MathsCBSE Class 11 · Chapter 13

Statistics Class 11: Mean Deviation, Variance and Solved Questions

Updated 14 October 2026 · Fact-checked

Statistics is Chapter 13 of CBSE Class 11 Maths. It covers measures of dispersion: the range, the mean deviation about the mean and about the median, and the variance and standard deviation, each for ungrouped data, discrete frequency distributions and continuous (grouped) frequency distributions, including the shortcut (step-deviation) method.

What Statistics is about

Two cricketers can have the same batting average and yet be very different players: one scores 45 to 55 every innings, the other swings between 5 and 95. The average (a measure of central tendency) cannot show this. We need a measure of dispersion: a number that tells how spread out the data are around the centre.

This chapter builds three such measures:

Range = largest − smallest value: quick, but it uses only two values.
Mean deviation about a centre a (the mean or the median): the average of the distances $∣ x_{i} - a ∣$ .
Variance $σ^{2}$ : the average of the squared deviations from the mean; its square root is the standard deviation $σ$ , which has the same units as the data.

You learn to compute each one for raw (ungrouped) data, for discrete frequency tables, and for continuous class intervals (using class mid-points), and a shortcut method that codes the data as $y_{i} = \frac{x _{i} - A}{h}$ to make the arithmetic easy. Standard deviation is the measure used most in later work, from probability to science. For 2026-27, the NCERT section on analysis of frequency distributions (coefficient of variation) has been removed.

Weightage: This chapter is part of Unit V, Statistics and Probability, which the syllabus gives 12 of the 80 theory marks in the school annual exam (CBSE adds that there is no fixed chapter-wise weightage).Source: CBSE Mathematics (041) syllabus 2026-27, Class XI course structure, Unit V Statistics and Probability

Key concepts

Dispersion and range

NCERT section 13.2 and 13.3

Dispersion (scatter, variability) describes how much the observations differ from each other or from a central value. Two data sets with the same mean can have very different dispersion.

Range = maximum value − minimum value.

Easy to find and easy to understand.
But it depends only on the two extreme values, so a single unusual value changes it a lot, and it ignores how the rest of the data are spread.

Better measures use every observation: mean deviation, variance and standard deviation.

Worked example

Find the range of 12, 7, 15, 9, 21, 10. Two data sets, A: 48, 50, 52 and B: 10, 50, 90, both have mean 50. Compare their ranges. [1 mark]

Try it first, then show the solutionSolution

Range = 21 − 7 = 14
Range of A = 4, range of B = 80: B is far more spread out although the means are equal

Answer: 14; ranges 4 and 80

Mean deviation for ungrouped data

NCERT section 13.4 and 13.4.1

The mean deviation about a of observations $x_{1}, x_{2}, \dots, x_{n}$ is
$M . D . (a) = \frac{1}{n} \sum_{i = 1}^{n} ∣ x_{i} - a ∣$
where a is usually the mean $\overset{x}{ˉ}$ or the median M.

Why the modulus? The plain deviations $x_{i} - \overset{x}{ˉ}$ always add up to 0, so their average says nothing. Taking absolute values measures distance regardless of direction.

Steps:
1. Find the centre (mean, or median after arranging in order).
2. Find each $∣ x_{i} - a ∣$ .
3. Add them and divide by n.

The mean deviation about the median is never larger than the mean deviation about any other point.

Worked example

Find the mean deviation about the median of 5, 9, 2, 14, 7, 11, 6, 8. [3 marks]

Try it first, then show the solutionSolution

In order: 2, 5, 6, 7, 8, 9, 11, 14; n = 8, so median $= \frac{7 + 8}{2} = 7.5$
$∣ x_{i} - 7.5∣$ : 5.5, 2.5, 1.5, 0.5, 0.5, 1.5, 3.5, 6.5; sum = 22
M.D.(M) $= \frac{22}{8} = 2.75$

Answer: 2.75

Mean deviation for grouped data

NCERT section 13.4.2

Discrete frequency distribution (values $x_{i}$ with frequencies $f_{i}$ , total $N = \sum f_{i}$ ):
$M . D . (a) = \frac{1}{N} \sum f_{i} ∣ x_{i} - a ∣$

About the mean: $\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{N}$ .
About the median: find cumulative frequencies; the median is the value whose cumulative frequency first reaches $\frac{N}{2}$ (for even N, NCERT uses the $\frac{N}{2}$ -th and next observation as needed).

Continuous frequency distribution (class intervals): replace each class by its mid-point $x_{i}$ and proceed as above.

Median of grouped data: $M = l + \frac{\frac{N}{2} - C}{f} \times h$ , where l is the lower limit of the median class, C the cumulative frequency before it, f its frequency and h the class width.

Limitation: the modulus makes mean deviation awkward for further algebra, which is why the variance is preferred.

Worked example

Find the mean deviation about the mean: x = 2, 4, 6, 8, 10 with f = 3, 5, 9, 5, 3. [3 marks]

Try it first, then show the solutionSolution

N = 25, $\sum f_{i} x_{i} = 6 + 20 + 54 + 40 + 30 = 150$ , so $\overset{x}{ˉ} = 6$
$∣ x_{i} - 6∣$ = 4, 2, 0, 2, 4; $\sum f_{i} ∣ x_{i} - 6∣ = 12 + 10 + 0 + 10 + 12 = 44$
M.D. $= \frac{44}{25} = 1.76$

Answer: 1.76

Variance and standard deviation: ungrouped data

NCERT section 13.5 and 13.5.1

Instead of absolute values, square the deviations (squares are never negative) and average them:
$σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2}$ (variance)
$σ = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2}$ (standard deviation, same units as the data)

Computing form (avoids decimals in $x_{i} - \overset{x}{ˉ}$ ):
$σ^{2} = \frac{1}{n} \sum x_{i}^{2} - \overset{x}{ˉ}^{2}$
('mean of squares minus square of mean').

Properties:

Adding a constant to every value leaves σ unchanged.
Multiplying every value by k multiplies σ by |k| (and the variance by $k^{2}$ ).
σ = 0 only when all observations are equal.
For the first n natural numbers, $σ^{2} = \frac{n ^{2} - 1}{12}$ .

Worked example

Find the variance and standard deviation of 6, 8, 10, 12, 14. [2 marks]

Try it first, then show the solutionSolution

$\overset{x}{ˉ} = 10$ ; squared deviations 16, 4, 0, 4, 16; sum 40
$σ^{2} = \frac{40}{5} = 8$ , $σ = 22 \approx 2.83$

Answer: Variance 8, SD ≈ 2.83

Standard deviation of frequency distributions

NCERT section 13.5.2 and 13.5.3

For values (or class mid-points) $x_{i}$ with frequencies $f_{i}$ and $N = \sum f_{i}$ :
$σ^{2} = \frac{1}{N} \sum f_{i} (x_{i} - \overset{x}{ˉ})^{2} = \frac{1}{N} \sum f_{i} x_{i}^{2} - \overset{x}{ˉ}^{2}$

Equivalent form used in NCERT:
$σ = \frac{1}{N} N \sum f_{i} x_{i}^{2} - (\sum f_{i} x_{i})^{2}$

Table layout: columns $x_{i}$ , $f_{i}$ , $f_{i} x_{i}$ , $f_{i} x_{i}^{2}$ (or $x_{i} - \overset{x}{ˉ}$ , $(x_{i} - \overset{x}{ˉ})^{2}$ , $f_{i} (x_{i} - \overset{x}{ˉ})^{2}$ ). Total the columns, then substitute. For classes, $x_{i}$ is the mid-point $\frac{lower + upper}{2}$ .

Worked example

Find the variance and SD of x = 2, 4, 6, 8 with frequencies 1, 4, 3, 2. [3 marks]

Try it first, then show the solutionSolution

N = 10; $\sum f_{i} x_{i} = 2 + 16 + 18 + 16 = 52$ , so $\overset{x}{ˉ} = 5.2$
$\sum f_{i} x_{i}^{2} = 4 + 64 + 108 + 128 = 304$
$σ^{2} = \frac{304}{10} - 5. 2^{2} = 30.4 - 27.04 = 3.36$ ; $σ \approx 1.83$

Answer: Variance 3.36, SD ≈ 1.83

Shortcut (step-deviation) method

NCERT section 13.5.4

When the $x_{i}$ are large or the classes have equal width h, code the data:
$y_{i} = \frac{x _{i} - A}{h}$
where A is an assumed mean (usually a middle value). Then
$\overset{x}{ˉ} = A + h \overset{y}{ˉ}, σ_{x}^{2} = h^{2} σ_{y}^{2} = h^{2} [\frac{\sum f _{i} y _{i}^{2}}{N} - (\frac{\sum f _{i} y _{i}}{N})^{2}]$

Why it works: subtracting A shifts the data (no change in spread); dividing by h scales the spread by $\frac{1}{h}$ , so we multiply back by h (or $h^{2}$ for the variance).

The $y_{i}$ are small whole numbers like −2, −1, 0, 1, 2, so the arithmetic is much lighter.

Worked example

Find the mean and SD of the classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 4, 6, 10, 6, 4, by the shortcut method. [3 marks]

Try it first, then show the solutionSolution

Mid-points 5, 15, 25, 35, 45; A = 25, h = 10: y = −2, −1, 0, 1, 2; N = 30
$\sum f_{i} y_{i} = - 8 - 6 + 0 + 6 + 8 = 0$ , so $\overset{x}{ˉ} = 25$ ; $\sum f_{i} y_{i}^{2} = 16 + 6 + 0 + 6 + 16 = 44$
$σ^{2} = 100 (\frac{44}{30} - 0) = 146.67$ ; $σ \approx 12.11$

Answer: Mean 25; SD ≈ 12.11

Maths in real life: Statistics

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Consistency

Suppose batter A scores 40, 50 and 60 in three matches, and batter B scores 10, 50 and 90.

Both average 50. Use the standard deviation to decide who is more consistent.

Try it first, then show the solutionSolution

A: deviations −10, 0, 10; $σ^{2} = \frac{200}{3}$ , $σ \approx 8.2$
B: deviations −40, 0, 40; $σ^{2} = \frac{3200}{3}$ , $σ \approx 32.7$
Smaller SD means less spread

Answer: A is more consistent (SD about 8.2 against 32.7)

Memory hook: Same mean, smaller SD = steadier.

Variance of daily temperatures

Imagine the maximum temperatures (°C) on five days in a hill town are 22, 25, 27, 24 and 22.

Find the variance and SD.

Try it first, then show the solutionSolution

Mean $= \frac{120}{5} = 24$ ; squared deviations 4, 1, 9, 0, 4; sum 18
$σ^{2} = 3.6$ , $σ \approx 1.9$ °C

Answer: Variance 3.6 °C²; SD ≈ 1.9 °C

Memory hook: SD has the same unit as the data; variance has the unit squared.

Effect of a pay rise

Suppose the monthly pocket money of a group of students has mean ₹12 hundred and SD ₹3 hundred, and every parent doubles it and then adds ₹5 hundred.

Find the new mean and SD.

Try it first, then show the solutionSolution

New mean $= 2 \times 12 + 5 = 29$ hundred
Adding 5 does not change spread; doubling doubles it: new SD $= 2 \times 3 = 6$ hundred

Answer: Mean ₹2900, SD ₹600

Memory hook: Shift moves the mean only; scaling stretches both.

Topics in this chapter

Statistics formula sheet

Range: $Range = x_{ma x} - x_{min}$
Mean deviation: $M . D . (a) = \frac{1}{n} \sum ∣ x_{i} - a ∣, M . D . (a) = \frac{1}{N} \sum f_{i} ∣ x_{i} - a ∣$
Median of grouped data: $M = l + \frac{\frac{N}{2} - C}{f} \times h$
Variance (ungrouped): $σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2} = \frac{1}{n} \sum x_{i}^{2} - \overset{x}{ˉ}^{2}$
Variance (frequency distribution): $σ^{2} = \frac{1}{N} \sum f_{i} x_{i}^{2} - (\frac{\sum f _{i} x _{i}}{N})^{2}$
Shortcut method: $y_{i} = \frac{x _{i} - A}{h}, \overset{x}{ˉ} = A + h \overset{y}{ˉ}, σ_{x}^{2} = h^{2} σ_{y}^{2}$
Effect of changing data: $x \to a x + b : \overset{x}{ˉ} \to a \overset{x}{ˉ} + b, σ \to ∣ a ∣ σ$

How the school exam asks Statistics

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Mean deviation of raw data	Mean deviation about the mean or median of a list of numbers (Exercise 13.1, first questions).	Find the centre, list absolute deviations, average them.
Mean deviation of grouped data	Discrete tables and class intervals, about mean or median.	Columns $x_{i}$ , $f_{i}$ , $∣ x_{i} - a ∣$ , $f_{i} ∣ x_{i} - a ∣$ ; mid-points for classes; median formula for grouped data.
Variance and SD	Raw data, the first n natural numbers, discrete and continuous tables (Exercise 13.2).	Use $\frac{\sum x ^{2}}{n} - \overset{x}{ˉ}^{2}$ or the shortcut with coded values.
Missing or corrected observations	Find two missing values from the mean and variance; correct the mean and SD after a wrong entry (Miscellaneous).	Rebuild $\sum x$ and $\sum x^{2}$ from the given mean and variance, adjust, recompute.
Effect of transformations	New mean and SD after multiplying or adding a constant to each observation.	Mean follows the transformation; SD ignores shifts and scales by \|a\|.

Tricks to remember Statistics

Short memory aids. Say each one aloud once, then use it on a question the same day.

Mean of squares minus square of mean
$σ^{2} = \overline{x^{2}} - (\overset{x}{ˉ})^{2}$ : 'MOSS minus SOM'.
Use it when: Quick variance.
Modulus for MD, square for variance
Both stop positive and negative deviations cancelling: MD uses |…|, variance uses (…)².
Use it when: Remembering why each measure exists.
Shift doesn't spread
Adding a constant moves the data but not its spread: SD unchanged.
Use it when: Transformation questions.
Code, compute, multiply back
Shortcut: $y = \frac{x - A}{h}$ , compute, then multiply the variance by $h^{2}$ .
Use it when: Step-deviation method.
Mid-points stand in for classes
For class intervals, every observation in a class is treated as its mid-point.
Use it when: Grouped data.

Statistics: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Effect of a shift · 1 mark · MCQExam level

If each observation of a data set is increased by 10, the standard deviation:
(a) increases by 10
(b) is multiplied by 10
(c) does not change
(d) becomes 0

Try it first, then show the solutionSolution

All deviations from the mean stay the same.

Answer: (c)

Example 2: Sum of deviations · 1 mark · Assertion-reasonExam level

Assertion (A): For any data, $\sum (x_{i} - \overset{x}{ˉ}) = 0$ .
Reason (R): This is why mean deviation uses absolute values of deviations.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

A is a property of the mean; R is a true statement about mean deviation, but it is a consequence of A, not an explanation of it.

Answer: (b)

Example 3: Mean deviation about the median (discrete) · 3 marks · Short answerHarder exam level

Find the mean deviation about the median: x = 3, 6, 9, 12, 15 with f = 2, 4, 6, 4, 2. [3 marks]

Try it first, then show the solutionSolution

N = 18; cumulative frequencies 2, 6, 12, 16, 18. The 9th and 10th values lie in x = 9, so M = 9
$∣ x_{i} - 9∣$ = 6, 3, 0, 3, 6; $\sum f_{i} ∣ x_{i} - 9∣ = 12 + 12 + 0 + 12 + 12 = 48$
M.D.(M) $= \frac{48}{18} \approx 2.67$

Answer: $\frac{8}{3} \approx 2.67$

Example 4: Grouped median and mean deviation · 5 marks · Long answerHarder exam level

For classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 5, 8, 12, 10, 5, find the median and the mean deviation about the median. [5 marks]

Try it first, then show the solutionSolution

N = 40, $\frac{N}{2} = 20$ ; cumulative frequencies 5, 13, 25, 35, 40, so the median class is 20-30
$M = 20 + \frac{20 - 13}{12} \times 10 = 25.83$
Mid-points 5, 15, 25, 35, 45; $∣ x_{i} - M ∣$ ≈ 20.83, 10.83, 0.83, 9.17, 19.17
$\sum f_{i} ∣ x_{i} - M ∣ \approx 104.17 + 86.67 + 10.00 + 91.67 + 95.83 = 388.33$
M.D.(M) $\approx \frac{388.33}{40} \approx 9.71$

Answer: Median ≈ 25.83; M.D. ≈ 9.71

Example 5: Variance of natural numbers · 3 marks · Short answerHarder exam level

Show that the variance of the first n natural numbers is $\frac{n ^{2} - 1}{12}$ . [3 marks]

Try it first, then show the solutionSolution

$\overset{x}{ˉ} = \frac{n + 1}{2}$ and $\frac{\sum x ^{2}}{n} = \frac{( n + 1 ) ( 2 n + 1 )}{6}$
$σ^{2} = \frac{( n + 1 ) ( 2 n + 1 )}{6} - \frac{( n + 1 ) ^{2}}{4} = \frac{( n + 1 ) [ 2 ( 2 n + 1 ) - 3 ( n + 1 )]}{12}$
$= \frac{( n + 1 ) ( n - 1 )}{12} = \frac{n ^{2} - 1}{12}$

Answer: Proved.

Example 6: Discrete SD · 3 marks · Short answerHarder exam level

Find the SD of x = 2, 4, 6, 8 with frequencies 1, 4, 3, 2 using deviations from the mean. [3 marks]

Try it first, then show the solutionSolution

$\overset{x}{ˉ} = 5.2$ ; deviations −3.2, −1.2, 0.8, 2.8; squares 10.24, 1.44, 0.64, 7.84
$\sum f_{i} d_{i}^{2} = 10.24 + 5.76 + 1.92 + 15.68 = 33.6$
$σ^{2} = 3.36$ , $σ \approx 1.83$ (same as the computing formula)

Answer: ≈ 1.83

Example 7: Scaling proof · 3 marks · Short answerHOTS / competency

If the mean and variance of $x_{1}, \dots, x_{n}$ are $\overset{x}{ˉ}$ and $σ^{2}$ , prove that the mean and variance of $a x_{1} + b, \dots, a x_{n} + b$ are $a \overset{x}{ˉ} + b$ and $a^{2} σ^{2}$ . [3 marks]

Try it first, then show the solutionSolution

New mean $= \frac{1}{n} \sum (a x_{i} + b) = a \overset{x}{ˉ} + b$
Each new deviation: $(a x_{i} + b) - (a \overset{x}{ˉ} + b) = a (x_{i} - \overset{x}{ˉ})$
New variance $= \frac{1}{n} \sum a^{2} (x_{i} - \overset{x}{ˉ})^{2} = a^{2} σ^{2}$

Answer: Proved.

Example 8: When is SD zero? · 2 marks · Very short answerHOTS / competency

Prove that the standard deviation of a data set is 0 if and only if all observations are equal. [2 marks]

Try it first, then show the solutionSolution

If all equal c, then $\overset{x}{ˉ} = c$ and every deviation is 0, so σ = 0
If σ = 0, then $\sum (x_{i} - \overset{x}{ˉ})^{2} = 0$ ; a sum of non-negative squares is 0 only if each is 0, so every $x_{i} = \overset{x}{ˉ}$

Answer: Proved.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Two shops' daily sales

Suppose two snack stalls record daily sales (in hundreds of rupees) for five days. Stall P: 8, 10, 12, 10, 10. Stall Q: 4, 16, 10, 6, 14.

(i) · 1 mark
Find the mean sales of each stall.
Answer
Answer: 10 and 10
Both totals are 50.
(ii) · 1 mark
Find the variance of P.
Answer
Answer: 1.6
Squared deviations 4, 0, 4, 0, 0; sum 8; ÷5
(iii) · 1 mark
Find the variance of Q.
Answer
Answer: 20.8
Squared deviations 36, 36, 0, 16, 16; sum 104; ÷5
(iv) · 1 mark
Which stall has steadier sales?
Answer
Answer: P
Smaller variance and SD.

Exam marks

Suppose a class of 30 has marks grouped as 0-10: 4, 10-20: 6, 20-30: 10, 30-40: 6, 40-50: 4.

(i) · 1 mark
Find the mean mark.
Answer
Answer: 25
Symmetric distribution about the middle class.
(ii) · 1 mark
Find the variance using the shortcut method (A = 25, h = 10).
Answer
Answer: ≈ 146.67
$100 \times \frac{44}{30}$
(iii) · 1 mark
Find the SD.
Answer
Answer: ≈ 12.11
$146.67$
(iv) · 1 mark
If the teacher adds 5 grace marks to everyone, what are the new mean and SD?
Answer
Answer: 30 and ≈ 12.11
A shift changes the mean only.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find the mean deviation about the mean of 3, 5, 8, 10, 12, 16, 16. [2 marks] (2 marks)

Step	Marks
$\overset{x}{ˉ} = \frac{70}{7} = 10$ ; $∣ x_{i} - 10∣$ : 7, 5, 2, 0, 2, 6, 6	1
Sum = 28; M.D. $= \frac{28}{7} = 4$	1

The mean and variance of five observations are 6 and 2. Three of them are 4, 6 and 8. Find the other two. [3 marks] (3 marks)

Step	Marks
$\sum x = 30$ , so the other two (x, y) have $x + y = 12$	1
$\frac{\sum x ^{2}}{5} - 36 = 2 \Rightarrow \sum x^{2} = 190$ ; $16 + 36 + 64 = 116$ , so $x^{2} + y^{2} = 74$	1
$2 x y = 144 - 74 = 70$ , xy = 35; x and y are roots of $t^{2} - 12 t + 35 = 0$ : 5 and 7	1

The mean and SD of 25 observations are 20 and 2. Later it was found that one observation was recorded as 15 instead of 25. Find the correct mean and SD. [5 marks] (5 marks)

Step	Marks
Incorrect $\sum x = 25 \times 20 = 500$ ; correct $\sum x = 500 - 15 + 25 = 510$	1
Correct mean $= \frac{510}{25} = 20.4$	1
Incorrect $\sum x^{2}$ : $\frac{\sum x ^{2}}{25} - 400 = 4 \Rightarrow \sum x^{2} = 10100$	1
Correct $\sum x^{2} = 10100 - 225 + 625 = 10500$	1
Correct variance $= \frac{10500}{25} - 20. 4^{2} = 420 - 416.16 = 3.84$ ; SD $= 3.84 \approx 1.96$	1

Where students lose marks in Statistics

The slips examiners see most often, and what to write instead.

Mistake: Forgetting the modulus in mean deviation
Adding signed deviations.
Wrong working
Σ(x − x̄) = 0, so M.D. = 0
Right working
Use Σ|x − x̄|
Do this: Write absolute values in a separate column.
Usually costs: Whole question.
Mistake: Finding the median without ordering
Taking the middle of the list as given.
Do this: Arrange in ascending order first.
Usually costs: 1-2 marks.
Mistake: Not multiplying back by $h^{2}$ in the shortcut method
Stopping at $σ_{y}^{2}$ .
Do this: $σ_{x}^{2} = h^{2} σ_{y}^{2}$ and $σ_{x} = h σ_{y}$ .
Usually costs: 1-2 marks.
Mistake: Dividing by n instead of N for frequency tables
Counting the number of rows.
Do this: Divide by the total frequency $N = \sum f_{i}$ .
Usually costs: 1 mark.
Mistake: Writing SD with squared units
Mixing up variance and SD.
Do this: Variance has squared units; SD has the units of the data.
Usually costs: Usually ½ to 1 mark.
Mistake: Arithmetic slips with $\sum f_{i} x_{i}^{2}$
Squaring $f_{i} x_{i}$ instead of multiplying $f_{i}$ by $x_{i}^{2}$ .
Do this: Compute $x_{i}^{2}$ first, then multiply by $f_{i}$ .
Usually costs: Final answer.

NCERT exercise guide: Statistics

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 13.1 (mean deviation)

What it tests: Mean deviation about the mean and the median for raw data, discrete distributions and class intervals.

How to approach it: Centre first, then the $f_{i} ∣ x_{i} - a ∣$ column.

Same kind as Exercise 13.1

Find the mean deviation about the mean for the classes 0-10, 10-20, 20-30, 30-40, 40-50 with frequencies 4, 6, 10, 6, 4.

Show the solutionSolution

Mid-points 5, 15, 25, 35, 45; $\overset{x}{ˉ} = \frac{20 + 90 + 250 + 210 + 180}{30} = 25$
$\sum f_{i} ∣ x_{i} - 25∣ = 80 + 60 + 0 + 60 + 80 = 280$ ; M.D. $= \frac{280}{30} \approx 9.33$

Answer: About 9.33

Exercise 13.2 (variance and standard deviation)

What it tests: Variance and SD of raw data, of the first n natural numbers, of discrete and continuous distributions, using direct and shortcut methods.

How to approach it: Table with $f_{i} x_{i}$ and $f_{i} x_{i}^{2}$ , or coded $y_{i}$ .

Same kind as Exercise 13.2

Find the variance of the first 10 natural numbers, and of the first 10 even natural numbers.

Show the solutionSolution

$\frac{n ^{2} - 1}{12} = \frac{99}{12} = 8.25$
Doubling every value multiplies the variance by 4: $4 \times 8.25 = 33$

Answer: 8.25 and 33

Miscellaneous Exercise on Chapter 13

What it tests: Finding missing observations from mean and variance, effect of multiplying observations by a constant, correcting mean and SD after wrong or omitted entries.

How to approach it: Rebuild $\sum x$ and $\sum x^{2}$ , adjust, recompute.

Same kind as Miscellaneous Exercise

The mean and SD of some observations are 12 and 3. Each observation is multiplied by 2 and then 5 is added. Find the new mean and SD.

Show the solutionSolution

New mean $= 2 \times 12 + 5 = 29$
New SD $= ∣2∣ \times 3 = 6$ (adding 5 has no effect on spread)

Answer: Mean 29, SD 6

Practice questions on Statistics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The variance of 2, 4, 6, 8, 10 is:
(a) 6
(b) 8
(c) 10
(d) $22$
Hint
Mean 6.
Answer
(b) 8
Q2 · 1 mark · MCQExam level
If every observation is multiplied by −2, the standard deviation is:
(a) multiplied by −2
(b) multiplied by 2
(c) multiplied by 4
(d) unchanged
Hint
SD scales by |a|.
Answer
(b)
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The mean deviation of 5, 5, 5, 5 about the mean is 0.
Reason (R): All observations equal the mean.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Every deviation is 0.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Find the mean deviation about the mean of 10, 12, 14, 16, 18.
Hint
Mean 14.
Answer
Deviations 4, 2, 0, 2, 4; M.D. $= \frac{12}{5} = 2.4$ .
Q5 · 2 marks · Very short answerHarder exam level
Find the SD of 12, 15, 9, 18, 6.
Hint
Mean 12.
Answer
$σ^{2} = \frac{0 + 9 + 9 + 36 + 36}{5} = 18$ ; $σ = 32 \approx 4.24$ .
Q6 · 2 marks · Very short answerHarder exam level
The mean and SD of 8 observations are 15 and 3. If each is multiplied by 4, find the new mean and SD.
Hint
Scale both.
Answer
Mean 60, SD 12.
Q7 · 3 marks · Short answerHarder exam level
Find the mean and variance of x = 10, 20, 30, 40, 50 with frequencies 2, 3, 5, 3, 2 using the shortcut method.
Hint
A = 30, h = 10.
Answer
y = −2, −1, 0, 1, 2; N = 15; $\sum f y = 0$ , so mean = 30; $\sum f y^{2} = 8 + 3 + 0 + 3 + 8 = 22$ ; variance $= 100 \times \frac{22}{15} \approx 146.67$ .
Q8 · 3 marks · Short answerHarder exam level
The mean and SD of 50 observations are 30 and 4. Later, two observations recorded as 34 and 32 were found to be wrong and are removed. Find the new mean.
Hint
Subtract from the total.
Answer
$\sum x = 1500 - 66 = 1434$ for 48 observations: mean $= \frac{1434}{48} = 29.875$ .
Q9 · 3 marks · Short answerHOTS / competency
For the same data, find the new SD after removing 34 and 32.
Hint
Rebuild $\sum x^{2}$ .
Answer
$\sum x^{2} = 50 (16 + 900) = 45800$ ; remove $1156 + 1024 = 2180$ : 43620. Variance $= \frac{43620}{48} - 29.87 5^{2} = 908.75 - 892.52 \approx 16.23$ ; SD ≈ 4.03.
Q10 · 3 marks · Short answerHOTS / competency
Two data sets each have 10 observations. Set A has mean 5 and variance 4; set B has mean 5 and variance 9. Find the variance of the combined 20 observations.
Hint
Combine $\sum x^{2}$ .
Answer
$\sum x^{2}$ : A: 10(4 + 25) = 290; B: 10(9 + 25) = 340. Combined mean 5, $\sum x^{2} = 630$ , variance $= \frac{630}{20} - 25 = 6.5$ .
Q11 · 2 marks · Very short answerHOTS / competency
Without calculating, explain why the mean deviation about the median of 1, 2, 3, 4, 100 is less than its mean deviation about the mean.
Hint
Property of the median.
Answer
The mean deviation is smallest when taken about the median. Here the mean (22) is pulled far from most values by 100, while the median (3) sits among them, so distances to the median are smaller overall (M.D. about median 20.2, about mean 31.2).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Why is the range a weak measure?Show answer
It uses only the two extreme values.
2.Formula for mean deviation about a?Show answer
$\frac{1}{n} \sum ∣ x_{i} - a ∣$ .
3.About which point is the mean deviation least?Show answer
The median.
4.Relation between SD and variance?Show answer
SD = √variance.
5.Variance of the first n natural numbers?Show answer
$\frac{n ^{2} - 1}{12}$ .
6.Effect on SD of adding 10 to every value?Show answer
None.
7.Effect on variance of multiplying every value by 3?Show answer
Multiplied by 9.
8.In the shortcut method, $σ_{x} =$ ?Show answer
$h σ_{y}$ .

Best order to study it

Understand why dispersion matters and learn the range.
Learn mean deviation about the mean for raw data.
Learn mean deviation about the median, including ordering and even n.
Extend mean deviation to discrete and continuous tables.
Learn variance and SD for raw data with both formulae.
Learn SD for frequency distributions with the table method.
Learn the shortcut (step-deviation) method.
Finish with missing-value, correction and transformation problems.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain dispersion with an example?
Can you find the range and say why it is limited?
Can you find mean deviation about the mean and the median of raw data?
Can you find mean deviation for a discrete table?
Can you find mean deviation for class intervals using mid-points?
Can you compute variance as 'mean of squares minus square of mean'?
Can you compute SD for a frequency table?
Can you use the shortcut method correctly?
Can you find missing observations from a given mean and variance?
Can you state the effect of adding or multiplying by a constant?

Statistics: frequently asked questions

Is the coefficient of variation in the 2026-27 syllabus?

No. The NCERT section 'analysis of frequency distributions', which introduces the coefficient of variation for comparing variability, was removed. The syllabus covers range, mean deviation, variance and standard deviation.

Should I use the mean or the median for mean deviation?

Use whichever the question asks for. If it just says 'mean deviation', NCERT usually means about the mean. The mean deviation about the median is the smallest possible.

Why do we prefer standard deviation to mean deviation?

Squares are easier to handle algebraically than absolute values, so variance and SD behave well in formulas (for combining groups, in probability and in later statistics). SD also has the same units as the data.

Do I have to use the shortcut method?

No, any correct method earns full marks. The shortcut method just reduces arithmetic when values are large or classes have equal width.

How is the Class 11 Maths exam set?

It is a school annual exam of 80 marks (3 hours) plus 20 marks of internal assessment; Unit V, Statistics and Probability, which includes this chapter, carries 12 marks. CBSE gives no Class 11 sample paper, so schools usually mix MCQs, assertion-reason, short and long answers and case-based questions.

Based on: CBSE Mathematics (041) syllabus 2026-27 (Classes XI-XII); NCERT Mathematics, Textbook for Class XI, Chapter 13 (Reprint 2026-27)