By the end of this chapter you'll be able to…

  • 1Compute the range of a dataset and explain why it uses only two data points
  • 2Compute mean deviation about the mean and about the median, for both ungrouped and grouped (discrete or continuous) data
  • 3Compute variance and standard deviation for ungrouped and grouped data using the direct formula
  • 4Apply the shortcut (step-deviation) method to compute variance and standard deviation when values are large
  • 5Use the scaling property (multiplying every observation by a constant scales the mean by that constant and the variance by its square) to solve correction and transformation problems
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Why this chapter matters
Two datasets can share the same mean and median yet behave completely differently — this chapter builds the tools (mean deviation, variance, standard deviation) that capture how spread out data actually is, for both raw lists and frequency tables. It's the direct foundation for Class 12's probability distributions and every applied field that reports a result as 'mean plus or minus a spread.'

Statistics

1. Check this before you revise anything

Coefficient of Variation is not part of the current book — and not part of the syllabus either. Coaching material commonly teaches CV () as this chapter's payoff, since it directly answers the "which of these two datasets is more consistent" question the chapter opens with.

But the word never appears anywhere in the book (confirmed by a full-text search — zero hits), and the syllabus line lists only "Range, Mean deviation, variance and standard deviation" — CV isn't named even as a formative-only topic. It simply isn't part of this chapter.

Quartile Deviation is named but explicitly excluded by the book's own words. The book lists four measures of dispersion — range, quartile deviation, mean deviation, standard deviation — then states directly: "we shall study all of these measures of dispersion except the quartile deviation." Don't expect a formula or a worked example for it here.

The old stub's exercise count was badly wrong. It listed four exercises (13.1–13.4, 34 questions total, with 13.3 and 13.4 entirely invented) and never mentioned the real Miscellaneous Exercise. The current book has exactly two numbered exercises plus a Miscellaneous Exercise — 28 questions in total.


2. What this chapter covers

Textbook sectionTopic
13.2–13.3Measures of dispersion; Range
13.4Mean deviation (about mean and about median) for ungrouped and grouped data
13.5Variance and standard deviation, including the shortcut (step-deviation) method

3. Range and mean deviation

Range is the simplest measure of spread: . It uses only the two extreme values and ignores everything in between.

Mean deviation about a value is the average of the absolute deviations from :

Absolute values are essential here — deviations from the mean always sum to zero, so a plain average of signed deviations is useless as a measure of spread. In practice is either the mean or the median .

For grouped data (discrete or continuous, using class midpoints for continuous data), weight each deviation by its frequency:

For a continuous distribution's median, use the same interpolation formula as earlier classes: , where is the median class's lower boundary, the cumulative frequency before it, its own frequency, and its width.

Worked, mirroring the textbook's own Example 1. Find the mean deviation about the mean for . Mean . Absolute deviations: , summing to . .


4. Variance and standard deviation

Squaring the deviations instead of taking absolute values gives a measure that's easier to work with algebraically — this is variance, and its square root is the standard deviation (taken in the original units, since variance is in squared units):

Shortcut (step-deviation) method, useful when the values or class midpoints are large. With assumed mean , class width , and :

A useful scaling property, worth knowing even though the book states it only through a worked example, not a boxed formula. If every observation is multiplied by a constant , the new mean is and the new variance is — squaring the deviations means the scale factor gets squared too.

Worked, mirroring the textbook's own Example 8. Find the variance of (step , ten terms). Using the step-deviation method with assumed mean , : the deviations run , giving and . Mean , and variance , so .


Summary

  • Range Maximum Minimum; the simplest but crudest measure of spread.
  • Mean deviation (or for grouped data), computed about the mean or the median.
  • Variance and standard deviation ; for grouped data, weight by frequency and use class midpoints for continuous distributions.
  • Shortcut method: with .
  • Scaling every observation by a constant multiplies the mean by and the variance by .
  • Quartile Deviation is named as a measure of dispersion but explicitly excluded by the book's own words; Coefficient of Variation is absent from both the book and the syllabus entirely.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Range
Range = Maximum value - Minimum value
Uses only the two extreme values; a rough, quick measure of spread
Mean deviation about a value a
M.D.(a) = sum|xi-a| / n (ungrouped); M.D.(a) = sum fi|xi-a| / N (grouped, N=sum fi)
a is usually the mean or the median; absolute values are essential since signed deviations from the mean always sum to zero
Median of a continuous distribution
M = l + [(N/2 - C)/f] . h
l = median class's lower boundary, C = cumulative frequency before it, f = its own frequency, h = its width; needed before computing M.D.(M) or median for grouped continuous data
Variance and standard deviation, ungrouped
sigma^2 = sum(xi-x_bar)^2 / n, sigma = sqrt(sigma^2)
Squaring (instead of absolute value) makes this measure algebraically tractable; sigma is in the same units as the original data
Variance and standard deviation, grouped
sigma^2 = sum fi(xi-x_bar)^2 / N
Use class midpoints as xi for continuous data; identical in structure to the ungrouped formula with frequency weighting
Shortcut (step-deviation) method
sigma^2 = (h^2/N^2)[N.sum(fi.yi^2) - (sum fi.yi)^2], where yi = (xi-A)/h
A is an assumed mean, h the class width; avoids large-number arithmetic when xi values are big
Scaling property of mean and variance
If yi = a.xi for all i, then y_bar = a.x_bar and variance(y) = a^2 . variance(x)
Shown via a worked example in the book rather than a boxed formula, but directly needed for several Miscellaneous Exercise questions
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Computing mean deviation using signed deviations instead of absolute values
Sum(xi - x_bar) always equals exactly 0, since the mean is the balance point of the data. Mean deviation must use |xi-a|, the absolute value, or the whole calculation collapses to zero.
WATCH OUT
Dividing by (n-1) instead of n when computing variance
This chapter's variance is sigma^2 = sum(xi-x_bar)^2/n, dividing by n. The (n-1) version (sample variance, Bessel's correction) belongs to inferential statistics, not this syllabus.
WATCH OUT
Stopping at variance and forgetting to take the square root for standard deviation
Standard deviation is sigma = sqrt(variance), not the variance itself. A question that asks for standard deviation is not answered by variance alone.
WATCH OUT
Using the raw class boundaries directly for a discrete-looking age or count distribution with gaps (like 16-20, 21-25, ...)
Classes like 16-20 and 21-25 have a gap (20 to 21), so they must first be converted to continuous form by subtracting 0.5 from each lower limit and adding 0.5 to each upper limit, exactly as the book's own hint on this type of question states.
WATCH OUT
Expecting quartile deviation to be tested from this chapter
The book explicitly lists quartile deviation as a measure of dispersion but then states it studies every measure except this one. No formula or worked example for it appears anywhere in the chapter.
WATCH OUT
Expecting Coefficient of Variation as this chapter's way of comparing two datasets
Coefficient of Variation doesn't appear anywhere in the current book or the syllabus line for this chapter at all — not even as a formative-only topic. It's genuinely useful statistics, but not examinable from this specific chapter.
WATCH OUT
Recomputing an entire mean and variance from scratch after a single wrong observation is corrected
Use the running totals instead: recover the original sum (n times mean) and sum of squares (n times [variance + mean^2]), adjust by removing or replacing the incorrect value, then recompute the mean and variance from the adjusted totals and count.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Statistics?

8 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

8 questions~6 min worth ~12 marks in CBSE exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Range = Maximum - Minimum, the simplest and crudest measure of spread
  • Mean deviation M.D.(a) = sum|xi-a|/n (or frequency-weighted for grouped data), using absolute values since signed deviations from the mean sum to zero
  • For a continuous distribution's median: M = l + [(N/2-C)/f].h
  • Variance sigma^2 = sum(xi-x_bar)^2/n; standard deviation sigma = sqrt(sigma^2), in the original units
  • Grouped-data variance uses frequency weighting and, for continuous data, class midpoints
  • Shortcut method: sigma^2 = (h^2/N^2)[N.sum(fi.yi^2)-(sum fi.yi)^2], yi=(xi-A)/h
  • Scaling every observation by a constant a multiplies the mean by a and the variance by a^2; adding a constant shifts the mean but leaves variance unchanged
  • Quartile Deviation is named but explicitly excluded by the book; Coefficient of Variation is absent from both the book and the syllabus for this chapter

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Part of Unit V's 12-mark Statistics and Probability block (shared with Probability; no chapter-wise split, per CBSE)

Question typeMarks eachTypical countWhat it tests
Range and Mean Deviation2-41Mean deviation about the mean or median, for ungrouped or grouped data
Variance and Standard Deviation, Including the Shortcut Method4-61Direct or step-deviation computation of variance and standard deviation for ungrouped or grouped data
Scaling Property and Correcting a Wrong Observation3-50-1Recovering unknown observations, applying the scaling property, or correcting a mean/variance after a misrecorded value is found
Prep strategy
  • Build a computation table with columns for xi (or class midpoint), fi, fi.xi, and either |xi-a| or (xi-x_bar)^2 before doing any arithmetic — this is what most board-exam mark schemes expect to see
  • For continuous data with gaps between class limits (like 16-20, 21-25), apply the continuity correction (subtract/add 0.5) before computing the median or midpoints
  • For 'find the remaining observations' problems, set up the sum and sum-of-squares equations first, then solve the resulting quadratic

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Investment risk in finance

The standard deviation of a stock's historical returns is a standard measure of its volatility (risk) — a higher standard deviation means a wider range of likely outcomes around the average return.

Manufacturing quality control

Production processes are monitored by tracking whether measurements (like component dimensions) stay within a small number of standard deviations of the target value, flagging a process as 'out of control' when spread grows too large.

Comparing exam performance across sections

Two class sections can have identical average marks yet very different standard deviations — one section performing consistently, the other having a wide spread of both very high and very low scorers, exactly the batsman-comparison idea this chapter opens with.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

1
Compute and double-check the mean first — every other formula in this chapter depends on it being correct
2
For grouped data, always set up the full computation table (xi or midpoint, fi, fi.xi, deviation, squared or absolute deviation) rather than trying to track running sums mentally
3
For continuous classes with gaps between them, apply the 0.5 continuity correction before finding the median class or computing midpoints
4
For 'correct a wrong observation' problems, work from the recovered total sum and total sum of squares rather than trying to adjust the mean or variance directly

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH
Coefficient of Variation (CV = sigma/x_bar x 100) is absent from this chapter and its syllabus entirely, but is the standard way to compare relative variability across datasets with different means — see the Advanced JEE problem above for a worked comparison
STRETCH
The combined variance of two or more groups (accounting for both each group's own spread and how far its mean sits from the overall combined mean) generalises the single-group variance formula in this chapter and is a common JEE Main extension — see the JEE section above
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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainRecovering two unknown observations from mean and varianceSum and sum-of-squares system

The mean and variance of 6 observations are 8 and 16 respectively. If four of the observations are 2, 6, 10, 12, find the remaining two.

Stuck? Show the approach

Use the mean to find the total sum, and the variance to find the total sum of squares; subtract the four known values' contributions to get a system in the two unknowns.

Show the full solution

Total sum = 6(8)=48. Known sum = 2+6+10+12=30, so the two unknowns sum to 18. Total sum of squares = 6(16+64)=480. Known sum of squares = 4+36+100+144=284, so the two unknowns' squares sum to 196. Let the unknowns be a,b: a+b=18, a^2+b^2=196. Squaring the first equation: (a+b)^2=324=a^2+2ab+b^2=196+2ab, so 2ab=128, giving ab=64. So a and b are the two roots of t^2-18t+64=0. By the quadratic formula: t=(18 +- sqrt(324-256))/2=(18 +- sqrt(68))/2=9 +- sqrt(17).

Answer: The two remaining observations are 9+sqrt(17) and 9-sqrt(17), approximately 13.12 and 4.88
The trap

Assuming the two unknowns are always nice integers — unlike the book's own Miscellaneous Exercise Q1 and Q2 (which are constructed to give integer answers), a JEE-style version of this problem may not factor neatly, and the quadratic formula is the reliable fallback.

JEE MainCombined variance of two groupsBeyond this chapter's own scope — a standard JEE extension

Group A has 10 observations with mean 20 and variance 4. Group B has 15 observations with mean 25 and variance 9. Find the variance of all 25 observations combined.

Stuck? Show the approach

This combined-groups formula isn't in the current book, but it's a standard extension: the combined mean is a weighted average, and the combined variance adds each group's own variance to a term accounting for how far each group's mean sits from the combined mean.

Show the full solution

Combined mean = (10(20)+15(25))/25 = (200+375)/25 = 23. Combined variance = [n1(var1+(mean1-combined)^2) + n2(var2+(mean2-combined)^2)] / (n1+n2) = [10(4+9) + 15(9+4)] / 25 = [10(13)+15(13)]/25 = 25(13)/25 = 13.

Answer: 13
The trap

Simply averaging the two variances (getting (4+9)/2=6.5) ignores both the different group sizes and the spread caused by the two groups having different means — both must be accounted for.

JEE MainEffect of adding a constant to every observationShift invariance of variance

If the mean and variance of 8 observations are 15 and 25, find the new mean and variance if 5 is added to every observation.

Stuck? Show the approach

Adding a constant shifts every observation and the mean by the same amount, but leaves every deviation from the mean — and hence the variance — unchanged.

Show the full solution

New mean = 15+5=20. Since (xi+5)-(x_bar+5) = xi - x_bar for every i, the deviations are identical to before, so the variance is unchanged.

Answer: New mean = 20, variance unchanged at 25
The trap

Confusing this with the scaling property — adding a constant (shifting) leaves variance unchanged, while multiplying by a constant (scaling) multiplies variance by the square of that constant. The two operations affect variance completely differently.

JEE MainMean deviation compared to standard deviation for the same dataComputing both measures on one dataset

For the data 3, 6, 9, 12, 15, compute both the mean deviation about the mean and the standard deviation, and verify that mean deviation is less than or equal to the standard deviation.

Stuck? Show the approach

Compute the mean once, then find both the average absolute deviation and the square root of the average squared deviation.

Show the full solution

Mean = 45/5=9. Deviations: -6,-3,0,3,6. Absolute deviations: 6,3,0,3,6, summing to 18, so M.D.=18/5=3.6. Squared deviations: 36,9,0,9,36, summing to 90, so variance=90/5=18, and SD=sqrt(18)=3(sqrt2), approximately 4.24.

Answer: Mean deviation = 3.6, standard deviation is approximately 4.24, confirming M.D. <= SD
The trap

Assuming the two measures should be numerically close for every dataset — the gap between them depends on how the data is distributed, and standard deviation is always at least as large as mean deviation about the mean.

JEE AdvancedCoefficient of Variation — beyond this chapter's own scopeComparing relative variability across two different means

Batsman A has mean score 53 with standard deviation 26.7. Batsman B has mean score 53 with standard deviation 5.4 (values close to the chapter's own opening example). If a third batsman C has mean score 80 and standard deviation 12, can standard deviation alone tell you whether C is more or less consistent than A and B?

Stuck? Show the approach

Coefficient of Variation isn't covered anywhere in this book or its syllabus line, but it directly resolves this: CV = (standard deviation / mean) x 100 normalises spread relative to the mean, making comparisons meaningful across different means.

Show the full solution

Comparing raw standard deviations (26.7 for A, 5.4 for B, 12 for C) is misleading since C's mean (80) is much larger than A and B's (53) — a bigger absolute spread is less surprising around a bigger mean. Computing CV instead: CV(A) = (26.7/53)x100 = 50.4%. CV(B) = (5.4/53)x100 = 10.2%. CV(C) = (12/80)x100 = 15%. Relative to their own means, C (15%) is more consistent than A (50.4%) but less consistent than B (10.2%).

Answer: No — standard deviation alone cannot be compared meaningfully across different means; Coefficient of Variation is needed, and it shows B is most consistent, then C, then A
The trap

Ranking datasets by raw standard deviation when their means differ substantially — this is exactly the comparison Coefficient of Variation exists to make fair, even though it's not part of this specific chapter's own content.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardHigh
JEE MainMedium
JEE AdvancedLow

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

No. It doesn't appear anywhere in the current book, and the syllabus line for this chapter names only range, mean deviation, variance, and standard deviation — not even as a formative-only topic.

No. The book names it as one of the four measures of dispersion but then explicitly says it will study every measure except quartile deviation. No formula or example for it appears anywhere in the chapter.

Both exist to avoid the problem that signed deviations from the mean always sum to zero. Squaring is more useful algebraically (differentiable, decomposes nicely for grouped or combined data) than taking absolute values, which is why variance and standard deviation are the more widely used measures.

Whenever the class midpoints or data values are large, making direct computation of (xi-x_bar)^2 tedious. The shortcut method rescales around an assumed mean A and class width h, keeping the arithmetic in small numbers throughout.
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