Chapter Introduction
Statistics is Chapter 13 of CBSE Class 11 Mathematics, falling under the Statistics and Probability unit. In Class 10, you learned about mean, median, and mode — measures of central tendency. This chapter goes further to study measures of dispersion: how spread out the data is. You will learn about range, mean deviation, variance, standard deviation, and coefficient of variation. This chapter typically carries 6-8 marks in board exams and the calculation-heavy nature means practice is essential.
The key insight: two distributions can have the same mean but very different spreads. Measures of dispersion tell you how much the data varies from the centre.
Key Definitions
| Term | Definition |
|---|---|
| Dispersion | The extent to which values in a distribution are spread out or scattered |
| Range | The difference between the maximum and minimum values: R = L - S |
| Mean Deviation | The average of absolute deviations from a central value (mean or median) |
| Variance | The average of squared deviations from the mean; denoted sigma^2 |
| Standard Deviation | The positive square root of variance; denoted sigma |
| Coefficient of Variation (CV) | The standard deviation expressed as a percentage of the mean: CV = (sigma/x-bar) x 100 |
| Ungrouped Data | Raw data listed as individual observations |
| Grouped Data | Data organized into classes (intervals) with frequencies |
| Class Mark | The midpoint of a class interval: (upper limit + lower limit)/2 |
Detailed Concepts
1. Measures of Dispersion — Overview
| Measure | What it measures | Advantage | Limitation |
|---|---|---|---|
| Range | Max - Min | Simple to compute | Uses only two values; affected by outliers |
| Mean Deviation | Average of absolute deviations | Uses all values; easy to understand | Absolute values make algebra difficult |
| Variance | Average of squared deviations | Mathematically tractable; uses all values | Units are squared; not in original units |
| Standard Deviation | Square root of variance | Same units as data; most widely used | More computation than range |
2. Mean Deviation for Ungrouped Data
About the mean (x-bar):
| MD(x-bar) = (1/n) * Sigma | x_i - x-bar |
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About the median (M):
| MD(M) = (1/n) * Sigma | x_i - M |
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Steps: 1. Calculate the mean (or median) of the data. 2. Find the deviation of each value from the mean (or median). 3. Take the absolute value of each deviation. 4. Sum all absolute deviations. 5. Divide by the number of observations.
Note: Mean deviation about the median is always less than or equal to mean deviation about the mean.
3. Mean Deviation for Grouped Data
Discrete frequency distribution:
| MD(x-bar) = (1/N) Sigma f_i | x_i - x-bar | , where N = Sigma f_i |
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Continuous frequency distribution:
Same formula, but x_i is replaced by the class mark of each class interval.
Steps: 1. Find class marks (midpoints of each class). 2. Calculate the mean x-bar.
4. Multiply by frequency f_i. 5. Sum and divide by N.
| 3. Find | x_i - x-bar | for each class. |
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4. Variance and Standard Deviation — Ungrouped Data
Variance:
sigma^2 = (1/n) * Sigma (x_i - x-bar)^2
Standard Deviation:
sigma = sqrt(sigma^2) = sqrt[(1/n) * Sigma (x_i - x-bar)^2]
5. Variance and Standard Deviation — Grouped Data
Variance:
sigma^2 = (1/N) Sigma f_i (x_i - x-bar)^2
Standard Deviation:
sigma = sqrt[(1/N) Sigma f_i (x_i - x-bar)^2]
Where N = Sigma f_i and x_i is the class mark for continuous distributions.
6. Shortcut Method for Variance
When values are large, use an assumed mean A to simplify calculations:
Let d_i = x_i - A
*sigma^2 = (1/N) Sigma f_i d_i^2 - [(1/N) Sigma f_i d_i]^2*
Steps: 1. Choose A close to the actual mean (usually the middle class mark). 2. Compute d_i = x_i - A for each value/class mark. 3. Compute f_i d_i and f_i d_i^2. 4. Sum both columns. 5. Apply the formula.
Why it works: This formula is derived from expanding (x_i - x-bar)^2 and simplifying using the algebraic identity. The result does not depend on the choice of A.
7. Coefficient of Variation
CV = (sigma / x-bar) x 100
Purpose: To compare the variability of two distributions with different means or different units.
- If CV(A) < CV(B): Distribution A is more consistent (less variable).
- If CV(A) > CV(B): Distribution A is more variable (less consistent).
Direct vs Shortcut Method — Comparison
| Aspect | Direct Method | Shortcut Method |
|---|---|---|
| Computed from | Actual mean x-bar | Assumed mean A |
| Formula (ungrouped) | sigma^2 = (1/n) Sigma(x_i - x-bar)^2 | sigma^2 = (1/n) Sigma d_i^2 - [(1/n) Sigma d_i]^2 |
| Best when | Values are small | Values are large |
| Accuracy | Exact | Exact (same result) |
| Computational effort | Higher (finding deviations from mean) | Lower (deviations from assumed mean are simple integers) |
Important Numericals
Example 1: Mean deviation for ungrouped data
Find the mean deviation about the mean for: 4, 6, 8, 10, 12.
Mean = (4 + 6 + 8 + 10 + 12)/5 = 40/5 = 8
| Deviations: | 4-8 | = 4, | 6-8 | = 2, | 8-8 | = 0, | 10-8 | = 2, | 12-8 | = 4 |
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MD = (4 + 2 + 0 + 2 + 4)/5 = 12/5 = 2.4
Example 2: Variance and SD for ungrouped data
Find the variance and standard deviation for: 6, 8, 10, 12, 14.
Mean = 50/5 = 10
Deviations squared: (6-10)^2 = 16, (8-10)^2 = 4, (10-10)^2 = 0, (12-10)^2 = 4, (14-10)^2 = 16
Variance = (16 + 4 + 0 + 4 + 16)/5 = 40/5 = 8
SD = sqrt(8) = 2*sqrt(2) = 2.83 (approx.)
Example 3: Grouped data — variance and SD
| Class | f_i | x_i (class mark) | f_i * x_i | x_i - x-bar | (x_i - x-bar)^2 | f_i * (x_i - x-bar)^2 |
|---|---|---|---|---|---|---|
| 0-10 | 5 | 5 | 25 | -15 | 225 | 1125 |
| 10-20 | 8 | 15 | 120 | -5 | 25 | 200 |
| 20-30 | 15 | 25 | 375 | 5 | 25 | 375 |
| 30-40 | 7 | 35 | 245 | 15 | 225 | 1575 |
N = 35, Sigma f_i * x_i = 765
x-bar = 765/35 = 21.86 (approx.)
Sigma f_i * (x_i - x-bar)^2 = 1125 + 200 + 375 + 1575 = 3275
Variance = 3275/35 = 93.57 (approx.)
SD = sqrt(93.57) = 9.67 (approx.)
Example 4: Shortcut method
For data with class marks 5, 15, 25, 35 and frequencies 5, 8, 15, 7, take A = 25.
| x_i | f_i | d_i = x_i - 25 | f_i * d_i | f_i * d_i^2 |
|---|---|---|---|---|
| 5 | 5 | -20 | -100 | 2000 |
| 15 | 8 | -10 | -80 | 800 |
| 25 | 15 | 0 | 0 | 0 |
| 35 | 7 | 10 | 70 | 700 |
N = 35, Sigma f_i d_i = -110, Sigma f_i d_i^2 = 3500
sigma^2 = 3500/35 - (-110/35)^2 = 100 - (110/35)^2 = 100 - 9.88 = 90.12 (approx.)
Example 5: Coefficient of variation comparison
Team A scores: mean = 50, SD = 10. Team B scores: mean = 40, SD = 8.
CV(A) = (10/50) x 100 = 20%
CV(B) = (8/40) x 100 = 20%
Both teams have equal consistency since CV is the same.
Common Mistakes to Avoid
| 1. Forgetting absolute values in mean deviation: Mean deviation uses | x_i - x-bar | , not (x_i - x-bar). You must take the absolute value before summing. |
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2. Confusing n and N: For ungrouped data, divide by n (number of observations). For grouped data, divide by N = Sigma f_i (total frequency), not by the number of classes.
3. Variance vs standard deviation: Variance is in squared units. Always take the square root to get SD. Do not report variance when the question asks for SD, or vice versa.
4. Class mark calculation: Class mark = (upper + lower)/2. A common error is using the class boundaries incorrectly, especially with continuous classes.
5. Shortcut method sign errors: d_i = x_i - A can be negative. When computing d_i^2, square the negative value correctly. The formula uses d_i^2 (always positive) and (Sigma d_i)^2 (square of the sum, which may have a different sign).
6. CV comparison direction: Smaller CV means more consistent. Students sometimes state the opposite.
7. Mean deviation about mean vs median: These are two different measures. Read the question carefully to determine which central value to use.
Exam Tips
- For 1-mark questions, know: definitions of variance, SD, CV; formula for CV.
- For 3-mark questions, practice: mean deviation for ungrouped data, variance/SD for small datasets.
- For 5-mark questions, practice: variance and SD for grouped data (with frequency table), shortcut method, CV comparison.
- In calculation questions, create a table with columns: x_i, f_i, f_ix_i, (x_i - x-bar), f_i(x_i - x-bar)^2. This is essential for organized work and partial marks.
- Always show the mean calculation first before computing deviations.
- For shortcut method, clearly state the assumed mean A you have chosen.
- When comparing two distributions using CV, clearly state which is more consistent and why.
Quick Revision Notes
NCERT Exercise Problem Patterns
| Exercise | Topic | Typical Question Type |
|---|---|---|
| 13.1 | Range, mean deviation | Find range, MD about mean and median (ungrouped) |
| 13.2 | Mean deviation for grouped data | MD for discrete/continuous distributions |
| 13.3 | Variance and SD | Variance, SD for ungrouped and grouped data |
| 13.4 | Shortcut method, CV | Compute variance by shortcut, compare using CV |
Derivations and Key Proofs
Derivation 1: Shortcut method for variance
Let d_i = x_i - A (deviation from assumed mean A).
x_i - x-bar = (x_i - A) - (x-bar - A) = d_i - d-bar
(x_i - x-bar)^2 = (d_i - d-bar)^2 = d_i^2 - 2d_id-bar + d-bar^2
Sigma f_i(x_i - x-bar)^2 = Sigma f_id_i^2 - 2d-barSigma f_id_i + d-bar^2*Sigma f_i
Since d-bar = (1/N)Sigma f_id_i:
= Sigma f_id_i^2 - 2Nd-bar^2 + Nd-bar^2 = Sigma f_id_i^2 - Nd-bar^2
sigma^2 = (1/N)Sigma f_id_i^2 - d-bar^2 = (1/N)Sigma f_id_i^2 - [(1/N)Sigma f_id_i]^2
Derivation 2: Why MD about median is less than MD about mean
For any set of data, the sum of absolute deviations is minimized about the median. This is a property of the median — it is the point that minimizes the sum of absolute deviations, just as the mean minimizes the sum of squared deviations.
Additional Solved Examples
Example 6: Mean deviation about median
Find the mean deviation about the median for: 3, 9, 5, 3, 12, 10, 18, 5.
Arranged: 3, 3, 5, 5, 9, 10, 12, 18. Median = (5 + 9)/2 = 7
| Deviations from 7: | 3-7 | = 4, | 3-7 | = 4, | 5-7 | = 2, | 5-7 | = 2, | 9-7 | = 2, | 10-7 | = 3, | 12-7 | = 5, | 18-7 | = 11 |
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Sum = 4 + 4 + 2 + 2 + 2 + 3 + 5 + 11 = 33
MD = 33/8 = 4.125
Example 7: Shortcut method step by step
Data: 4, 7, 10, 13, 16. Take A = 10.
| x_i | d_i = x_i - 10 | d_i^2 |
|---|---|---|
| 4 | -6 | 36 |
| 7 | -3 | 9 |
| 10 | 0 | 0 |
| 13 | 3 | 9 |
| 16 | 6 | 36 |
Sigma d_i = 0, Sigma d_i^2 = 90
sigma^2 = 90/5 - (0/5)^2 = 18 - 0 = 18
SD = sqrt(18) = 3*sqrt(2) = 4.24 (approx.)
Example 8: CV for consistency comparison
Batsman A: mean = 60, SD = 12. Batsman B: mean = 48, SD = 8.
CV(A) = (12/60) x 100 = 20%
CV(B) = (8/48) x 100 = 16.67%
Batsman B is more consistent (smaller CV), even though A scores more on average.
Practice Checklist
- Calculate mean deviation (about mean and median) for 3 ungrouped datasets
- Compute variance and SD for 2 grouped data distributions
- Use the shortcut method for at least 2 problems
- Compare consistency of 2 distributions using CV
- Practice creating frequency tables with all required columns
- Verify shortcut method results match direct method results
Quick Revision Notes
1. Range: R = Max - Min (simplest but least informative)
3. Mean deviation about median: Replace x-bar with Median M in the above formula 4. Variance (ungrouped): sigma^2 = (1/n) Sigma (x_i - x-bar)^2 5. Variance (grouped): sigma^2 = (1/N) Sigma f_i(x_i - x-bar)^2 6. Standard deviation: sigma = sqrt(variance) 7. Shortcut method: sigma^2 = (1/N) Sigma f_i d_i^2 - [(1/N) Sigma f_i d_i]^2, where d_i = x_i - A 8. Coefficient of variation: CV = (sigma/x-bar) x 100 9. CV comparison: Smaller CV = more consistent; larger CV = more variable 10. Class mark: (upper + lower)/2 11. N = Sigma f_i (total frequency, not number of classes) 12. Mean deviation about median is always <= mean deviation about mean
| 2. Mean deviation about mean: MD = (1/n) Sigma | x_i - x-bar | (ungrouped); (1/N) Sigma f_i | x_i - x-bar | (grouped) |
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