Variance and standard deviation: raw data

The variance is the average of the squared deviations from the mean. The standard deviation is its square root. Why squared deviations? They:

1. Treat positive and negative deviations symmetrically.
2. Are differentiable (no absolute values) , convenient for algebra.
3. Amplify large deviations, which is often what we want to detect.

Definitions

For raw data $x_1, x_2, \dots, x_n$ with mean $\bar{x}$: $\boxed{\sigma^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2.}$

This is the variance. The standard deviation is $\boxed{\sigma = \sqrt{\sigma^2}.}$

The standard deviation has the same units as the data. Variance has units squared.

Alternative formula

Expanding the square: $\sigma^2 = \frac{1}{n} \sum (x_i^2 - 2 \bar{x} x_i + \bar{x}^2) = \frac{\sum x_i^2}{n} - 2 \bar{x} \cdot \bar{x} + \bar{x}^2 = \frac{\sum x_i^2}{n} - \bar{x}^2.$

So $\boxed{\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2.}$

This is often easier to compute , you only need $\sum x_i$ and $\sum x_i^2$, no pre-computed $\bar{x}$ needed inside the sum.

Properties

Non-negativity. $\sigma^2 \ge 0$, with equality iff all $x_i$ are equal.

Shift invariance. If $y_i = x_i + a$, then $\sigma_y = \sigma_x$.

Scale. If $y_i = b x_i$, then $\sigma_y = |b| \sigma_x$, and $\sigma_y^2 = b^2 \sigma_x^2$.

Combining: if $y_i = a + b x_i$, then $\sigma_y = |b| \sigma_x$.

Worked examples

Example 1. Variance and SD of $4, 6, 8, 10, 12$.

$\bar{x} = 40/5 = 8$. Deviations: $-4, -2, 0, 2, 4$. Squared: $16, 4, 0, 4, 16$. Sum: $40$. $\sigma^2 = 40/5 = 8$. $\sigma = \sqrt{8} = 2\sqrt{2} \approx 2.83$.

Example 2. Variance of $2, 4, 6, 8, 10$ using alternative formula.

$\sum x = 30 \Rightarrow \bar{x} = 6$. $\sum x^2 = 4 + 16 + 36 + 64 + 100 = 220$. $\sigma^2 = 220/5 - 36 = 44 - 36 = 8$. $\sigma = 2\sqrt{2}$.

Example 3. SD of $1, 3, 5, 7, 9, 11$.

$\bar{x} = 36/6 = 6$. $\sum x^2 = 1 + 9 + 25 + 49 + 81 + 121 = 286$. $\sigma^2 = 286/6 - 36 = 47.\overline{6} - 36 = 11.\overline{6}$. $\sigma \approx 3.42$.

(Or directly: deviations $-5, -3, -1, 1, 3, 5$; squares $25, 9, 1, 1, 9, 25$; sum $70$; $\sigma^2 = 70/6 \approx 11.67$.)

Example 4. Find the SD of the first $n$ natural numbers.

$\bar{x} = (n+1)/2$. $\sum x_i^2 = n(n+1)(2n+1)/6$.

$\sigma^2 = \dfrac{n(n+1)(2n+1)/6}{n} - \left(\dfrac{n+1}{2}\right)^2 = \dfrac{(n+1)(2n+1)}{6} - \dfrac{(n+1)^2}{4}$.

Common denominator $12$: $\dfrac{2(n+1)(2n+1) - 3(n+1)^2}{12} = \dfrac{(n+1)[2(2n+1) - 3(n+1)]}{12} = \dfrac{(n+1)(n - 1)}{12} = \dfrac{n^2 - 1}{12}$.

So $\sigma = \sqrt{\dfrac{n^2 - 1}{12}}$.

Example 5. If $\sigma_x = 5$ and $y = 3 x + 7$, find $\sigma_y$.

$\sigma_y = |3| \cdot \sigma_x = 15$.

Try it yourself

1. SD of $2, 4, 4, 4, 5, 5, 7, 9$.
2. Variance of $7, 7, 7, 7$.
3. SD of $1, 2, 3, 4, 5, 6, 7$.
4. Variance and SD of $10, 12, 14, 16, 18$.
5. If $\sum x_i = 50$, $\sum x_i^2 = 500$, $n = 10$, find $\sigma$.
6. Find SD of first $20$ natural numbers.
7. If $y = 2 x - 5$ and $\sigma_x = 4$, find $\sigma_y$.
8. The mean of $5$ numbers is $20$ and the variance is $4$. Add a number equal to the mean. What is the new variance?
9. Show that if all $x_i$ are equal, $\sigma = 0$.
10. SD of $3, 5, 7, 9, 11, 13, 15$.
11. The variance of $25$ observations is $4$. If each is multiplied by $2$, find the new variance.
12. The mean of $10$ numbers is $24$, SD is $6$. If $3$ is added to each, what are the new mean and SD?

Pitfalls / Tricks

• Variance has units squared (e.g., cm²); standard deviation has the same units as data.
• Use the alternative formula when raw sums are easy to compute.
• Shifting data does not change SD; scaling multiplies it.
• Insight. Variance is the square of standard deviation. They contain the same information; choose the one that's algebraically convenient.