Statistics Class 11 Notes | CBSE Maths Chapter 13

Statistics is Chapter 13 of CBSE Class 11 Maths — and it is the chapter that finally answers a question you have been asking since school: how spread out is my data? Two batsmen can have the same average score, yet one is wildly inconsistent and the other is rock-steady. Measures of dispersion are the tools that tell them apart, and this chapter hands you every one of them.

By the end of these notes you will be able to compute range, mean deviation, variance, and standard deviation for both ungrouped and grouped data, use the lightning-fast shortcut (step-deviation) method, and compare two data sets using the coefficient of variation. This is a scoring chapter carrying roughly 6–8 marks in boards, and it is the direct foundation for Probability and all of Class 12 statistics.


Table of Contents


Key Concepts

1. Measures of Dispersion

A measure of central tendency (mean, median, mode) tells you where the centre of the data lies, but it hides how scattered the values are. A measure of dispersion tells you how far the observations are spread out from that centre.

The four measures of dispersion you study in this chapter are range, mean deviation, variance, and standard deviation. The last three measure scatter about a central value, which is why they are far more useful than range alone.

Key idea: Less dispersion means the data is more consistent and the average is more reliable.


2. Range

The range is the simplest measure of dispersion — it is just the difference between the largest and the smallest observation.

Range = Maximum value − Minimum value

  • The coefficient of range = (Max − Min)/(Max + Min), a unit-free comparison.
  • It uses only two extreme values, so it ignores how the rest of the data is distributed.
  • It is highly affected by outliers, which is its biggest weakness.

3. Mean Deviation

Mean deviation is the average of the absolute distances of all observations from a central value (the mean or the median). We take absolute values so that positive and negative deviations do not cancel out.

For ungrouped data with n observations:

  • About the mean: M.D.(x̄) = (1/n) Σ |xᵢ − x̄|
  • About the median: M.D.(M) = (1/n) Σ |xᵢ − M|

For grouped (frequency) data with total frequency N = Σfᵢ:

  • About the mean: M.D.(x̄) = (1/N) Σ fᵢ|xᵢ − x̄|
  • About the median: M.D.(M) = (1/N) Σ fᵢ|xᵢ − M|

Important property: Mean deviation is least when taken about the median. So if minimising deviation matters, the median is the best central value to use.


4. Variance

The drawback of mean deviation is the awkward modulus sign. Variance fixes this by squaring the deviations instead of taking their absolute value — squares are always positive and far easier to handle algebraically.

Variance is the mean of the squared deviations from the mean. It is denoted by σ² (sigma squared).

  • Ungrouped data: σ² = (1/n) Σ (xᵢ − x̄)²
  • Grouped data: σ² = (1/N) Σ fᵢ(xᵢ − x̄)²

Note: Because deviations are squared, variance has the square of the original units (e.g., if data is in cm, variance is in cm²). That is why we usually take its square root.


5. Standard Deviation

Standard deviation (σ) is the positive square root of the variance. It brings the measure of spread back to the original units of the data, which makes it the most widely used measure of dispersion.

  • Ungrouped data: σ = √[(1/n) Σ (xᵢ − x̄)²]
  • Grouped data: σ = √[(1/N) Σ fᵢ(xᵢ − x̄)²]

A handy computational form that avoids first finding the mean is:

σ = √[ (Σxᵢ²)/n − ( (Σxᵢ)/n )² ]

For grouped data this becomes σ = (1/N) √[ N Σfᵢxᵢ² − (Σfᵢxᵢ)² ].


6. Shortcut (Step-Deviation) Method

When the values xᵢ are large, computing deviations from the mean is tedious. The shortcut method shifts the origin to an assumed mean A and scales by the class width h, which keeps the arithmetic small.

Let dᵢ = (xᵢ − A)/h. Then the standard deviation is:

σ = h × √[ (Σfᵢdᵢ²)/N − ( (Σfᵢdᵢ)/N )² ]

  • A is the assumed mean (usually a middle class mark).
  • h is the common class width.
  • The answer is exactly the same as the direct method — only the computation is lighter.

7. Coefficient of Variation

Standard deviation has units, so you cannot fairly compare the spread of, say, heights (in cm) with weights (in kg). The coefficient of variation (C.V.) solves this by expressing standard deviation as a percentage of the mean — a pure, unit-free number.

C.V. = (σ / x̄) × 100  (x̄ ≠ 0)

  • It measures relative variability or consistency.
  • The series with the greater C.V. is more variable (less consistent).
  • The series with the smaller C.V. is more consistent (more stable / uniform).

8. Comparing Two Frequency Distributions

When two data sets have the same mean, you compare them directly by standard deviation — the one with the smaller σ is more consistent.

When the two data sets have different means, standard deviation alone is misleading. You must use the coefficient of variation, because it accounts for the size of the mean.

Rule of thumb: For consistency or stability questions (most consistent player, most stable share price), always reach for C.V. — smaller C.V. wins.


Weightage in Board & Entrance Exams

ExamTypical WeightageMost-Tested Areas
CBSE Board (Class 11)6–8 marksMean deviation, standard deviation (grouped), coefficient of variation
JEE Main1 questionVariance/SD of ungrouped data, effect of change of scale, combined SD
CUET / School tests2–3 questionsRange, mean deviation about median, C.V. comparison

[TABLE: Question-type split — VSA (1 mark): definitions, range, coefficient of range; SA (2–3 marks): mean deviation about mean/median; LA (5 marks): standard deviation of grouped data by shortcut method, comparing two distributions using C.V.]


Important Definitions

TermDefinition
DispersionThe extent to which observations are scattered about a central value
RangeDifference between the largest and smallest observation: Max − Min
Mean deviationMean of the absolute deviations from a central value: (1/N) Σ fᵢ|xᵢ − a|
VarianceMean of the squared deviations from the mean: σ² = (1/N) Σ fᵢ(xᵢ − x̄)²
Standard deviationPositive square root of the variance: σ = √(variance)
Coefficient of rangeRelative range: (Max − Min)/(Max + Min)
Coefficient of variationRelative variability: C.V. = (σ/x̄) × 100
ConsistencyA data set with smaller C.V. is more consistent (less variable)

Solved Examples

Example 1

Find the range and the coefficient of range for the data: 24, 18, 32, 41, 16, 29.

Answer: Max = 41, Min = 16. Range = 41 − 16 = 25. Coefficient of range = (41 − 16)/(41 + 16) = 25/57 ≈ 0.439.

Example 2

Find the mean deviation about the mean for: 6, 8, 10, 12, 14.

Answer: x̄ = (6 + 8 + 10 + 12 + 14)/5 = 50/5 = 10. Absolute deviations: 4, 2, 0, 2, 4; their sum = 12. M.D.(x̄) = 12/5 = 2.4.

Example 3

Find the mean deviation about the median for: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.

Answer: Arrange: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 (n = 11). Median = 6th term = 9. Σ|xᵢ − 9| = 6+6+5+4+2+0+1+3+9+10+12 = 58. M.D.(M) = 58/11 ≈ 5.27.

Example 4

Find the variance and standard deviation of: 2, 4, 6, 8, 10.

Answer: x̄ = 30/5 = 6. Squared deviations: 16, 4, 0, 4, 16; sum = 40. σ² = 40/5 = 8. σ = √8 = 2√2 ≈ 2.83.

Example 5

Using the computational formula, find the standard deviation of: 4, 7, 8, 9, 10, 12, 14 (n = 7).

Answer: Σxᵢ = 64, so x̄ = 64/7. Σxᵢ² = 16+49+64+81+100+144+196 = 650. σ² = 650/7 − (64/7)² = 92.857 − 83.592 = 9.265. σ = √9.265 ≈ 3.04.

Example 6

The mean and standard deviation of the runs scored by two batsmen are: Batsman A — mean 50, σ = 5; Batsman B — mean 40, σ = 5. Who is more consistent?

Answer: C.V.(A) = (5/50) × 100 = 10%; C.V.(B) = (5/40) × 100 = 12.5%. Since A has the smaller C.V., Batsman A is more consistent even though both have the same standard deviation.


Important Questions for Board Exams

1-Mark Questions (VSA)

  1. Define dispersion. Name any two measures of dispersion.
  2. Write the formula for the coefficient of range.
  3. About which central value is the mean deviation least?
  4. What is the relationship between variance and standard deviation?
  5. If every observation of a data set is increased by 5, what happens to its standard deviation?

2–3-Mark Questions (SA)

  1. Find the mean deviation about the mean for: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
  2. Find the mean deviation about the median for: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.
  3. Calculate the variance and standard deviation of the first n natural numbers and state the formula.
  4. Explain, with an example, why the coefficient of variation is preferred over the standard deviation when comparing two distributions with different means.

5-Mark Questions (LA)

  1. Find the mean, variance, and standard deviation of the following grouped data using the shortcut (step-deviation) method, given classes 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 5, 8, 15, 16, 6.
  2. The means and standard deviations of two distributions of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the standard deviation of the combined distribution of 250 items.
  3. From the prices of shares X and Y given over several days, determine which share is more stable in value using the coefficient of variation.

Quick Revision Points

  • Dispersion = scatter of data about a central value; measures: range, mean deviation, variance, SD
  • Range = Max − Min; coefficient of range = (Max − Min)/(Max + Min)
  • M.D. about mean = (1/N) Σ fᵢ|xᵢ − x̄|; about median = (1/N) Σ fᵢ|xᵢ − M|
  • Mean deviation is least about the median
  • Variance σ² = (1/N) Σ fᵢ(xᵢ − x̄)²; units are squared
  • Standard deviation σ = √(variance); same units as the data
  • Computational form: σ = √[ (Σxᵢ²)/n − ((Σxᵢ)/n)² ]
  • Shortcut method: σ = h √[ (Σfᵢdᵢ²)/N − ((Σfᵢdᵢ)/N)² ], where dᵢ = (xᵢ − A)/h
  • Coefficient of variation C.V. = (σ/x̄) × 100 — unit-free
  • Smaller C.V. ⇒ more consistent; greater C.V. ⇒ more variable
  • Adding a constant to all data leaves σ unchanged; multiplying by k multiplies σ by |k|

Next Chapter: Chapter 14 — Probability

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