Statistics

Two batsmen, A and B, each play ten matches. Batsman A scores 30, 91, 0, 64, 42, 80, 30, 5, 117, 71; batsman B scores 53, 46, 48, 50, 53, 53, 58, 60, 57, 52. Work out the mean of each series and you get exactly the same number, 53, for both — and the median, too, comes out 53 for both. By every measure of central tendency you learned in earlier classes, these two batsmen look identical. Yet anyone glancing at the raw numbers can see they are nothing alike: A veers from a duck to a century, while B never strays far from the low fifties. A single average has quietly erased the most important difference between a reliably consistent player and a wildly unpredictable one. This chapter exists to recover that missing information. The quantity that measures how spread out or bunched together data is, once you already know its centre, is called dispersion, and a single number that captures it is a measure of dispersion. Class 11 Statistics builds four such measures in increasing order of sophistication: the range, the crudest but quickest; the mean deviation, which averages how far each observation strays from a chosen centre; and variance and its square root, the standard deviation, which square the deviations instead of taking absolute values, buying a measure that is far more amenable to further algebra and is, for that reason, the standard tool of statistics, probability, and quality control from here through JEE, NEET, and beyond. The chapter's real engine is a repeating three-beat rhythm: define a measure for raw, ungrouped data; extend it to discrete frequency distributions where each value carries a weight $f_i$ ; then extend it again to continuous (grouped) distributions where class mid-points stand in for the values. A step-deviation shortcut, built by shifting and rescaling the origin, tames the arithmetic at every stage. By the end you should be able to compute and interpret the range, mean deviation about the mean or median, and variance and standard deviation for any of these three data forms — and to explain, using the algebra of deviations, exactly why standard deviation, not mean deviation, became statistics's workhorse measure of spread.

Key takeaways

Dispersion measures how scattered data is around a central value; two data sets can share the same mean and median yet have wildly different spread, as the batsman A vs batsman B example shows.
Range = Maximum value minus Minimum value — the simplest measure of dispersion, but it ignores every observation except the two extremes.
Mean deviation about $a$ , written $\text{M.D.}(a)$ , is the mean of the absolute deviations $|x_i-a|$ from a chosen central value $a$ (usually the mean $\bar x$ or the median $M$ ); absolute values are essential because the plain deviations from the mean always sum to zero.
For ungrouped data, $\text{M.D.}(\bar x)=\dfrac{1}{n}\sum|x_i-\bar x|$ and $\text{M.D.}(M)=\dfrac{1}{n}\sum|x_i-M|$ ; for a frequency distribution with $N=\sum f_i$ , the same formulas gain an $f_i$ weight: $\text{M.D.}(\bar x)=\dfrac{1}{N}\sum f_i|x_i-\bar x|$ .
Variance $\sigma^2=\dfrac{1}{n}\sum(x_i-\bar x)^2$ replaces absolute values with squares, avoiding the zero-cancellation problem while staying algebraically well-behaved; standard deviation $\sigma=+\sqrt{\sigma^2}$ restores the original units, since variance is measured in squared units.
A computational shortcut avoids ever subtracting the mean observation by observation: $\sigma^2=\dfrac{1}{N}\sum f_ix_i^2-\bar x^2=\dfrac{1}{N}\sqrt{N\sum f_ix_i^2-\left(\sum f_ix_i\right)^2}$ in standard-deviation form.
The step-deviation method takes an assumed mean $A$ and class width $h$ , defines $y_i=\dfrac{x_i-A}{h}$ , and recovers $\bar x=A+h\bar y$ and $\sigma_x=h\sigma_y$ — turning large or awkward numbers into small integers before any arithmetic is done.
If every observation is scaled by a constant $a$ and shifted by a constant $b$ (i.e. $y_i=ax_i+b$ ), the new mean is $a\bar x+b$ but the new variance is $a^2\sigma^2$ — adding a constant never changes the spread, while multiplying by $a$ scales the variance by $a^2$ and the standard deviation by $|a|$ .
Mean deviation is intuitive but algebraically awkward (absolute value functions cannot be differentiated or manipulated cleanly), and mean deviation about mean is always at least as large as mean deviation about median is not generally true either way in general — this is why standard deviation, despite squaring and then re-rooting, is preferred for further statistical theory.
Key idea: dispersion measures how far data is spread from its centre, and while mean deviation averages absolute deviations directly, variance and standard deviation square the deviations first and take a square root at the end, which is what makes them algebraically tractable enough to survive into every later chapter of statistics and probability.

The core insight

Every measure of dispersion in this chapter — range, mean deviation, variance, standard deviation — is built by taking the observations' deviations from a chosen centre and combining them into one number, differing only in whether that combination uses the two extremes, an absolute value, or a square, with variance and its square root, standard deviation, winning out as the standard tool because squaring avoids the zero-sum problem of raw deviations while remaining fully compatible with algebraic manipulation.

What this chapter covers

Why central tendency is not enough — two data sets with identical mean and median can have completely different spread; the batsman A vs batsman B example motivates every measure that follows.
Range — the crudest measure of dispersion, maximum minus minimum, and its obvious limitation of ignoring every observation in between.
Mean deviation — the mean of absolute deviations from a chosen central value $a$ , defined first for ungrouped data, then extended to discrete and continuous frequency distributions, with a worked step-deviation shortcut for the mean itself.
Limitations of mean deviation — why the absolute-value operation, despite being intuitive, is algebraically inconvenient and is not carried forward into deeper statistical theory.
Variance and standard deviation — replacing absolute values with squares to avoid the zero-cancellation problem, first motivated with two data sets of equal mean but different spread (sets A and B of the chapter), then defined precisely.
Variance and SD for discrete and continuous frequency distributions, including the algebraically shortcut computational formula $\sigma^2=\frac{1}{N}\sum f_ix_i^2-\bar x^2$ .
The step-deviation (shortcut) method for both mean deviation and variance, using an assumed mean $A$ and common factor $h$ .
Effect of linear transformations — what happens to the mean and variance when every observation is shifted by a constant, scaled by a constant, or both; this is the tool behind several of the chapter's miscellaneous examples.

The core ideas

Why a measure of central tendency alone is not enough

A measure of central tendency — mean, median, or mode — gives a single number around which data clusters, but says nothing about how tightly or loosely the data clusters around that number. The batsman example in the overview makes this concrete: both series have mean and median $53$ , yet A's range of scores runs from $0$ to $117$ while B's runs only from $46$ to $60$ . Plotting both series as dots on a number line shows B's dots huddled close to $53$ while A's dots are scattered across the whole range — a picture worth keeping in mind throughout this chapter, since every measure of dispersion defined below is, at bottom, an attempt to turn that picture into a single descriptive number. This single number is called a measure of dispersion, and this chapter develops four of them in order: range, mean deviation, variance, and standard deviation.

Range

The simplest measure of dispersion looks only at the two most extreme values in a data set.

\text{Range} = \text{Maximum value} - \text{Minimum value} \qquad (1)

For batsman A, Range $=117-0=117$ ; for batsman B, Range $=60-46=14$ . Since Range of A is far larger than Range of B, the range correctly signals that A's scores are more scattered. where: the range is measured in the same units as the raw data and depends only on the two extreme observations — every other observation in the data set is completely ignored, which is exactly the range's weakness: a single unusually large or small value (an outlier) can distort it badly, and two data sets with very different internal spread can still share the same range if their extremes happen to match.

Mean deviation: averaging distance from a centre

To use more than just the two extremes, define the deviation of an observation $x_i$ from a fixed value $a$ as $x_i-a$ . If $a$ is a measure of central tendency such as the mean or median, then some deviations are positive (observations above $a$ ) and some are negative (observations below $a$ ), and — as the next derivation proves — the deviations from the mean always sum to exactly zero. Averaging the raw deviations is therefore useless; instead, take the absolute value of each deviation, $|x_i-a|$ , which measures pure distance from $a$ regardless of direction, and average those. This average is the mean deviation about $a$ , denoted $\text{M.D.}(a)$ :

\text{M.D.}(a)=\frac{\text{sum of absolute values of deviations from }a}{\text{number of observations}} \qquad (2)

where: $a$ can in principle be any fixed number, but mean deviation about the mean $\bar x$ and mean deviation about the median $M$ are the two versions used in practice. For $n$ ungrouped observations $x_1,x_2,\ldots,x_n$ ,

\text{M.D.}(\bar x)=\frac{1}{n}\sum_{i=1}^{n}|x_i-\bar x|, \qquad \text{M.D.}(M)=\frac{1}{n}\sum_{i=1}^{n}|x_i-M| \qquad (3)

For a discrete frequency distribution, where distinct values $x_1,\ldots,x_n$ occur with frequencies $f_1,\ldots,f_n$ and $N=\sum f_i$ , every sum above simply picks up an $f_i$ weight:

\text{M.D.}(\bar x)=\frac{1}{N}\sum_{i=1}^n f_i|x_i-\bar x|, \qquad \text{M.D.}(M)=\frac{1}{N}\sum_{i=1}^n f_i|x_i-M| \qquad (4)

For a continuous frequency distribution (class intervals with no gaps), the same formulas (4) apply directly, with $x_i$ now standing for the mid-point of the $i$ -th class — exactly as when computing the mean of grouped data, the entire frequency of a class is treated as concentrated at its mid-point. Finding the median for a continuous distribution uses the usual formula $M=l+\dfrac{\frac{N}{2}-C}{f}\times h$ , where $l$ is the lower limit of the median class (the class whose cumulative frequency first equals or exceeds $N/2$ ), $f$ its frequency, $h$ its width, and $C$ the cumulative frequency of the class just before it.

Shifting the origin: the step-deviation shortcut

When the mid-points $x_i$ are large or inconveniently spaced, computing $\bar x$ directly is tedious. The step-deviation method fixes this by picking an assumed mean $A$ — any convenient value near the middle of the data — and a common factor $h$ (usually the class width), then working with the small numbers $d_i=x_i-A$ or, after dividing by $h$ , the even smaller step-deviations $y_i=\dfrac{x_i-A}{h}$ . This is simply a change of origin (shifting zero to $A$ ) followed by a change of scale (measuring in units of $h$ instead of $1$ ), and it recovers the true mean by

\bar x = A + \frac{\sum f_i d_i}{N}, \qquad \text{or equivalently} \qquad \bar x = A + \left(\frac{\sum f_i y_i}{N}\right)h \qquad (5)

The rest of the mean-deviation calculation — finding $|x_i-\bar x|$ and then $\frac{1}{N}\sum f_i|x_i-\bar x|$ — proceeds exactly as before; the step-deviation trick only simplifies the computation of $\bar x$ itself, not the deviations taken from it afterward.

Figure

Two dot plots on parallel number lines comparing the scattered runs of batsman A with the tightly clustered runs of batsman B around the shared mean of fifty three

Variance and standard deviation

Mean deviation's absolute values, while intuitive, are hard to differentiate and manipulate algebraically. An alternative way to stop positive and negative deviations from cancelling is to square each deviation instead: $(x_i-\bar x)^2$ is always non-negative regardless of the sign of $x_i-\bar x$ , and it is zero only when $x_i=\bar x$ exactly. But the raw sum of squared deviations, $\sum(x_i-\bar x)^2$ , is not by itself a fair measure of dispersion, because it grows simply by adding more observations — a set of $31$ numbers can have a larger sum of squared deviations than a set of $6$ numbers even though the $6$ -number set is actually more spread out (this is worked through explicitly for two such sets, of size $6$ and $31$ , both with mean $30$ , in the chapter's own motivating example). The fix is to average the squared deviations rather than sum them; this average is called the variance, denoted $\sigma^2$ (sigma squared):

\sigma^2=\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar x)^2 \qquad (6)

Because variance involves squared deviations, it is measured in squared units (runs $^2$ , rupees $^2$ , and so on) — units that do not match the original data. Taking the positive square root restores comparability with the original data and gives the standard deviation, denoted $\sigma$ :

\sigma=\sqrt{\frac{1}{n}\sum_{i=1}^{n}(x_i-\bar x)^2} \qquad (7)

where: $\sigma$ is always taken as the non-negative square root, since a "negative spread" has no meaning; standard deviation is the most widely used measure of dispersion in statistics precisely because squaring (rather than taking absolute values) keeps the whole calculation inside ordinary algebra, letting it be manipulated, differentiated, and combined across data sets in ways mean deviation cannot be. For a discrete frequency distribution with values $x_i$ , frequencies $f_i$ , and $N=\sum f_i$ ,

\sigma^2=\frac{1}{N}\sum_{i=1}^n f_i(x_i-\bar x)^2, \qquad \sigma=\sqrt{\frac{1}{N}\sum_{i=1}^n f_i(x_i-\bar x)^2} \qquad (8)

For a continuous frequency distribution, the same formula (8) applies with $x_i$ read as class mid-points, exactly as for mean deviation. An equivalent computational formula, derived in full below, avoids computing every individual deviation $x_i-\bar x$ by working instead with $\sum f_ix_i$ and $\sum f_ix_i^2$ directly:

\sigma=\frac{1}{N}\sqrt{N\sum_{i=1}^n f_ix_i^2-\left(\sum_{i=1}^n f_ix_i\right)^2} \qquad (9)

Just as with mean deviation, a step-deviation (shortcut) method simplifies variance for awkward numbers. With assumed mean $A$ , common factor $h$ , and $y_i=\dfrac{x_i-A}{h}$ , one can show $\sigma_x^2=h^2\sigma_y^2$ , which combines with (9) applied to the $y$ -values to give

\sigma=\frac{h}{N}\sqrt{N\sum_{i=1}^n f_iy_i^2-\left(\sum_{i=1}^n f_iy_i\right)^2} \qquad (10)

where: $y_i$ are small step-deviations, so $\sum f_iy_i$ and $\sum f_iy_i^2$ involve only small integers, and the final multiplication by $h$ restores the correct scale.

Figure

A number line first with the origin at zero and raw class mid points then shifted so the origin sits at the assumed mean A with small step deviations y i marked in units of the class width h

Three results worth proving carefully

Derivation AWhy the deviations of any data set from its own mean always add up to zero

The goal is to prove, rigorously and in general, that for any $n$ observations $x_1,x_2,\ldots,x_n$ with mean $\bar x$ , the sum $\sum_{i=1}^n(x_i-\bar x)$ is identically zero — not just true in one worked example — because this single fact is the entire reason mean deviation must use absolute values rather than plain deviations, and it silently underlies several shortcuts used later in the chapter, including why $\sum f_ix_i=N\bar x$ can be substituted freely into other formulas.

Step 1. Start from the definition of the mean of $n$ observations: $\bar x=\dfrac{1}{n}\sum_{i=1}^n x_i$ , which on rearranging gives $\sum_{i=1}^n x_i = n\bar x$ . This single rearrangement is the only fact needed for the rest of the proof.

Step 2. Expand the sum of deviations using linearity of summation: $\sum_{i=1}^n(x_i-\bar x)=\sum_{i=1}^n x_i-\sum_{i=1}^n\bar x$ . The second sum adds the constant $\bar x$ to itself $n$ times, so $\sum_{i=1}^n\bar x=n\bar x$ .

Step 3. Substitute the Step 1 result into Step 2: $\sum_{i=1}^n(x_i-\bar x)=n\bar x-n\bar x=0$ . This holds for any data set whatsoever, with no restriction on the individual values $x_i$ — it is a direct consequence of how the mean itself is defined, not a coincidence of any particular example.

Step 4. The consequence for mean deviation follows immediately: since the raw (signed) deviations always cancel to exactly zero, their mean is always $0/n=0$ , which carries no information about spread whatsoever. To recover a genuine measure of dispersion, the deviations must first be converted to non-negative quantities — either by taking absolute values, which produces mean deviation, or by squaring, which produces variance — before averaging; this is precisely why both remaining measures of dispersion in the chapter are built on a non-linear transformation of the deviations rather than on the deviations themselves.

Go deeper · Derivation B · 1 minThe computational shortcut formula for variance and standard deviation

Derivation BThe computational shortcut formula for variance and standard deviation

The goal is to derive the alternative formula $\sigma^2=\dfrac{1}{N}\sum f_ix_i^2-\bar x^2$ from the defining formula $\sigma^2=\dfrac{1}{N}\sum f_i(x_i-\bar x)^2$ , because expanding a squared bracket and simplifying is exactly what turns a tedious observation-by-observation subtraction into a calculation using only the two running sums $\sum f_ix_i$ and $\sum f_ix_i^2$ , both of which are easy to accumulate directly from a frequency table without ever computing an individual deviation.

Step 1. Begin from the definition and expand the square algebraically: $(x_i-\bar x)^2=x_i^2+\bar x^2-2\bar xx_i$ , so $\sigma^2=\dfrac{1}{N}\sum_{i=1}^n f_i\left(x_i^2+\bar x^2-2\bar xx_i\right)$ .

Step 2. Distribute the sum across the three terms inside the bracket, treating $\bar x$ as a constant that can be pulled outside each partial sum: $\sigma^2=\dfrac{1}{N}\left[\sum_{i=1}^n f_ix_i^2+\bar x^2\sum_{i=1}^n f_i-2\bar x\sum_{i=1}^n f_ix_i\right]$ .

Step 3. Simplify each piece using known facts: $\sum_{i=1}^n f_i=N$ by definition, and $\sum_{i=1}^n f_ix_i=N\bar x$ from the definition of the mean of a frequency distribution (the frequency-weighted analogue of the fact proved in the previous derivation). Substituting, $\sigma^2=\dfrac{1}{N}\left[\sum f_ix_i^2+\bar x^2\cdot N-2\bar x\cdot N\bar x\right]=\dfrac{1}{N}\sum f_ix_i^2+\bar x^2-2\bar x^2$ .

Step 4. Combine the remaining $\bar x^2$ terms: $\bar x^2-2\bar x^2=-\bar x^2$ , giving the compact result $\sigma^2=\dfrac{1}{N}\sum f_ix_i^2-\bar x^2$ . Substituting $\bar x=\dfrac{1}{N}\sum f_ix_i$ once more and clearing the outer fraction produces the fully expanded form used for hand calculation, $\sigma=\dfrac{1}{N}\sqrt{N\sum f_ix_i^2-\left(\sum f_ix_i\right)^2}$ — algebraically identical to the defining formula, but requiring only the two running totals $\sum f_ix_i$ and $\sum f_ix_i^2$ rather than a fresh subtraction $x_i-\bar x$ for every single data point.

Go deeper · Derivation C · 1 minHow shifting and scaling every observation changes the mean and the variance

Derivation CHow shifting and scaling every observation changes the mean and the variance

The goal is to determine, in full generality, what happens to the mean and variance of a data set when every observation $x_i$ is replaced by $y_i=ax_i+b$ for fixed constants $a$ and $b$ — a single unified result that explains both why adding a constant to every observation leaves the variance completely unchanged, and why multiplying every observation by a constant $a$ scales the variance by $a^2$ , two facts that appear as separate miscellaneous examples in this chapter but are really one theorem viewed from two angles.

Step 1. Find the new mean $\bar y$ in terms of the old mean $\bar x$ . By definition, $\bar y=\dfrac{1}{n}\sum_{i=1}^n y_i=\dfrac{1}{n}\sum_{i=1}^n(ax_i+b)=\dfrac{a}{n}\sum_{i=1}^n x_i+\dfrac{1}{n}\sum_{i=1}^n b=a\bar x+b$ , since $\frac1n\sum x_i=\bar x$ and averaging the constant $b$ over $n$ terms just returns $b$ .

Step 2. Compute the deviation of each new observation from the new mean: $y_i-\bar y=(ax_i+b)-(a\bar x+b)=a(x_i-\bar x)$ . Notice that the additive constant $b$ has cancelled completely — it shifts every observation and the mean by exactly the same amount, leaving every deviation, and hence the entire spread of the data, untouched.

Step 3. Square this deviation and average over all observations to get the new variance: $\sigma_y^2=\dfrac{1}{n}\sum_{i=1}^n(y_i-\bar y)^2=\dfrac{1}{n}\sum_{i=1}^n a^2(x_i-\bar x)^2=a^2\cdot\dfrac{1}{n}\sum_{i=1}^n(x_i-\bar x)^2=a^2\sigma_x^2$ , since $a^2$ is a constant that factors outside the summation.

Step 4. Read off the two named special cases. Setting $a=1$ (pure shift, no scaling) gives $\sigma_y^2=\sigma_x^2$ — adding or subtracting the same number from every observation never changes the variance, because a shift moves the whole data set and its mean together, leaving every gap between them fixed. Setting $b=0$ (pure scaling, no shift) gives $\sigma_y^2=a^2\sigma_x^2$ , so $\sigma_y=|a|\sigma_x$ — multiplying every observation by $a$ multiplies the standard deviation by $|a|$ and the variance by $a^2$ , exactly matching the chapter's own numerical illustration where doubling twenty observations of variance $5$ produces a new variance of $2^2\times5=20$ .

Worked examples

Board level - mean deviation about the mean for ungrouped data (NCERT Example 1)

Find the mean deviation about the mean for the data 6, 7, 10, 12, 13, 4, 8, 12.

Step 1. Compute the mean: $\bar x=\dfrac{6+7+10+12+13+4+8+12}{8}=\dfrac{72}{8}=9$ .

Step 2. Find the deviation of each observation from the mean, $x_i-\bar x$ : $-3,-2,1,3,4,-5,-1,3$ . Notice these sum to zero, exactly as the first derivation proves they must.

Step 3. Take the absolute value of each deviation: $3,2,1,3,4,5,1,3$ , and add them: $3+2+1+3+4+5+1+3=22$ .

Step 4. Divide by the number of observations, $n=8$ : $\text{M.D.}(\bar x)=\dfrac{22}{8}$ . Answer: the mean deviation about the mean is $2.75$ .

Board level - variance and standard deviation of a continuous distribution by the step-deviation method (NCERT Example 12)

Calculate the mean, variance, and standard deviation for the distribution with classes 30-40, 40-50, 50-60, 60-70, 70-80, 80-90, 90-100 and respective frequencies 3, 7, 12, 15, 8, 3, 2.

Step 1. Take mid-points $x_i=35,45,55,65,75,85,95$ and choose assumed mean $A=65$ (the mid-point of the modal-looking middle class) with $h=10$ . Compute step-deviations $y_i=\dfrac{x_i-65}{10}$ : these are $-3,-2,-1,0,1,2,3$ .

Step 2. Tabulate $f_iy_i$ : $-9,-14,-12,0,8,6,6$ , summing to $\sum f_iy_i=-15$ ; and $f_iy_i^2$ : $27,28,12,0,8,12,18$ , summing to $\sum f_iy_i^2=105$ . The total frequency is $N=3+7+12+15+8+3+2=50$ .

Step 3. Find the mean using formula (5): $\bar x=A+\left(\dfrac{\sum f_iy_i}{N}\right)h=65+\left(\dfrac{-15}{50}\right)\times10=65-3=62$ .

Step 4. Find the variance using the step-deviation shortcut formula (10) squared, $\sigma^2=\dfrac{h^2}{N^2}\left[N\sum f_iy_i^2-\left(\sum f_iy_i\right)^2\right]=\dfrac{100}{2500}\left[50\times105-(-15)^2\right]=\dfrac{100}{2500}\left[5250-225\right]=\dfrac{100\times5025}{2500}=201$ . Answer: mean $=62$ , variance $=201$ , and standard deviation $\sigma=\sqrt{201}\approx14.18$ .

Competitive level (JEE / NEET) - recovering two missing observations from mean and variance, then rescaling (extends NCERT Miscellaneous Example 14)

The mean of five observations is $4.4$ and their variance is $8.24$ . Three of the observations are $1,2,6$ ; find the other two observations $x$ and $y$ (with $x>y$ ), and hence state the new variance of all five observations if every observation is first multiplied by $3$ .

Step 1. Use the mean condition. With five observations $1,2,6,x,y$ , the mean gives $\dfrac{1+2+6+x+y}{5}=4.4$ , so $9+x+y=22$ , i.e. $x+y=13$ .

Step 2. Use the variance condition. Since $\sigma^2=\dfrac{1}{5}\sum(x_i-\bar x)^2=8.24$ , the sum of squared deviations is $5\times8.24=41.2$ . Writing $\sum x_i^2$ directly (using $\sigma^2=\dfrac1n\sum x_i^2-\bar x^2$ from the shortcut derivation, applied here to ungrouped data) gives $\dfrac{1}{5}\sum x_i^2-(4.4)^2=8.24$ , so $\sum x_i^2=5(8.24+19.36)=5\times27.6=138$ . Since $1^2+2^2+6^2=1+4+36=41$ , the two unknowns satisfy $x^2+y^2=138-41=97$ .

Step 3. Combine the two conditions. From $x+y=13$ , squaring gives $x^2+y^2+2xy=169$ ; subtracting $x^2+y^2=97$ gives $2xy=72$ , so $xy=36$ . Then $(x-y)^2=(x+y)^2-4xy=169-144=25$ , so $x-y=\pm5$ . Taking $x>y$ , combine with $x+y=13$ to get $x=9,\ y=4$ .

Step 4. Apply the scaling result from the third derivation: multiplying every observation by $a=3$ multiplies the variance by $a^2=9$ , regardless of what the individual observations are. Answer: the missing observations are $x=9$ and $y=4$ , and after multiplying every observation by $3$ the new variance is $9\times8.24=74.16$ . Timing note: about two minutes. Setting up $x+y$ from the mean and $x^2+y^2$ from the variance is the standard two-equation pattern for "find the missing observations" questions — always convert the variance condition into a statement about $\sum x_i^2$ using the shortcut formula rather than expanding $(x_i-\bar x)^2$ term by term, since that avoids carrying decimals through several steps. The final rescaling step is a free mark if the linear-transformation result ( $a^2\sigma^2$ ) is recognised immediately rather than recomputed from scratch — JEE frequently tacks such one-line extensions onto a classic NCERT miscellaneous problem.

Common traps

Forgetting that deviations from the mean always sum to zero, and mistakenly trying to use that sum, rather than the sum of absolute or squared deviations, as a measure of dispersion.
Computing mean deviation about the median using the mean's deviations, or vice versa — always recompute deviations from the specific central value ( $\bar x$ or $M$ ) that the question asks for; the two mean deviations are generally different numbers for the same data.
Forgetting the $f_i$ weights when a discrete or continuous frequency distribution is involved — every sum ( $\sum x_i$ , $\sum|x_i-\bar x|$ , $\sum(x_i-\bar x)^2$ ) must be replaced by its frequency-weighted version $\sum f_ix_i$ , $\sum f_i|x_i-\bar x|$ , $\sum f_i(x_i-\bar x)^2$ , and divided by $N=\sum f_i$ , not by the number of classes.
Using the class boundaries instead of the class mid-point for continuous data — mean, mean deviation, and variance for grouped data are always computed by treating the entire frequency of a class as concentrated at its mid-point, not its lower or upper limit.
Dropping the square root at the end, reporting variance when standard deviation was asked for (or vice versa) — remember $\sigma$ is in the same units as the raw data, while $\sigma^2$ is in squared units.
Sign errors in step-deviations: forgetting that $y_i=\dfrac{x_i-A}{h}$ can be negative for classes below the assumed mean, and that $\sum f_iy_i$ can therefore itself be negative — this negative sum is exactly what shifts $\bar x$ below $A$ , so a positive-only running total is a red flag to recheck the arithmetic.
Misremembering the shortcut formula's sign: it is $\sigma^2=\dfrac{1}{N}\sum f_ix_i^2-\bar x^2$ with a minus, not a plus — a common transcription slip that silently inflates the computed variance.
Believing that multiplying all observations by $a$ scales the standard deviation by $a$ , not $a^2$ — the variance scales by $a^2$ , so the standard deviation scales by $|a|$ , and this distinction is exactly what the third derivation exists to nail down.

Common trapThe single biggest misconception in this chapter is treating mean deviation and standard deviation as interchangeable measures that will always agree with each other

Mean deviation about the mean and standard deviation are built from the same raw material — deviations of each observation from a central value — but they combine those deviations in fundamentally different ways, one using absolute values and the other using squares followed by a square root, and there is no algebraic reason the two resulting numbers should be equal, or even close, for a given data set. Squaring gives disproportionately more weight to observations that are far from the mean than taking an absolute value does, so standard deviation is systematically more sensitive to outliers than mean deviation is; two data sets can have identical mean deviation yet noticeably different standard deviation, or the reverse. The deeper reason standard deviation is preferred throughout the rest of statistics is not that it is "more correct," but that squares — unlike absolute values — can be expanded, differentiated, and combined algebraically (exactly what the second and third derivations exploit), which is what lets variance support the far more advanced machinery built on top of it in probability and inferential statistics. Never assume a small mean deviation guarantees a small standard deviation, or treat the two numbers as reporting the same fact about a data set in different units.

You should be able to…

State the definition of range, mean deviation about a value $a$ , variance, and standard deviation, and explain in words what each one measures.
Prove that the sum of deviations of any data set from its own mean is always zero, and explain why this forces mean deviation to use absolute values.
Compute mean deviation about the mean and about the median for ungrouped data, discrete frequency distributions, and continuous frequency distributions.
Compute variance and standard deviation for the same three data forms, using both the direct definition and the computational shortcut formula $\sigma^2=\frac1N\sum f_ix_i^2-\bar x^2$ .
Apply the step-deviation method, with an assumed mean $A$ and common factor $h$ , to simplify mean, mean deviation, and variance calculations for large or awkward numbers.
Derive and apply the rule that adding a constant to every observation leaves variance unchanged, while multiplying every observation by $a$ scales variance by $a^2$ and standard deviation by $|a|$ .
Recover missing observations from given values of the mean and variance by translating both conditions into equations in $\sum x_i$ and $\sum x_i^2$ .
Correct a previously computed mean and standard deviation when one recorded observation turns out to have been wrong, without recomputing the entire data set from scratch.

Ask the tutor: if the shortcut variance formula, the step-deviation method, or the missing-observations style of problem still feels shaky, ask the AI tutor to walk through a fresh mean-deviation and variance calculation with you, one step at a time, until the pattern is automatic.

Why do we bother with variance and standard deviation when mean deviation already measures spread?

Mean deviation is simple and intuitive, but the absolute value function inside it is algebraically awkward — it cannot be expanded, differentiated, or combined across data sets the way a square can. Variance, built from squared deviations, avoids this problem entirely: the derivation of the shortcut formula $\sigma^2=\frac1N\sum f_ix_i^2-\bar x^2$ only works because squares expand into simple polynomial terms that absolute values do not. This algebraic convenience is exactly why standard deviation, not mean deviation, becomes the foundation for every later topic in statistics and probability, including the normal distribution and hypothesis testing you will meet after Class 11.

Is mean deviation about the median always smaller than mean deviation about the mean?

For many data sets, yes — the median minimises the sum of absolute deviations, so mean deviation about the median is generally the smallest possible mean deviation about any fixed value, including the mean. But this is a property of absolute deviations specifically, not a rule you should treat as automatic without checking, and the chapter itself notes that when data is highly skewed, the median stops being a good representative centre, so mean deviation about the median calculated for such a series cannot be fully relied upon either. Always compute the one the question actually asks for rather than assuming they are close.

Does adding the same number to every observation change the mean deviation, variance, or standard deviation?

No — none of them change. Adding a constant $b$ to every observation shifts the mean by exactly $b$ as well, so every deviation $x_i-\bar x$ stays exactly the same as before (the shift cancels out completely). This is proved in general in the third derivation of this chapter, using the linear transformation $y_i=ax_i+b$ with $a=1$ . It only fails to hold if you multiply, rather than add — multiplying by $a$ scales variance by $a^2$ and standard deviation by $|a|$ , which is the other half of that same derivation.

How do I decide whether to use the direct formula or the step-deviation shortcut for variance?

Use the step-deviation method whenever the mid-points or raw values are large, unevenly convenient numbers, or when the class width $h$ is constant across a continuous distribution — subtracting a nearby assumed mean $A$ and dividing by $h$ turns the arithmetic into small integers that are far easier to square and sum by hand. If the values are already small (single digits, for instance), the direct formula is just as fast and skips the extra bookkeeping of choosing $A$ and $h$ . Both methods give exactly the same final answer; the step-deviation method is a computational convenience, not a different definition.

If a data value was recorded incorrectly and I already computed mean and standard deviation, do I have to start over from scratch?

Not from scratch — you can correct the running totals directly. Since $\bar x=\frac1N\sum x_i$ , an incorrect sum can be fixed by subtracting the wrong value and adding the correct one, then dividing by $N$ again for the corrected mean. For standard deviation, use the shortcut formula's building block $\sum x_i^2$ : subtract the square of the wrong value and add the square of the correct value, then reapply $\sigma=\sqrt{\frac1N\sum x_i^2-\bar x^2}$ with the corrected mean. This running-total correction, rather than recomputing every individual deviation, is exactly the technique used in the chapter's own example about a student who mistakenly recorded one observation as $50$ instead of $40$ .