Discrete random variables: probability distribution, expected value, variance and standard deviation

What this dot point is asking

NESA wants you to recognise a discrete random variable, check that its probability distribution is valid, compute the expected value and variance from the distribution, and apply the linear-transformation rules to $a X + b$. Everything starts from the probability distribution, the list of values with their probabilities, so reading and validating that table is the first marked move in almost every question.

The answer

Discrete random variables and their distributions

A discrete random variable $X$ takes a countable list of values $x_1, x_2, \dots, x_n$ with probabilities $p_i = P(X = x_i)$. The list of values with their probabilities is the probability distribution of $X$. For it to be valid, two conditions must hold:

• $0 \le p_i \le 1$ for every $i$ (each is a genuine probability),
• $\sum_i p_i = 1$ (something must happen).

The spike graph above is the natural picture: each value sits on the horizontal axis and the height of its spike is its probability, so the heights are the $p_i$ and they must add to $1$. The probability that $X$ falls in some set is the sum of $p_i$ for the values in that set. For example, $P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2)$ if $X$ takes integer values from $0$.

Expected value

The expected value (or mean) of $X$ is the long-run average value if we repeated the experiment many times. It is the weighted sum

The expected value need not be one of the values $X$ can actually take; it is a balance point, not an outcome.

Expected value of a function of $X$

For any function $g$,

The most common case is $g(x) = x^2$, which gives

This is the quantity you build to find the variance, so it is worth setting up as its own column of working.

Variance and standard deviation

The variance of $X$ measures spread around the mean. By definition it is the expected squared deviation,

which is algebraically equivalent (and almost always easier to compute) to

The standard deviation is $\sigma = \sqrt{\text{Var}(X)}$, in the same units as $X$, which is why it is the spread measure you can compare directly against the mean.

Linear transformations

If $Y = a X + b$ for constants $a$ and $b$,

Shifting $X$ by $b$ slides the mean but leaves the spread untouched; scaling by $a$ multiplies the mean by $a$ and the standard deviation by $|a|$ (and the variance by $a^2$). These rules let you find the mean and variance of $Y$ without rebuilding any sums.

Reading a distribution and finding its mean, stage by stage

The two diagrams here use the distribution $P(X=0)=0.1$, $P(X=1)=0.4$, $P(X=2)=0.3$, $P(X=3)=0.2$.

Stage 1, read the distribution and check it is valid. Whether it arrives as a two-row table or as the spike graph above, the first move is the same: confirm the probabilities are between $0$ and $1$ and sum to $1$. Here $0.1 + 0.4 + 0.3 + 0.2 = 1$, so the distribution is valid and you can build calculation columns from it. (If a constant were involved, you would solve $\sum p_i = 1$ for it first.)

Stage 2, find the expected value as the balance point. The mean is the weighted sum $E(X) = 0(0.1) + 1(0.4) + 2(0.3) + 3(0.2) = 1.6$. Picture the probabilities as weights placed along the axis: $E(X)$ is the point where the bar would balance, marked by the fulcrum below. Note that $1.6$ is not one of the values $X$ can take, which is exactly what "balance point, not an outcome" means.

Presenting a distribution as a table

In the exam a discrete distribution is usually laid out as a two-row table: the values $x_i$ on top and the probabilities $p_i$ underneath. Reading it correctly is the first marked step. Check the probabilities sum to $1$ (solve for any unknown if a constant is involved), then build the calculation columns you need: $x_i p_i$ for the mean and $x_i^2 p_i$ for $E(X^2)$. Laying the work out in columns keeps the arithmetic tidy and is exactly what markers look for.

Interpreting expected value and variance

The expected value is the balance point of the distribution: if you placed the probabilities as weights along a number line, $E(X)$ is where it would balance, as the fulcrum in the diagram shows. The variance measures how widely the values spread around that balance point, in squared units, and the standard deviation brings it back to the original units so it can be compared with the mean. A small standard deviation means the outcomes cluster tightly around the mean; a large one means they are spread out. This interpretation is what justifies the linear-transformation rules: shifting every value left or right slides the balance point but leaves the spread untouched, while stretching the scale stretches both.

Why $E(X^2) - \mu^2$ is the practical formula

The definition $\text{Var}(X) = \sum (x_i - \mu)^2 p_i$ is conceptually clear but arithmetically painful because it subtracts $\mu$ inside every term. The equivalent $E(X^2) - \mu^2$ is almost always faster: build one extra column of $x_i^2 p_i$, sum it, and subtract the square of the mean. The two formulas are algebraically identical, so use the second to compute and quote the first to explain.

How exam questions ask about discrete random variables

• "Show that the table is a valid probability distribution" or "find the value of $k$." Check $0 \le p_i \le 1$ and solve $\sum p_i = 1$ for any unknown.
• "Find $P(X \le 2)$" or "$P(X \ge 1)$." Add the relevant $p_i$; for "at least" it is often quicker to use $1 - P(\text{the rest})$.
• "Find the expected value / mean." Compute the weighted sum $\sum x_i p_i$, showing the products.
• "Find the variance / standard deviation." Build $E(X^2) = \sum x_i^2 p_i$, then $\text{Var}(X) = E(X^2) - \mu^2$, then square-root for $\sigma$.
• "Let $Y = aX + b$. Find $E(Y)$ and $\text{Var}(Y)$ / $\sigma_Y$." Apply $E(Y) = aE(X) + b$, $\text{Var}(Y) = a^2\text{Var}(X)$, $\sigma_Y = |a|\sigma_X$.

Edge cases worth knowing

• An unknown probability via the sum. If one entry is missing or given as $k$, find it from $\sum p_i = 1$ before any mean or variance work.
• The mean is not an attainable value. $E(X)$ is a balance point, so a fair die has mean $3.5$ even though you can never roll $3.5$. Do not "round it to a face".
• A symmetric distribution. If the probabilities are symmetric about a central value, that value is the mean immediately, with no weighted sum required.
• Negative-looking variance. Variance is a sum of squared terms times probabilities, so it can never be negative; a negative result signals an arithmetic slip, usually $E(X)^2$ confused with $E(X^2)$.