NCERT Solutions for Class 11 Maths Chapter 15 – Statistics
Subject: Mathematics | Class: 11 | Chapter: 15 | Board: CBSE | Curriculum: New NCERT
What Are NCERT Solutions for Class 11 Maths Chapter 15 – Statistics?
Numbers collected without context tell you very little. Statistics is the mathematical toolkit for extracting meaning from data — summarising it, identifying patterns, measuring how spread out it is, and comparing different data sets. Chapter 15 of Class 11 Mathematics focuses on the two central pillars of descriptive statistics: measures of central tendency (where is the data centred?) and measures of dispersion (how spread out is the data?).
You have already computed means and medians in earlier classes. Chapter 15 deepens this considerably. It introduces three measures of dispersion — Range, Mean Deviation, and Standard Deviation — and works through each for both ungrouped and grouped data. Standard Deviation, in particular, is the most important statistical measure you will encounter in the entire Class 11 curriculum. It is the foundation of variance analysis, quality control, financial risk analysis, and virtually every branch of applied statistics. For all Chapters must, read NCERT Solutions for Class 11 Maths and subject-wise NCERT Solutions for Class 11.
The NCERT solutions for chapter also introduces the concept of the coefficient of variation — a dimensionless measure that allows you to compare the variability of two data sets even when they have different units or scales. This is a CBSE favourite in 5-mark questions. The solutions here are structured to show the full tabulation method that CBSE mark schemes award step marks for, so every intermediate column in the table is shown and explained.
Download PDF – All Exercises of NCERT Solutions for Class 11 Maths Chapter 15 Statistics
| Exercise | Topic Covered | Number of Questions |
|---|---|---|
| Exercise 15.1 | Mean Deviation – Ungrouped and Grouped Data | 12 Questions |
| Exercise 15.2 | Variance and Standard Deviation – Ungrouped Data | 10 Questions |
| Exercise 15.3 | Variance and Standard Deviation – Grouped Data | 5 Questions |
| Miscellaneous | Comparison of Data Sets; Coefficient of Variation | 7 Questions |
Chapter 15 – Statistics: Concepts, Explanation and Key Tables
Measures of Central Tendency — Recap and Extension
| Measure | Formula (Ungrouped) | When It Is Most Useful |
|---|---|---|
| Mean (x̄) | x̄ = Σxᵢ / n | When data has no extreme outliers; most mathematical measure |
| Median | Middle value after sorting; average of two middle if n is even | When data has outliers or is skewed |
| Mode | Most frequently occurring value | When the most common value matters (e.g. shoe sizes) |
Measures of Dispersion
| Measure | What It Tells You | Limitation |
|---|---|---|
| Range | Difference between maximum and minimum value | Affected entirely by two extreme values; ignores all other data |
| Mean Deviation | Average of absolute deviations from mean (or median) | Does not use actual signs; less mathematically convenient |
| Variance (σ²) | Average of squared deviations from mean | Units are squared (e.g. cm²), not same as original data |
| Standard Deviation (σ) | Square root of variance; same unit as data | Most reliable and widely used measure of dispersion |
Mean Deviation Formulas
Mean Deviation is calculated about the mean or about the median.
| Data Type | Mean Deviation About Mean | Mean Deviation About Median |
|---|---|---|
| Ungrouped (individual) | MD = Σ | xᵢ – x̄ |
| Discrete frequency distribution | MD = Σfᵢ | xᵢ – x̄ |
| Continuous frequency distribution | Use class midpoints as xᵢ, then apply discrete formula | Same approach with midpoints |
Variance and Standard Deviation Formulas
| Data Type | Variance Formula | Standard Deviation |
|---|---|---|
| Ungrouped | σ² = Σ(xᵢ – x̄)² / n | σ = √[Σ(xᵢ – x̄)² / n] |
| Ungrouped (shortcut) | σ² = Σxᵢ²/n – (x̄)² | σ = √[Σxᵢ²/n – (x̄)²] |
| Discrete grouped | σ² = Σfᵢ(xᵢ – x̄)² / N | σ = √[Σfᵢ(xᵢ – x̄)² / N] |
| Grouped (shortcut) | σ² = Σfᵢxᵢ²/N – (x̄)² | σ = √[Σfᵢxᵢ²/N – (x̄)²] |
| Step deviation method | σ = h × √[Σfᵢdᵢ²/N – (Σfᵢdᵢ/N)²] | where dᵢ = (xᵢ – A)/h, A is assumed mean |
Coefficient of Variation
The Coefficient of Variation (CV) allows comparison of variability between two different data sets (even with different units):
CV = (σ / x̄) × 100 %
| Interpretation | Meaning |
|---|---|
| Lower CV | Data is more consistent (less variable relative to its mean) |
| Higher CV | Data is more spread out relative to its mean |
| CV = 0 | All values are identical |
Relationship Between Variance, SD, and the Data
| Key Property | Statement |
|---|---|
| If all values are equal | Variance = 0, SD = 0 |
| If a constant is added to all values | Mean shifts by that constant; SD and variance unchanged |
| If all values are multiplied by constant k | Mean multiplies by k; SD multiplies by |
| SD is never negative | σ ≥ 0 always |
| Variance = SD squared | σ² = (σ)² by definition |
Study Tips for Chapter 15
- Always present your solution as a table with columns for xᵢ, fᵢ, fᵢxᵢ, xᵢ – x̄, (xᵢ – x̄)², and fᵢ(xᵢ – x̄)². CBSE mark schemes award marks for each column, not just the final answer.
- The step deviation method is the fastest approach for grouped continuous data with a common class width — use it for Exercise 15.3 to save time.
- Coefficient of Variation questions (in the miscellaneous exercise) always require computing both mean and SD for each data set first — do not skip those steps.
- Remember: adding a constant to all data points shifts the mean but does not change dispersion. Multiplying by a constant changes both. These properties appear as short-answer questions.