NDA Maths · Statistics

Dispersion — Standard Deviation, Variance, Mean Deviation

How spread out the data is around its centre — mean deviation, variance and standard deviation each measure spread on a different scale.

All Statistics notes NDA Maths strategy

Why this matters

44 PYQs across 2017–2026. 22 EASY + 18 MODERATE + 4 HARD — almost every paper has one. The favourite shapes are linear-transformation effects on SD, the computational identity that links mean-of-squares to mean-squared and variance, and coefficient-of-variation comparisons. Master the six concepts below and dispersion becomes formulaic, not intimidating.

Test yourself — a quick 12-question Statistics recall quiz, instant score.

Concept 1 of 7: Mean Deviation

Average distance between each observation and a chosen centre — almost always the mean or the median. It is the simplest measure of spread, but its absolute-value formula makes algebra clunky, which is why SD is preferred in higher statistics.

Definition

For $n$ observations $x_1,\ldots,x_n$ and a reference value $A$ , the mean deviation about $A$ is the average of the absolute deviations from $A$ . When $A$ is the median, the mean deviation is minimum among all possible $A$ .

Mean Deviation about A

\text{MD}(A) = \dfrac{1}{n}\sum_{i=1}^{n}|x_i - A|

$A$ reference point (typically mean or median)
$|x_i - A|$ absolute deviation of $x_i$ from $A$

Diagram · mean deviation = average distance from the centre

MD = (3 + 1 + 0 + 4) / 4 = 2

Take each value's distance from the centre (the red segments, signs dropped) and average them. Mean deviation can be taken about the mean or the median; about the median it is smallest.

Worked example

Find the mean deviation of

2, 4, 6, 8, 10

about the mean.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2024 · Q110Easy

Example 1 · Statistics · Dispersion — Standard Deviation, Variance, Mean Deviation

What is the mean deviation of the first 10 natural numbers?

Mean deviation about median is always $\leq$ about the mean

Among all reference points

A

, the median minimises

\sum|x_i - A|

. If a PYQ asks for the minimum mean deviation, the answer uses the median, not the mean.

Drill 10 more on mean deviation

Concept 2 of 7: Variance

Average of the squared distances from the mean. Squaring kills the absolute-value bracket and gives variance much better algebraic properties — but the unit becomes the data's unit squared (cm becomes cm², rupees becomes rupees²), which is why we usually report its square root, the standard deviation.

Definition

For $n$ observations with mean $\bar{x}$ , the variance is the average of the squared deviations from $\bar{x}$ . It can also be computed using the identity $\sigma^2 = \overline{x^2} - \bar{x}^2$ , where $\overline{x^2}$ is the mean of the squares.

Variance — two equivalent forms

\sigma^2 = \dfrac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = \dfrac{\sum x_i^2}{n} - \bar{x}^2

$\sigma^2$ variance
$\bar{x}$ arithmetic mean
$\sum x_i^2$ sum of squares of observations

Visualization · move the points, watch the squared deviations

x1: 2.0x2: 4.0x3: 6.0x4: 8.0x5: 10.0

Variance σ² = 8.00SD σ = 2.83Σ(xᵢ − x̄)² = 40.00

Each red square has side |xᵢ − x̄|, so its AREA is the squared deviation. Variance is the AVERAGE area. Pull all points toward the mean — every square shrinks. Pull them apart — they grow.

Worked example

Find the variance of

2, 4, 6, 8, 10

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2021 · Q116Moderate

Example 2 · Statistics · Dispersion — Standard Deviation, Variance, Mean Deviation

\sum_{i=1}^{10}x_{i}=110

and

\sum_{i=1}^{10}x_{i}^{2}=1540

, then what is the variance?

Computational form saves time on $\Sigma x_i^2$ -style PYQs

When you are given

\sum x_i

and

\sum x_i^2

directly, use

\sigma^2 = \overline{x^2} - \bar{x}^2

— not the original definition. NDA papers favour this shape because it tests whether you remember the identity.

Drill 6 more on variance

Concept 3 of 7: Standard Deviation

Square root of the variance — brings the unit back to the data's original unit. SD is what statisticians actually report when they talk about "spread". If the data is in centimetres, SD is in centimetres; if in rupees, SD is in rupees. Variance lives in squared units.

Definition

The standard deviation is the non-negative square root of the variance. It has the same unit as the original observations.

Standard Deviation

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

$\sigma$ standard deviation (always $\geq 0$ )

Worked example

Find the standard deviation of

2, 4, 6, 8, 10

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2023 · Q103Moderate

Example 3 · Statistics · Dispersion — Standard Deviation, Variance, Mean Deviation

If the mean and sum of squares of 10 observations are 40 and 16160 respectively, then what is the standard deviation?

SD and mean deviation share units; variance does not

If the data is measured in cm, SD and MD are also in cm but variance is in cm². A PYQ asks "which has the same unit as the mean?" — answer is SD or MD, not variance.

Drill 5 more on standard deviation

Concept 4 of 7: Linear Transformation of SD and Variance

Adding a constant shifts every value but the spread between them is unchanged — so shift has zero effect on SD or variance. Multiplying by

a

stretches all distances by

|a|

, so SD scales by

|a|

and variance by

a^2

Definition

If $Y = aX + b$ is a linear transformation of $X$ , then the variance of $Y$ is $a^2$ times the variance of $X$ , and the standard deviation of $Y$ is $|a|$ times the SD of $X$ . The shift $b$ has no effect on either. The same rule covers every other measure of spread: $\text{Range}(aX+b) = |a|\,\text{Range}(X)$ , $\text{MD}(aX+b) = |a|\,\text{MD}(X)$ , quartile deviation likewise — shift-invariant, scaled by $|a|$ . Only the coefficient of variation behaves differently, because its denominator (the mean) shifts with $b$ .

Variance and SD under Y = aX + b

\text{Var}(Y) = a^2\,\text{Var}(X) \qquad \sigma_Y = |a|\,\sigma_X

$a$ scale factor
$b$ shift — irrelevant for dispersion

Worked example

The SD of a dataset

X

is 4. Find the SD of

Y = 3X + 7

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2026 · Q115Easy

Example 4 · Statistics · Dispersion — Standard Deviation, Variance, Mean Deviation

The standard deviation of 100 observations is 10. If 5 is multiplied to each of the observations, then what is the new standard deviation?

Squaring $a$ for variance, taking absolute value for SD

Students often write

\sigma_Y^2 = a\,\sigma_X^2

(forgetting the square) or

\sigma_Y = a\,\sigma_X

(forgetting the modulus). If

a

is negative,

|a|

is the correct scale for SD — SD is non-negative by definition.

Drill 10 more on linear transformation of sd and variance

Concept 5 of 7: Special-Case Variance & SD — Natural Numbers and AP

Just as there are closed-form MEANS for structured sequences, there is a closed-form VARIANCE for the first n natural numbers — and, by the linear-transformation rule, for any AP. Recognise the structure and skip the term-by-term sum.

Definition

For the first $n$ natural numbers $1, 2, \ldots, n$ :

Variance: $\sigma^2 = \dfrac{n^2 - 1}{12}$
SD: $\sigma = \sqrt{\dfrac{n^2 - 1}{12}}$
Any AP with common difference $d$ and $n$ terms is $d \times (\text{natural numbers}) + \text{shift}$ ; the shift leaves variance unchanged and the scale squares it, so $\sigma^2_{\text{AP}} = d^2\,\dfrac{n^2 - 1}{12}$ .

Variance of the first n natural numbers and of an AP

\sigma^2_{1..n} = \dfrac{n^2 - 1}{12} \qquad \sigma^2_{\text{AP}} = d^2\,\dfrac{n^2 - 1}{12}

$n$ number of terms
$d$ common difference of the AP ( $=1$ for natural numbers)

Worked example

Find the variance of the first

7

natural numbers.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2019 · Q118Moderate

Example 5 · Statistics · Dispersion — Standard Deviation, Variance, Mean Deviation

V

is the variance and

M

is the mean of first 15 natural numbers, then what is

V + M^{2}

equal to ?

It is $(n^2 - 1)/12$ , not $n^2/12$ or $(n^2 + 1)/12$

The exact constant is

\dfrac{n^2 - 1}{12}

— memorise the

-1

. A frequent slip is writing

n^2/12

(forgetting the

-1

) or confusing it with the mean-of-squares formula.

For an AP, only the common difference matters — not the starting value

Adding a constant shifts every term but not the spread, so

\{3, 6, \ldots, 60\}

and the natural numbers

\{1, \ldots, 20\}

scaled by 3 have the SAME variance

\left(3^2 \times \dfrac{20^2-1}{12} = 299.25\right)

. Only

d

and

n

enter the formula.

Drill 1 more on special-case variance & sd — natural numbers and ap

Concept 6 of 7: Coefficient of Variation (CV)

SD measures absolute spread, but a SD of 10 means different things on a salary scale (small) versus a marks scale (large). CV normalises SD by the mean and reports a unitless percentage, so you can compare variability across totally different datasets.

Definition

The coefficient of variation is the ratio of the standard deviation to the arithmetic mean, expressed as a percentage. The dataset with the higher CV is considered more variable.

Coefficient of Variation

\text{CV} = \dfrac{\sigma}{\bar{x}} \times 100 \%

$\sigma$ standard deviation
$\bar{x}$ arithmetic mean

Worked example

A dataset has mean

50

and standard deviation

10

. Find its coefficient of variation.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Apr 2021 · Q109Easy

Example 6 · Statistics · Dispersion — Standard Deviation, Variance, Mean Deviation

If the mean of a frequency distribution is 100 and the coefficient of variation is 45%, then what is the value of the variance?

CV is unitless — that's the entire point

Some answers give CV with a unit attached. Wrong. CV is a percentage. If the question compares two datasets with different units (e.g. height in cm vs weight in kg), only CV makes a fair comparison — not SD.

Drill 5 more on coefficient of variation (cv)

Concept 7 of 7: Computational Identity & Minimum-SSE Property

Two identities that NDA papers exploit relentlessly. First: the mean of the squares equals the variance plus the square of the mean. Second: among all reference points, the sum of squared deviations is minimised when the reference is the mean (the median minimises absolute deviations; the mean minimises squared deviations).

Definition

From $\sigma^2 = \overline{x^2} - \bar{x}^2$ follows the identity $\sum x_i^2 = n(\sigma^2 + \bar{x}^2)$ . Also, the function $f(a) = \sum (x_i - a)^2$ is minimised when $a = \bar{x}$ . The reason, and a formula in its own right: $\sum (x_i - A)^2 = \sum (x_i - \bar x)^2 + n(\bar x - A)^2$ for ANY reference $A$ (the cross term vanishes because $\sum(x_i - \bar x) = 0$ ). So a sum of squared deviations about a wrong centre exceeds the one about the mean by exactly $n(\bar x - A)^2$ — use it to convert " $\sum(x_i - 50)^2 = 5000$ " into a variance without recomputing.

Two load-bearing identities

\dfrac{\sum x_i^2}{n} = \bar{x}^2 + \sigma^2 \qquad \text{and} \qquad \arg\min_{a}\sum_{i}(x_i - a)^2 = \bar{x}

Worked example

Given

\sum x_i = 50

and

\sum x_i^2 = 530

for

n = 10

observations, find the variance.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2024 · Q111Moderate

Example 7 · Statistics · Dispersion — Standard Deviation, Variance, Mean Deviation

Let

\displaystyle\sum_{i=1}^{9}x_i^2=855

. If

M

is the mean and

\sigma

is the standard deviation of

x_1,x_2,\ldots,x_9

, then what is the value of

M^2+\sigma^2

$\overline{x^2} \neq (\bar{x})^2$ — they differ by exactly $\sigma^2$

Mean of the squares is NOT the square of the mean. Their difference is the variance:

\overline{x^2} - \bar{x}^2 = \sigma^2 \geq 0

. PYQs plant this trap by asking for "mean of squares" or "

M^2 + \sigma^2

" and expecting you to recognise it as

\sum x_i^2/n

Scaling inside the deviation moves the minimiser too

For

S(a) = \sum (c\,x_i - a)^2

, expand and minimise: the minimum is at

a = c\,\bar{x}

, NOT

a = \bar{x}

. PYQs commonly use

c = 2

(e.g.

S = \sum (2x_i - a)^2

) and expect you to identify the minimiser as

2\bar{x}

— twice the mean, not the mean.

Drill 5 more on computational identity & minimum-sse property

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

Mean Deviation
Mean Deviation about A
$\text{MD}(A) = \dfrac{1}{n}\sum_{i=1}^{n}|x_i - A|$
Variance
Variance — two equivalent forms
$\sigma^2 = \dfrac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = \dfrac{\sum x_i^2}{n} - \bar{x}^2$
Standard Deviation
Standard Deviation
$\sigma = \sqrt{\sigma^2} = \sqrt{\dfrac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2}$
Linear Transformation of SD and Variance
Variance and SD under Y = aX + b
$\text{Var}(Y) = a^2\,\text{Var}(X) \qquad \sigma_Y = |a|\,\sigma_X$
Special-Case Variance & SD — Natural Numbers and AP
Variance of the first n natural numbers and of an AP
$\sigma^2_{1..n} = \dfrac{n^2 - 1}{12} \qquad \sigma^2_{\text{AP}} = d^2\,\dfrac{n^2 - 1}{12}$
Coefficient of Variation (CV)
Coefficient of Variation
$\text{CV} = \dfrac{\sigma}{\bar{x}} \times 100 \%$
Computational Identity & Minimum-SSE Property
Two load-bearing identities
$\dfrac{\sum x_i^2}{n} = \bar{x}^2 + \sigma^2 \qquad \text{and} \qquad \arg\min_{a}\sum_{i}(x_i - a)^2 = \bar{x}$

Watch out for (9)

Mean deviation about median is always $\leq$ about the mean
→ Mean Deviation
Computational form saves time on $\Sigma x_i^2$ -style PYQs
→ Variance
SD and mean deviation share units; variance does not
→ Standard Deviation
Squaring $a$ for variance, taking absolute value for SD
→ Linear Transformation of SD and Variance
It is $(n^2 - 1)/12$ , not $n^2/12$ or $(n^2 + 1)/12$
→ Special-Case Variance & SD — Natural Numbers and AP
For an AP, only the common difference matters — not the starting value
→ Special-Case Variance & SD — Natural Numbers and AP
CV is unitless — that's the entire point
→ Coefficient of Variation (CV)
$\overline{x^2} \neq (\bar{x})^2$ — they differ by exactly $\sigma^2$
→ Computational Identity & Minimum-SSE Property
Scaling inside the deviation moves the minimiser too
→ Computational Identity & Minimum-SSE Property

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