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How do we measure the centre and spread of a discrete probability distribution?

The expected value is the long-run mean of a discrete random variable; the variance and standard deviation measure how spread out its values are.

How to compute the expected value, variance and standard deviation of a discrete random variable, including the shortcut variance formula and linear transformation rules.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Expected value
Variance and standard deviation
Linear transformations
Decision-making with expected value
Common errors
Why it matters

What this dot point is asking

The expected value is the average value of $X$ you would see over very many repetitions - the "centre" of the distribution. The variance and standard deviation measure how far, on average, the outcomes fall from that centre.

Expected value

You multiply each value by its probability and add the results. The expected value need not be a value the variable can actually take (a die's mean is $3.5$ ).

Expected value of a biased spinner

A spinner pays out $\$1$ , $\$2$ or $\$5$ with the distribution below.

$x$	1	2	5
$P(X=x)$	$0.5$	$0.3$	$0.2$

Find $E(X)$ .

Step 1: Recall the expected value formula

The expected value is the probability-weighted average of all outcomes. Each value is multiplied by how likely it is, then the products are summed.

E(X) = \sum_{\text{all }x} x\,P(X=x).

Step 2: Multiply each value by its probability

Work through each outcome in the table.

$x = 1$ : contributes $1 \times 0.5 = 0.5$
$x = 2$ : contributes $2 \times 0.3 = 0.6$
$x = 5$ : contributes $5 \times 0.2 = 1.0$

Step 3: Add the contributions

E(X) = 0.5 + 0.6 + 1.0 = 2.1.

Final answer: The expected payout per spin is $\$2.10$ .

Variance and standard deviation

Variance is the expected squared distance from the mean. The definition and the much faster computational form are:

The shortcut $E(X^2)-\mu^2$ is almost always quicker than working with $(x-\mu)^2$ directly.

Variance using the shortcut formula

For the spinner above (with $E(X) = 2.1$ ), find $\operatorname{Var}(X)$ and the standard deviation $\sigma$ .

Step 1: Compute $E(X^2)$

The shortcut formula for variance requires $E(X^2)$ , which is the expected value of $X^2$ . This is calculated the same way as $E(X)$ , but each value is squared before multiplying by its probability.

E(X^2) = 1^2(0.5) + 2^2(0.3) + 5^2(0.2) = 0.5 + 1.2 + 5.0 = 6.7.

Step 2: Apply the shortcut formula

Variance is the difference between the mean of the squared values and the square of the mean. Using $\mu = 2.1$ from the previous example:

\operatorname{Var}(X) = E(X^2) - \mu^2 = 6.7 - (2.1)^2 = 6.7 - 4.41 = 2.29.

Step 3: Find the standard deviation

The standard deviation is the square root of the variance, bringing the measure of spread back to the same units as $X$ .

\sigma = \sqrt{2.29} \approx 1.51.

Final answer: $\operatorname{Var}(X) = 2.29$ and $\sigma \approx 1.51$ .

Linear transformations

If you scale and shift a random variable, $Y=aX+b$ , the mean and variance transform predictably:

For example, if each spin's payout is doubled and a $\$1$ entry fee is subtracted, $Y=2X-1$ , then $E(Y)=2(2.1)-1=3.2$ and $\operatorname{Var}(Y)=2^2(2.29)=9.16$ .

The reason $b$ disappears from the variance is intuitive: adding a fixed amount to every outcome slides the whole distribution sideways without changing how spread out the values are. Scaling by $a$ , on the other hand, stretches the distances from the mean by a factor of $a$ , and since variance is built from squared distances, those distances get squared too. This is why the standard deviation, which is in the original units, scales by $|a|$ while the variance scales by $a^2$ .

Decision-making with expected value

A frequent SACE application is judging whether a game is "fair". A game is fair if the expected net gain is zero once the cost of playing is subtracted. Suppose a game costs $\$2$ to play and pays out according to $X$ with $E(X)=2.1$ . The expected net gain is $E(X)-2=0.10$ dollars per play, so the game slightly favours the player. If instead the cost were $\$2.50$ , the expected net would be $-\$0.40$ , so over many plays you would expect to lose forty cents per game. Expected value is the standard tool for these long-run decision questions, and examiners expect you to interpret the sign of the result in context, not just compute it.

Common errors

Why it matters

Expected value underpins fair-game and decision questions, while variance and standard deviation describe risk and spread - concepts you reuse for the binomial distribution, the normal distribution in Topic 5, and the sampling distributions in Topic 6.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20233 marksCalculator-assumed. A discrete random variable

X

takes values

-1

0

and

2

with probabilities

0.2

0.7

and

0.1

respectively, so that

E(X) = 0

. Calculate the standard deviation of

X

Show worked answer →

The standard deviation is the square root of the variance. With $\mu = 0$ , the variance is $\sum (x - \mu)^2 P(X=x)$ :

\operatorname{Var}(X) = (-1)^2(0.2) + (0)^2(0.7) + (2)^2(0.1) = 0.2 + 0 + 0.4 = 0.6.

Standard deviation $= \sqrt{0.6} \approx 0.775$ (to three significant figures).

Marks: one for the variance set-up, one for evaluating it to $0.6$ , and one for the square root. Forgetting to square root the variance is a common slip.

SACE 20222 marksCalculator-assumed. The number

X

printed on a randomly selected duck takes the values

1

2

5

and

10

with probabilities

0.5

0.2

0.2

and

0.1

, and

E(X) = 2.9

. Calculate the standard deviation of

X

Show worked answer →

Use $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$ .

First $E(X^2) = (1)^2(0.5) + (2)^2(0.2) + (5)^2(0.2) + (10)^2(0.1) = 0.5 + 0.8 + 5 + 10 = 16.3$ .

Then $\operatorname{Var}(X) = 16.3 - (2.9)^2 = 16.3 - 8.41 = 7.89$ .

Standard deviation $= \sqrt{7.89} \approx 2.81$ (to three significant figures).

Marks: one for the variance, one for the square root. The large spread reflects the rare but high value of $10$ .

SACE 20213 marksCalculator-assumed. A random variable

X

has

E(X) = 4

and

\operatorname{Var}(X) = 9

. The variable

Y = 3X - 2

. Find

E(Y)

and

\operatorname{Var}(Y)

Show worked answer →

Apply the linear transformation rules. For $Y = aX + b$ with $a = 3$ , $b = -2$ :

E(Y) = aE(X) + b = 3(4) - 2 = 10.

\operatorname{Var}(Y) = a^2 \operatorname{Var}(X) = 3^2 (9) = 81.

Marks: one for $E(Y) = 10$ , one for using $a^2$ (not $a$ ) on the variance, and one for $\operatorname{Var}(Y) = 81$ . The constant $-2$ shifts the mean but leaves the spread unchanged.