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SAMath MethodsSyllabus dot point

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 4.78 min answer

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

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  1. What this dot point is asking
  2. Expected value
  3. Variance and standard deviation
  4. Linear transformations
  5. Common errors
  6. Why it matters

What this dot point is asking

The expected value is the average value of XX 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 3.53.5).

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(X2)−μ2E(X^2)-\mu^2 is almost always quicker than working with (x−μ)2(x-\mu)^2 directly.

Linear transformations

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

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

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.

2017 SACE Stage 23 marksA 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. Using the formula on the SACE sheet, the variance is the sum of (x - mean)^2 times p(x), with mean = 0:

Var(X) = (-1 - 0)^2 (0.2) + (0 - 0)^2 (0.7) + (2 - 0)^2 (0.1)
= (1)(0.2) + (0)(0.7) + (4)(0.1)
= 0.2 + 0 + 0.4 = 0.6.

Standard deviation = sqrt(0.6) = 0.775 (to three significant figures).

Marks: one for the correct variance set-up, one for evaluating it to 0.6, and one for taking the square root to get 0.775. Forgetting to square root the variance is a common slip.

2018 SACE Stage 22 marksThe 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 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 Var(X) = 16.3 - (2.9)^2 = 16.3 - 8.41 = 7.89.

Standard deviation = sqrt(7.89) = 2.81 (to three significant figures).

Marks: one for computing the variance (via E(X^2) minus mean squared, or via the sum of squared deviations), one for the square root. The large spread reflects the rare but high value of 10.