Variance and Standard Deviation

From Absolute Values to Squares

Mean deviation dodged the sign problem with absolute values; squaring the deviations dodges it more gracefully — every $(x_i - \bar{x})^2$ is non-negative, and squares welcome algebra.

But the raw sum of squares misleads. Compare set A ($5, 15, 25, 35, 45, 55$; six values, mean 30) with set B ($15, 16, \ldots, 45$; thirty-one values, mean 30):

Sum of squared deviations: A gives 1750, B gives 2480 — the sum crowns B as more scattered, yet A's values visibly stray farther from 30. Sums grow just because $n$ grows. Divide by $n$: A gives $\frac{1750}{6} \approx 291.7$, B gives $\frac{2480}{31} = 80$ — now A correctly shows more scatter.

Key Point (Variance and standard deviation):

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

The variance carries squared units (cm² for heights in cm); taking the positive square root — the standard deviation — returns to the data's own units, which is why $\sigma$ is the preferred measure of dispersion about the mean.

Computing Variance for Ungrouped Data

The direct method: for $6, 8, 10, \ldots, 24$ (ten terms), the mean is 15; the squared deviations $81, 49, 25, 9, 1, 1, 9, 25, 49, 81$ sum to 330, so

$$\sigma^2 = \frac{330}{10} = 33, \qquad \sigma = \sqrt{33} \approx 5.74$$

The computing identity — expand $(x_i - \bar{x})^2$ and simplify:

$$\sigma^2 = \frac{1}{n}\sum x_i^2 - \bar{x}^2$$

"Mean of squares minus square of mean." It avoids subtracting the mean from every observation and powers most JEE problems (finding missing observations, correcting statistics).

Two standard results worth memorising:

For the first $n$ natural numbers: $\bar{x} = \frac{n+1}{2}$ and $\sigma^2 = \frac{n^2 - 1}{12}$ — derived from $\sum x = \frac{n(n+1)}{2}$ and $\sum x^2 = \frac{n(n+1)(2n+1)}{6}$ via the identity.

For the first ten multiples of 3 ($3, 6, \ldots, 30$): each value is 3 times a natural number, so the mean triples ($16.5$) and the variance picks up $3^2$: $\sigma^2 = 9 \times \frac{99}{12} = 74.25$ — a first taste of the scaling law made precise in the JEE Corner.

Interpretation: small $\sigma$ means the data hugs the mean; $\sigma = 0$ forces every observation to equal the mean; large $\sigma$ means wide scatter.

[Board Tip] State variance and standard deviation as separate answers with units — variance in squared units, SD in original units. Writing $\sigma^2 = 5.74$ when the question wanted $\sigma$ is a silent full-mark loss.

Solved Examples

Example 1: The direct method

Find the variance of $6, 8, 10, 12, 14, 16, 18, 20, 22, 24$.

Solution:

Step 1 — Mean. $\bar{x} = \frac{150}{10} = 15$.

Step 2 — Squared deviations. $81, 49, 25, 9, 1, 1, 9, 25, 49, 81$ — symmetric about the centre, summing to 330.

Step 3 — Divide. $\sigma^2 = \frac{330}{10} = 33$; $\sigma = \sqrt{33} \approx 5.74$.

Takeaway: Evenly spaced data gives mirror-image squared deviations — compute half the column and double it.

Example 2: Mean and variance together

Find the mean and variance of $6, 7, 10, 12, 13, 4, 8, 12$.

Solution:

Step 1 — Mean. $\bar{x} = \frac{72}{8} = 9$.

Step 2 — Squared deviations. $9, 4, 1, 9, 16, 25, 1, 9$ — sum 74.

Step 3 — Divide. $\sigma^2 = \frac{74}{8} = 9.25$.

Takeaway: Tabulate $x_i$, $x_i - \bar{x}$, $(x_i - \bar{x})^2$ in three columns — the layout is half the accuracy.

Example 3: First $n$ natural numbers

Find the mean and variance of the first $n$ natural numbers.

Solution:

Step 1 — Mean from the power sum. $\bar{x} = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}$.

Step 2 — Computing identity. $\sigma^2 = \frac{1}{n}\cdot\frac{n(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$.

Step 3 — Simplify. Factor $\frac{n+1}{12}$: $\frac{n+1}{12}\left[2(2n+1) - 3(n+1)\right] = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}$.

Takeaway: $\bar{x} = \frac{n+1}{2}$, $\sigma^2 = \frac{n^2-1}{12}$ — memorise the pair; JEE quotes them without derivation time.

Example 4: First ten multiples of 3

Find the mean and variance of $3, 6, 9, \ldots, 30$.

Solution:

Step 1 — Recognise the scaling. The data is $3 \times (1, 2, \ldots, 10)$.

Step 2 — Scale the mean. Mean $= 3 \times \frac{11}{2} = 16.5$.

Step 3 — Scale the variance by $3^2$. $\sigma^2 = 9 \times \frac{10^2 - 1}{12} = 9 \times 8.25 = 74.25$.

Takeaway: Multiplying data by $k$ multiplies the mean by $k$ and the variance by $k^2$ — the scaling law, met here for the first time.

Example 5: The computing identity in action

Using $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$, recompute the variance of $6, 7, 10, 12, 13, 4, 8, 12$.

Solution:

Step 1 — The two sums. $\sum x_i = 72$; $\sum x_i^2 = 36+49+100+144+169+16+64+144 = 722$.

Step 2 — Apply. $\sigma^2 = \frac{722}{8} - 9^2 = 90.25 - 81$.

Step 3 — Conclude. $9.25$ ✓ — matching Example 2 without touching a single deviation.

Takeaway: "Mean of squares minus square of mean" needs only $\sum x_i$ and $\sum x_i^2$ — the identity JEE problems assume you use.

Example 6: When is the variance zero?

Show that $\sigma^2 = 0$ forces all observations to be equal.

Solution:

Step 1 — Write the condition. $\sum(x_i - \bar{x})^2 = 0$ — a sum of non-negative terms equal to zero.

Step 2 — Kill each term. Every $(x_i - \bar{x})^2 = 0$, so $x_i = \bar{x}$ for all $i$.

Step 3 — Conclude. Zero variance means zero scatter: a constant data set. ∎

Takeaway: Non-negative terms summing to zero must each vanish — a proof pattern that recurs across inequalities and statistics.

Example 7: Comparing two data sets

Set P: $48, 49, 50, 51, 52$. Set Q: $30, 40, 50, 60, 70$. Compare means and standard deviations.

Solution:

Step 1 — Means. Both are 50 — the centres agree.

Step 2 — Set P's spread. Squared deviations $4+1+0+1+4 = 10$: $\sigma^2 = 2$, $\sigma = \sqrt{2} \approx 1.41$.

Step 3 — Set Q's spread. $400+100+0+100+400 = 1000$: $\sigma^2 = 200$, $\sigma = 10\sqrt{2} \approx 14.14$ — ten times P's.

Takeaway: Q is P stretched tenfold about the same centre — and its SD is exactly ten times larger, previewing the scaling law.

Example 8: Variance from sums alone

For 10 observations, $\sum x_i = 60$ and $\sum x_i^2 = 1000$. Find $\sigma$.

Solution:

Step 1 — Mean. $\bar{x} = \frac{60}{10} = 6$.

Step 2 — Identity. $\sigma^2 = \frac{1000}{10} - 6^2 = 100 - 36 = 64$.

Step 3 — Root. $\sigma = 8$.

Takeaway: Two sums fully determine mean and SD — no individual observations needed; this is the format JEE uses for "corrected statistics" problems.

Example 9: An unknown observation from the mean

The observations $1, 2, 3, 4, x$ have mean 3. Verify $x$ and find the variance.

Solution:

Step 1 — Solve for $x$. $\frac{10 + x}{5} = 3$: $x = 5$.

Step 2 — Recognise the data. $1, 2, 3, 4, 5$ — the first five naturals.

Step 3 — Quote the standard result. $\sigma^2 = \frac{5^2 - 1}{12} = 2$.

Takeaway: Solve for unknowns from the mean FIRST — the completed data set often collapses onto a known formula.

Example 10: SD of a symmetric AP

Find the standard deviation of $a - 2d, a - d, a, a + d, a + 2d$.

Solution:

Step 1 — Mean by symmetry. The terms balance about $a$: $\bar{x} = a$.

Step 2 — Squared deviations. $4d^2, d^2, 0, d^2, 4d^2$ — sum $10d^2$.

Step 3 — Variance and SD. $\sigma^2 = 2d^2$: $\sigma = |d|\sqrt{2}$.

Takeaway: The centre $a$ never appears in the answer — dispersion sees only spacing, not location. That is the shift-invariance law in miniature.