Discrete random variables, their probability distributions, the expected value (mean) $E(X) = \sum x P(X = x)$, the variance $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$ and the standard deviation $\mathrm{sd}(X) = \sqrt{\mathrm{Var}(X)}$

What this dot point is asking

VCAA wants you to work with discrete random variables, their probability distributions, and the summary statistics (mean, variance, standard deviation) that describe them. These appear in Paper 1 calculations and underpin the binomial distribution.

Discrete random variables

A discrete random variable $X$ takes values from a countable set (typically a finite list of integers). Its probability distribution is a table or rule specifying $P(X = x)$ for each possible value $x$.

Two conditions every distribution must satisfy:

1. IMATH_10 for all $x$.
2. IMATH_12 (the probabilities sum to one).

Solving for an unknown probability often comes down to using condition 2.

Example distribution

A spinner is divided into four unequal sectors numbered 1, 2, 3, 4 with respective probabilities $0.2, 0.3, 0.3, 0.2$.

The sum is 1. The variable $X$ representing the outcome is a discrete random variable.

Expected value (mean)

The expected value of $X$ is the long-run average outcome if the experiment is repeated many times:

It is a weighted average: each possible value is multiplied by its probability.

For the spinner above: $E(X) = 1(0.2) + 2(0.3) + 3(0.3) + 4(0.2) = 0.2 + 0.6 + 0.9 + 0.8 = 2.5$.

Expected value of a function of IMATH_19

For any function $g$:

The most common application is $g(x) = x^2$, giving $E(X^2)$, which is needed for variance.

Variance and standard deviation

The variance measures the spread of $X$ around its mean. The computational formula (memorise this):

The equivalent definition is $\mathrm{Var}(X) = E\!\left((X - \mu)^2\right)$, but the computational formula is faster.

The standard deviation is the positive square root:

Standard deviation has the same units as $X$ itself, which makes it easier to interpret than variance.

For the spinner: $E(X^2) = 1(0.2) + 4(0.3) + 9(0.3) + 16(0.2) = 0.2 + 1.2 + 2.7 + 3.2 = 7.3$.

$\mathrm{Var}(X) = 7.3 - (2.5)^2 = 7.3 - 6.25 = 1.05$. So $\mathrm{sd}(X) = \sqrt{1.05} \approx 1.025$.

Linear transformations

For constants $a$ and $b$:

Key insight: an added constant shifts the mean but not the variance. A multiplicative constant scales the mean linearly and the variance by the square.

Worked example

A game pays \Y$where$Y = 10 X - 5$and$X$ is the spinner above.

E(Y) = 10 E(X) - 5 = 10(2.5) - 5 = \20$.

$\mathrm{Var}(Y) = 100 \mathrm{Var}(X) = 100(1.05) = 105$.

\mathrm{sd}(Y) = 10 \cdot 1.025 \approx \10.25$.

Cumulative distribution

The cumulative probability $P(X \leq x)$ is the sum of $P(X = k)$ for all $k \leq x$. Inequalities to watch:

• IMATH_40 for integer-valued $X$.
• IMATH_42 .
• IMATH_43 for integer-valued $X$.

Read the inequality direction carefully. Strict versus non-strict matters when $X$ takes the boundary value with positive probability.

Common Paper 1 traps

Forgetting that probabilities must sum to 1. Solving for an unknown probability in a distribution table almost always starts by setting the sum equal to 1.

Computing $E(X)^2$ instead of $E(X^2)$. These are different! $E(X^2)$ uses each $x^2$ times its probability; $[E(X)]^2$ squares the already-computed mean. Variance is the difference between them.

Adding instead of multiplying for variance under scaling. $\mathrm{Var}(2X) = 4 \mathrm{Var}(X)$, not $2 \mathrm{Var}(X)$.

Forgetting the absolute value in $\mathrm{sd}(aX + b) = |a| \cdot \mathrm{sd}(X)$. Standard deviation is always non-negative, so the coefficient is the absolute value of $a$.

Off-by-one on cumulative probabilities for integer $X$. $P(X < 3) = P(X \leq 2)$, not $P(X \leq 3)$.

In one sentence

A discrete random variable has a probability distribution summing to 1, an expected value $E(X) = \sum x P(X = x)$ that is the long-run average outcome, and a variance $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$ that measures spread, with linear transformations following $E(aX + b) = a E(X) + b$ and $\mathrm{Var}(aX + b) = a^2 \mathrm{Var}(X)$.