Statistics – Class 11 Maths Chapter 13 Notes & NCERT PDF

Class 11MathematicsChapter 13NCERT book: Mathematics - Class 11

Notes and a simple summary of Chapter 13 of the NCERT Class 11 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 13: Statistics

Introduction

Chapter 13, Statistics, builds on measures of central tendency (mean, median, mode) from earlier classes. The book explains that an average alone cannot describe data fully: two batsmen may have the same average but very different consistency. What we also need is a measure of how much the observations scatter around a central value — the dispersion. This chapter studies range, mean deviation, variance and standard deviation.

Key Concepts

13.2 Measures of Dispersion

Dispersion is the variability or scatter of data about a central value.

Measures studied: range, mean deviation, variance and standard deviation.

13.3 Range

Range = Maximum value − Minimum value. It gives a rough idea of spread but uses only the two extreme values.

For the data 6, 7, 10, 12, 13, 4, 8, 12 the range is 13 − 4 = 9.

13.4 Mean Deviation

Deviation of xᵢ from a central value a is (xᵢ − a); since deviations have signs and their sum about the mean is zero, we use absolute values |xᵢ − a|.

Mean deviation about a: M.D.(a) = (1/n) Σ|xᵢ − a|. It can be taken about the mean or the median.

#### 13.4.1 Ungrouped data — steps

Find the central value (mean or median).

Find deviations xᵢ − a and take absolute values.

Add them and divide by n.

#### 13.4.2 Grouped data

Discrete frequency distribution: M.D.(x̄) = (1/N) Σ fᵢ|xᵢ − x̄|, where N = Σfᵢ.

Continuous frequency distribution: use the mid-points of classes as xᵢ; for the median, first locate the median class using cumulative frequency, then apply M = l + [(N/2 − C)/f] × h.

Shortcut (step-deviation) for M.D. about the mean: when mid-points are large, take an assumed mean A and width h, compute dᵢ = (xᵢ − A)/h, find x̄ = A + h·(Σfᵢdᵢ/N), and then proceed with |xᵢ − x̄| as usual — this saves heavy multiplication.

#### 13.4.3 Limitations of mean deviation

For highly variable data, the median is not a good central value, so M.D. about the median is unreliable.

The sum of deviations about the mean is used with absolute values, which are not easy to handle algebraically — this motivates variance.

13.5 Variance and Standard Deviation

Variance: σ² = (1/n) Σ(xᵢ − x̄)², the mean of the squared deviations from the mean.

Standard deviation: σ = √variance. It has the same units as the data.

Discrete frequency distribution: σ² = (1/N) Σ fᵢ(xᵢ − x̄)².

Continuous frequency distribution: same formula using class mid-points; a shortcut is σ² = (1/N²)[N Σfᵢxᵢ² − (Σfᵢxᵢ)²], and the step-deviation method uses yᵢ = (xᵢ − A)/h, giving σₓ = h·σᵧ.

Why standard deviation is preferred

Squaring removes the sign problem without absolute values, so σ is easy to handle algebraically.

Squaring gives more weight to observations far from the mean, making σ sensitive to large deviations.

Taking the square root brings the measure back to the same units as the data, so σ can be compared with the mean directly.

Effect of changes in observations

Adding the same constant a to every observation does not change the variance or standard deviation (it only shifts the mean by a).

Multiplying every observation by k multiplies the variance by k² and the standard deviation by |k|.

Worked Examples

M.D. about mean (Example 1): data 6, 7, 10, 12, 13, 4, 8, 12 → mean = 72/8 = 9; absolute deviations 3, 2, 1, 3, 4, 5, 1, 3 sum to 22; M.D.(x̄) = 22/8 = 2.75.

M.D. about median (Example 3): data 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21 (n = 11). Arranged: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 → median = 6th value = 9. Σ|xᵢ − 9| = 58, so M.D.(M) = 58/11 ≈ 5.27.

M.D. about mean, larger data set (Example 2): 20 observations (12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5) have Σxᵢ = 200, so mean = 10; Σ|xᵢ − 10| = 124 → M.D.(x̄) = 124/20 = 6.2.

Discrete frequency distribution (Example 4): xᵢ: 2, 5, 6, 8, 10, 12 with fᵢ: 2, 8, 10, 7, 8, 5. N = 40, Σfᵢxᵢ = 300 → mean = 7.5. The |xᵢ − 7.5| column is 5.5, 2.5, 1.5, 0.5, 2.5, 4.5, so Σfᵢ|xᵢ − 7.5| = 92 and M.D.(x̄) = 92/40 = 2.3.

Variance (Example 8): data 6, 8, 10, …, 24 (10 values). Mean = 15; squared deviations 81, 49, 25, 9, 1, 1, 9, 25, 49, 81 sum to 330 → variance = 33, σ ≈ 5.74.

Discrete distribution (Example 9): xᵢ: 4, 8, 11, 17, 20, 24, 32 with fᵢ: 3, 5, 9, 5, 4, 3, 1. N = 30, Σfᵢxᵢ = 420 → mean = 14; Σfᵢ(xᵢ − 14)² = 1374 → variance = 45.8, σ ≈ 6.77.

Scaling (Example 13): variance of 20 observations is 5; multiplying each by 2 gives new variance 2² × 5 = 20.

Correcting an error (Example 16): n = 100, mean 40, σ = 5.1, but 50 was taken instead of 40. Correct sum = 4000 − 50 + 40 = 3990 → correct mean 39.9; recomputing Σxᵢ² and the variance gives the correct σ = 5.

Formula Sheet

Range = max − min

M.D.(a) = (1/n)Σ|xᵢ − a|; grouped: (1/N)Σfᵢ|xᵢ − a|

σ² = (1/n)Σ(xᵢ − x̄)²; σ = √σ²

Grouped: σ² = (1/N)Σfᵢ(xᵢ − x̄)² = (1/N²)[NΣfᵢxᵢ² − (Σfᵢxᵢ)²]

Step deviation: yᵢ = (xᵢ − A)/h, x̄ = A + h·ȳ, σₓ = h·σᵧ

Median (continuous): M = l + [(N/2 − C)/f] × h

Exam Focus

Mean deviation (Exercise 13.1): show the table — xᵢ, |xᵢ − a| (and fᵢ|xᵢ − a| for grouped data) — then compute.

- Outline: state mean/median → deviation column → sum → divide by n or N.

Variance and SD (Exercise 13.2): prefer the shortcut or step-deviation method for large numbers; always state units for σ.

- Outline: choose A and h → yᵢ column → Σfᵢyᵢ, Σfᵢyᵢ² → σᵧ → σₓ = hσᵧ.

Error correction problems (Miscellaneous exercise): recompute Σxᵢ and Σxᵢ² from the wrong mean and σ, replace the wrong value, then recalculate — the pattern of Example 16.

Key Terms

Dispersion (प्रकीर्णन) — spread of data about a central value

Range (परिसर) — max minus min

Mean deviation (माध्य विचलन)

Absolute value (निरपेक्ष मान)

Variance (प्रसरण)

Standard deviation (मानक विचलन)

Frequency distribution (बारंबारता बंटन)

Class mark / mid-point (वर्ग चिह्न)

Cumulative frequency (संचयी बारंबारता)

Median class (माध्यिका वर्ग)

Step-deviation method (पग-विचलन विधि)

Assumed mean (कल्पित माध्य)

Common Mistakes

Forgetting to take absolute values in mean deviation — signed deviations about the mean add up to zero.

Finding the median without first arranging the data in order.

Using class limits instead of mid-points for continuous distributions.

Forgetting to multiply by h (or h²) when using step deviations.

Thinking that adding a constant to every value changes the standard deviation — it does not.

Reporting variance in the data's units; variance is in squared units, σ is in the original units.

💡 Key Learning Points

✓Measures of central tendency are not enough; dispersion measures (range, mean deviation, variance, standard deviation) describe the spread of data.
✓Mean deviation about a is (1/n)Σ|xᵢ − a|, taken about the mean or the median, using absolute deviations.
✓For grouped data, mid-points are used for continuous classes, and the median class is located with cumulative frequency.
✓Variance σ² = (1/n)Σ(xᵢ − x̄)² and standard deviation σ = √σ², with shortcut and step-deviation methods for grouped data.
✓Adding a constant leaves σ unchanged, while multiplying every value by k multiplies the variance by k².

👨‍🏫 Teaching Tips

→Start with the book's idea of two data sets that share the same mean but have very different spreads, to motivate dispersion.
→Work Example 1 (M.D. about mean) as a class, building the deviation table column by column on the board.
→Contrast M.D. about the mean with M.D. about the median using Example 3 so students see why data must be ordered first.
→Teach step-deviation with one fully worked continuous table, then let students redo it with a different assumed mean to see the answer is unchanged.
→Use a spreadsheet or calculator activity where students add 5 to all values and then double all values, observing the effect on mean and σ.
→Assign the error-correction pattern (Example 16) as a structured three-step template: recover Σx and Σx², correct them, recompute.
→Collect a small real data set (e.g., students' travel times) and compute range, M.D. and σ in groups.

📋 Assessment Questions

Can the student calculate range and mean deviation about the mean and median for ungrouped data?
Can the student compute mean deviation and variance for discrete and continuous frequency distributions?
Can the student apply the shortcut and step-deviation methods correctly?
Can the student explain how adding or multiplying a constant affects mean, variance and standard deviation?
Can the student correct the mean and standard deviation when an observation was recorded wrongly?

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