What's the difference between variance and standard deviation?

Variance is \operatorname{Var}(X) = E(X^2) - [E(X)]^2, while standard deviation is \sigma = \sqrt{\operatorname{Var}(X)}. Standard deviation is in the same units as X, making it more interpretable; variance is squared, which is why it appears in the shortcut formula.

Why use the formula E(X^2) - [E(X)]^2 instead of computing deviations directly?

The shortcut formula E(X^2) - [E(X)]^2 is algebraically equivalent to the definition E[(X - E(X))^2] but often faster to compute by hand because you only need E(X) and E(X^2), not the squared deviations from the mean.

Can variance ever be negative?

No, variance is always non-negative because it sums squared deviations. If you calculate a negative variance, you have made an error—most likely reversing the subtraction order or forgetting to square the mean.

How do I know which formula to use, E(X^2) - [E(X)]^2 or the definition with deviations?

Both are correct and equivalent. Use the E(X^2) - [E(X)]^2 formula when you already know the probability distribution, as it requires fewer steps; use the deviation formula E[(X - \mu)^2] if you want to emphasize how variance measures spread around the mean.

Variance of a Random Variable

Variance measures how spread out a random variable's values are around its mean, using the formula Var(X) = E(X²) - [E(X)]².

\operatorname{Var}(X) = E(X^2) - \big[E(X)\big]^2

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What each symbol means

What Variance of a Random Variable takes

Symbol	Meaning
$X$	A random variable representing the outcomes of a random experiment; it can be discrete (like a die roll) or continuous (like a weight), and the variance tells you how far its values typically scatter from the mean.

When to use it

Use variance when you need to quantify how much a random variable typically deviates from its expected value.

Level

Usually taught in: Statistics

Worked examples

1. Heads in two fair coin flips

Problem

A fair coin is flipped twice. Let

X

be the number of heads. Find

\operatorname{Var}(X)

$X \in \{0, 1, 2\}, \quad P(X=0) = \tfrac{1}{4}, \quad P(X=1) = \tfrac{1}{2}, \quad P(X=2) = \tfrac{1}{4}$
The sample space has four equally likely outcomes: HH (2 heads), HT and TH (1 head each), and TT (0 heads).
$E(X) = 0 \cdot \tfrac{1}{4} + 1 \cdot \tfrac{1}{2} + 2 \cdot \tfrac{1}{4} = \tfrac{1}{2} + \tfrac{1}{2} = 1$
Calculate the expected value by multiplying each outcome by its probability and summing.
$E(X^2) = 0^2 \cdot \tfrac{1}{4} + 1^2 \cdot \tfrac{1}{2} + 2^2 \cdot \tfrac{1}{4} = \tfrac{1}{2} + 1 = \tfrac{3}{2}$
Calculate the expected value of $X^2$ by squaring each outcome first, then applying the same weighting.
$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \tfrac{3}{2} - 1^2 = \tfrac{1}{2}$
Apply the variance formula: subtract the square of the mean from the expected value of the square.

Answer: $\operatorname{Var}(X) = \tfrac{1}{2}$

This formula directly applies the definition of variance using the shortcut formula. The positive result confirms that the number of heads does vary from its expected value of 1.

2. Variance with negative values and fractions

Problem

A discrete random variable

X

takes values

-2

0

, and

3

with probabilities

\tfrac{1}{4}

\tfrac{1}{2}

, and

\tfrac{1}{4}

respectively. Find

\operatorname{Var}(X)

$E(X) = (-2) \cdot \tfrac{1}{4} + 0 \cdot \tfrac{1}{2} + 3 \cdot \tfrac{1}{4} = -\tfrac{1}{2} + 0 + \tfrac{3}{4} = \tfrac{1}{4}$
Multiply each value by its probability; negative values keep their sign.
$E(X^2) = (-2)^2 \cdot \tfrac{1}{4} + 0^2 \cdot \tfrac{1}{2} + 3^2 \cdot \tfrac{1}{4} = 1 + 0 + \tfrac{9}{4} = \tfrac{13}{4}$
When squaring negative values, they become positive: $(-2)^2 = 4$ , so negative outcomes still contribute positively to $E(X^2)$ .
$[E(X)]^2 = \left(\tfrac{1}{4}\right)^2 = \tfrac{1}{16}$
Square the mean—not the individual terms, but the result $E(X) = \tfrac{1}{4}$ .
$\operatorname{Var}(X) = \tfrac{13}{4} - \tfrac{1}{16} = \tfrac{52}{16} - \tfrac{1}{16} = \tfrac{51}{16}$
Convert to a common denominator: $\tfrac{13}{4} = \tfrac{52}{16}$ , then subtract.

Answer: $\operatorname{Var}(X) = \tfrac{51}{16} = 3.1875$

This example shows that variance is always non-negative, even when the random variable takes negative values, because we square all deviations in the formula.

3. Point spread in a basketball season

Problem

During a basketball season, a player scored the following points over 5 games: 12, 18, 15, 20, and 10. Treating these as equally likely outcomes, find the variance of the player's scoring.

$E(X) = \frac{12 + 18 + 15 + 20 + 10}{5} = \frac{75}{5} = 15$
The expected value is the mean of the five scores.
$E(X^2) = \frac{12^2 + 18^2 + 15^2 + 20^2 + 10^2}{5} = \frac{144 + 324 + 225 + 400 + 100}{5} = \frac{1193}{5}$
Square each score, sum them, then divide by the number of games to find the average of the squares.
$[E(X)]^2 = 15^2 = 225$
Square the mean score.
$\operatorname{Var}(X) = \frac{1193}{5} - 225 = \frac{1193}{5} - \frac{1125}{5} = \frac{68}{5} = 13.6$
Subtract: convert 225 to fifths ( $225 = \frac{1125}{5}$ ) so the subtraction is valid.

Answer: $\operatorname{Var}(X) = \frac{68}{5} = 13.6 \text{ points}^2$

The variance of 13.6 points² tells us that the player's scoring varies considerably from the average of 15 points per game. To describe the spread in the same units (points), we would take the square root to get the standard deviation: $\sqrt{13.6} \approx 3.69$ points.

Common mistakes

Where Variance of a Random Variable usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\operatorname{Var}(X) = E(X^2) + [E(X)]^2$ (adding instead of subtracting).	Addition would artificially inflate the variance and violate the mathematical definition; variance measures deviations from the mean, not a sum of them.	Always subtract: $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$ .
Computing $E(X^2) - E(X)$ instead of $E(X^2) - [E(X)]^2$ , forgetting to square the mean.	The second term must be the expected value squared, not just the expected value; omitting the square gives a numerically wrong and dimensionally incorrect answer.	Calculate $E(X)$ first, then square that single number to get $[E(X)]^2$ , not the expectation of $X$ without squaring.
Computing the formula backwards as $[E(X)]^2 - E(X^2)$ , which gives a negative result when the variable is spread out.	Variance is always non-negative because it measures spread; a negative result signals that the formula order was reversed.	Remember the order: $E(X^2)$ comes first, then subtract $[E(X)]^2$ .

Tips and when to use something else

Variance is always non-negative: $\operatorname{Var}(X) \geq 0$ , with equality only when $X$ is constant.
Variance has squared units: if $X$ is measured in dollars, $\operatorname{Var}(X)$ is in dollars²; use the standard deviation $\sigma = \sqrt{\operatorname{Var}(X)}$ to describe spread in the original units.
When you need to compare spread across random variables with different units or scales, variance is often less intuitive than the standard deviation or coefficient of variation.
For a linear transformation, $\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$ (constant $b$ disappears, and the coefficient $a$ is squared)—use this shortcut instead of recomputing from scratch.

Related formulas

Expected ValueExpected value is the average outcome you'd get if you repeated an experiment many times, weighted by how likely each outcome is.Law of Large NumbersThe Law of Large Numbers tells you that the sample mean gets arbitrarily close to the true population mean as your sample size increases.Binomial ProbabilityBinomial Probability finds the likelihood of exactly k successes in n independent identical trials, each with fixed probability p.

Frequently asked questions

What's the difference between variance and standard deviation?: Variance is $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$ , while standard deviation is $\sigma = \sqrt{\operatorname{Var}(X)}$ . Standard deviation is in the same units as $X$ , making it more interpretable; variance is squared, which is why it appears in the shortcut formula.
Why use the formula $E(X^2) - [E(X)]^2$ instead of computing deviations directly?: The shortcut formula $E(X^2) - [E(X)]^2$ is algebraically equivalent to the definition $E[(X - E(X))^2]$ but often faster to compute by hand because you only need $E(X)$ and $E(X^2)$ , not the squared deviations from the mean.
Can variance ever be negative?: No, variance is always non-negative because it sums squared deviations. If you calculate a negative variance, you have made an error—most likely reversing the subtraction order or forgetting to square the mean.
How do I know which formula to use, $E(X^2) - [E(X)]^2$ or the definition with deviations?: Both are correct and equivalent. Use the $E(X^2) - [E(X)]^2$ formula when you already know the probability distribution, as it requires fewer steps; use the deviation formula $E[(X - \mu)^2]$ if you want to emphasize how variance measures spread around the mean.

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Reviewed 2026-09-18