📐 NCERT Class 10 Mathematics

Chapter 13: Statistics

Complete theory, formulas for mean (direct method, assumed mean method, step-deviation method), mode, median of grouped data, cumulative frequency graphs (ogives), and exercise solutions for CBSE Class 10 Board Exam 2025-26.

📖 Introduction 📊 Mean of Grouped Data 🎯 Mode of Grouped Data 📈 Median of Grouped Data 📉 Ogive (Cumulative Frequency) 🔗 Empirical Relationship 📝 Solved Examples ⭐ Key Points

📖 Introduction to Statistics

Statistics is the branch of mathematics that deals with the collection, organization, analysis, and interpretation of data. In Class 10, we focus on finding three measures of central tendency for grouped data (data presented in class intervals):

Mean (Average): The sum of all observations divided by the number of observations.
Median: The middle value when data is arranged in order.
Mode: The value that occurs most frequently.

💡 Important Terms:

Class Interval: Group into which data is organized (e.g., 0-10, 10-20).
Class Mark (xi): Midpoint of class interval = (Upper limit + Lower limit)/2.
Frequency (fi): Number of observations in a class interval.
Cumulative Frequency: Running total of frequencies.

📊 Mean of Grouped Data

There are three methods to find the mean of grouped data:

Method 1: Direct Method

Mean (x̄) = (Σ fᵢxᵢ) / Σ fᵢ

where fᵢ = frequency of i-th class, xᵢ = class mark of i-th class.

Method 2: Assumed Mean Method

Mean (x̄) = a + (Σ fᵢdᵢ) / Σ fᵢ

where a = assumed mean, dᵢ = xᵢ - a (deviation).

Method 3: Step-Deviation Method (Easiest for large numbers)

Mean (x̄) = a + (Σ fᵢuᵢ) / Σ fᵢ × h

where uᵢ = (xᵢ - a)/h, h = class width.

📝 Example (Step-Deviation Method)

Find the mean of the following distribution:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 5, 8, 12, 7, 8

Solution: h = 10, let a = 25 (mid of 20-30). Calculate uᵢ = (xᵢ-25)/10. Σfᵢ = 40, Σfᵢuᵢ = -4. Mean = 25 + (-4/40)×10 = 25 - 1 = 24.

🎯 Mode of Grouped Data

Mode = L + [ (f₁ - f₀) / (2f₁ - f₀ - f₂) ] × h

where:
L = Lower limit of modal class (class with highest frequency)
f₁ = Frequency of modal class
f₀ = Frequency of class preceding modal class
f₂ = Frequency of class succeeding modal class
h = Class width

📝 Example

Find the mode of the distribution:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 4, 8, 12, 6, 3

Solution: Modal class = 20-30 (highest frequency 12).
L = 20, f₁ = 12, f₀ = 8, f₂ = 6, h = 10.
Mode = 20 + [(12-8)/(2×12-8-6)] × 10 = 20 + [4/(24-14)] × 10 = 20 + (4/10)×10 = 20 + 4 = 24.

📈 Median of Grouped Data

Median = L + [ (N/2 - cf) / f ] × h

where:
L = Lower limit of median class (class where cumulative frequency reaches N/2)
N = Total frequency (Σfᵢ)
cf = Cumulative frequency of class preceding median class
f = Frequency of median class
h = Class width

📝 Example

Find the median of the distribution:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 5, 8, 12, 7, 8

Solution: N = 40, N/2 = 20. Cumulative frequencies: 5, 13, 25, 32, 40. Median class = 20-30 (cf = 25 ≥ 20).
L = 20, cf = 13, f = 12, h = 10.
Median = 20 + [(20-13)/12] × 10 = 20 + (7/12)×10 = 20 + 70/12 = 20 + 5.83 = 25.83.

📉 Ogive (Cumulative Frequency Curve)

An ogive is a graph that represents the cumulative frequency distribution. There are two types:

Less than ogive: Plot points (upper limit, cumulative frequency).
More than ogive: Plot points (lower limit, cumulative frequency).

💡 Finding Median from Ogives:

From the point N/2 on the y-axis, draw a horizontal line to meet the ogive.
From that point, draw a vertical line down to the x-axis.
The x-coordinate gives the median.
The intersection of the "less than" and "more than" ogives also gives the median.

📝 Example

Draw a less than ogive for the distribution:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 2, 5, 8, 4, 1

Solution: Less than cumulative frequencies: 2, 7, 15, 19, 20. Plot points (10,2), (20,7), (30,15), (40,19), (50,20).

🔗 Empirical Relationship between Mean, Median, Mode

3 Median = Mode + 2 Mean

Mode = 3 Median - 2 Mean

Mean = (3 Median - Mode)/2

This relationship holds for a moderately skewed distribution. It is very useful when two of the three measures are known and we need to find the third.

📝 Example

If mean = 30 and median = 28, find mode.
Solution: Mode = 3 × 28 - 2 × 30 = 84 - 60 = 24.

📝 NCERT Solved Examples (Detailed)

🔹 Example 1 (Direct Method)

Question: Find the mean of the following distribution using direct method:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 2, 3, 5, 7, 3

Solution: Class marks: 5, 15, 25, 35, 45. Σfᵢ = 20, Σfᵢxᵢ = 2×5 + 3×15 + 5×25 + 7×35 + 3×45 = 10+45+125+245+135 = 560. Mean = 560/20 = 28.

🔹 Example 2 (Assumed Mean Method)

Question: Find the mean of the distribution using assumed mean method. Take a = 25.
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 4, 6, 10, 8, 2

Solution: xᵢ: 5,15,25,35,45. dᵢ = xᵢ-25: -20,-10,0,10,20. Σfᵢdᵢ = 4×(-20)+6×(-10)+10×0+8×10+2×20 = -80-60+0+80+40 = -20. Σfᵢ = 30. Mean = 25 + (-20/30) = 25 - 0.67 = 24.33.

🔹 Example 3 (Median)

Question: Find the median of the distribution:
Class: 0-10, 10-20, 20-30, 30-40, 40-50
Frequency: 6, 9, 15, 12, 8

Solution: N = 50, N/2 = 25. Cumulative: 6,15,30,42,50. Median class = 20-30. L=20, cf=15, f=15, h=10. Median = 20 + [(25-15)/15]×10 = 20 + (10/15)×10 = 20 + 6.67 = 26.67.

🔹 Example 4 (Mode - Board Exam PYQ)

Question: Find the mode of the distribution:
Class: 0-20, 20-40, 40-60, 60-80, 80-100
Frequency: 5, 8, 12, 7, 3

Solution: Modal class = 40-60 (f₁=12, L=40, f₀=8, f₂=7, h=20). Mode = 40 + [(12-8)/(2×12-8-7)]×20 = 40 + [4/(24-15)]×20 = 40 + (4/9)×20 = 40 + 80/9 = 40 + 8.89 = 48.89.

⭐ Key Points for Board Exam (Quick Revision)

📌 Formula Summary Table

Measure	Formula	When to Use
Direct Mean	x̄ = Σfᵢxᵢ/Σfᵢ	Small numbers, easy calculation
Assumed Mean	x̄ = a + Σfᵢdᵢ/Σfᵢ	Moderate-sized numbers
Step-Deviation	x̄ = a + (Σfᵢuᵢ/Σfᵢ) × h	Large numbers, multiples of h
Mode	L + [(f₁-f₀)/(2f₁-f₀-f₂)]×h	Find most frequent value
Median	L + [(N/2 - cf)/f]×h	Find middle value
Empirical	Mode = 3 Median - 2 Mean	When two measures are known

⚠️ Important Points to Remember:

Class mark (xᵢ) = (Upper limit + Lower limit)/2
Class width (h) = Upper limit - Lower limit (for equal classes)
For median, first find N/2, then locate median class using cumulative frequency.
For mode, modal class has the highest frequency.
Ogive (less than) is always increasing; (more than) is decreasing.
The x-coordinate of intersection of both ogives gives the median.
In step-deviation method, uᵢ = (xᵢ - a)/h, where h is class size.
Choose assumed mean (a) from the middle class for easier calculation.

📌 Mean: Σfᵢxᵢ/Σfᵢ | Mode: L + [(f₁-f₀)/(2f₁-f₀-f₂)]×h | Median: L + [(N/2 - cf)/f]×h

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