) in 50−(36+x)=14−x50 - (36+x) = 14 - x (sign error gives 50−36+x=14+x50 - 36 + x = 14 + x).
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Trap 3: Misapplying empirical relation (it is 3Median=Mode+2Mean3\text{Median} = \text{Mode} + 2\text{Mean}, not 3Mean=Mode+2Median3\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 π=22/7\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 (cm2,m3\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 (cm,cm2,m3\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: xˉ=∑fixi∑fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}
2. Assumed Mean Method: xˉ=a+∑fidi∑fi\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}, where di=xi−ad_i = x_i - a
3. Step-Deviation Method: xˉ=a+(∑fiui∑fi)h\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) h, where ui=xi−ahu_i = \frac{x_i - a}{h}

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