Use probability density functions and cumulative distribution functions to find probabilities, medians, modes, means and variances of continuous random variables

What this dot point is asking

NESA wants you to work with continuous random variables defined by a probability density function (pdf). You must find unknown constants by enforcing total probability $= 1$, compute probabilities as definite integrals, build the cumulative distribution function, and find the mean, variance, median and mode using integrals. The thread running through all of it is one picture: the pdf is a curve, and the probability of any interval is the area under that curve, so every task here is an area or an integral.

The answer

Probability density functions

A continuous random variable $X$ is described by a probability density function $f$ satisfying

• $f(x) \ge 0$ for all $x$ (a density cannot be negative),
• $\int_{-\infty}^{\infty} f(x) \, dx = 1$ (the total area under the curve is $1$).

In Maths Advanced, $f$ is non-zero only on a finite interval $[a, b]$ called the support, and the total integral is taken over that interval. The value $f(x)$ is a density, not a probability: it can exceed $1$, and on its own it tells you nothing until you integrate it over an interval. This is the key shift from the discrete case, where each value carried its own probability $p_i$.

The most important consequence is that for a continuous random variable, $P(X = c) = 0$ for any single value $c$. A single point has no width, so it has no area, so it has no probability. Probabilities live on intervals only.

Probabilities as integrals

For any interval $[c, d]$ inside the support, the probability is the area under the density over that interval:

Because single points have zero probability, $P(c \le X \le d) = P(c < X < d)$. The strict and non-strict inequalities give the same value, so you never have to worry about whether the endpoints are included.

Cumulative distribution function

The cumulative distribution function (cdf) accumulates probability from the left:

Useful properties:

• $F$ is non-decreasing, with $F(-\infty) = 0$ and $F(\infty) = 1$ (and $F = 0$ below the support, $F = 1$ above it).
• $P(c \le X \le d) = F(d) - F(c)$, so once you have $F$ every interval probability is a subtraction.
• Where $f$ is continuous, $F'(x) = f(x)$: the pdf is the derivative of the cdf, the cdf is the integral of the pdf. They are two views of the same distribution.

Mean, variance, median, mode

The mean (expected value) weights each value $x$ by its density and integrates:

The variance integrates the squared deviation from the mean, and is almost always computed via the shortcut on the right:

where $E(X^2) = \int x^2 f(x) \, dx$. The standard deviation is $\sigma = \sqrt{\text{Var}(X)}$.

The median $m$ splits the area in half:

The mode is the value of $x$ where $f$ is largest. If $f$ is differentiable on the interior of the support, look for a critical point of $f$; if $f$ is monotone on the support, the mode is at the endpoint where $f$ is highest.

Why every technique here is an integral

Continuous random variables are where the calculus and statistics strands of Maths Advanced meet. A probability is an integral, the cdf is an integral with a variable upper limit, the mean weights $x$ by the density and integrates, and the variance integrates the squared deviation. So the practical skill being tested is your integration: setting up the right definite integral over the support, finding the antiderivative, and evaluating cleanly. The statistics is the interpretation wrapped around the calculus, which is also why this dot point pairs so naturally with the integration techniques from the calculus strand.

The standard problem types

NESA questions on this dot point fall into a small number of recognisable shapes, and naming the type tells you the first move:

• "Find the value of the constant $k$": enforce $\int f = 1$ over the support.
• "Find $P(a \le X \le b)$": integrate $f$ from $a$ to $b$ (or use $F(b) - F(a)$).
• "Find the mean / expected value": integrate $x f(x)$.
• "Find the variance / standard deviation": compute $E(X^2)$, then use $\text{Var}(X) = E(X^2) - \mu^2$.
• "Find the median": solve $F(m) = \frac{1}{2}$.
• "Find the mode": maximise $f$ on the support.

Sketching the density

A quick sketch of $f$ over its support guides the work. The total area under the curve must be $1$, the median splits that area in half, and the mode sits under the highest point of the curve. For a symmetric density the mean, median and mode coincide at the centre of symmetry, which can save an integral if you spot the symmetry early.

How exam questions ask about continuous random variables

• "Show that $k = \dots$" or "find the value of $k$." Set $\int_{\text{support}} f = 1$ and solve.
• "Find $P(X > a)$ / $P(a \le X \le b)$." Integrate $f$ over the interval, or use $F$. For a "greater than" question, integrate up to the top of the support (or use $1 - F(a)$).
• "Find the expected value / mean." Integrate $x f(x)$ over the support.
• "Find the variance / standard deviation." Find $E(X^2) = \int x^2 f$, then $\text{Var}(X) = E(X^2) - \mu^2$, then square-root for $\sigma$.
• "Find the cumulative distribution function." Integrate $f$ with a variable upper limit; remember to state $F = 0$ below the support and $F = 1$ above it.
• "Find the median / mode." Solve $F(m) = \frac12$ for the median; maximise $f$ for the mode.

Edge cases worth knowing

• A piecewise density. If $f$ is defined in pieces, integrate each piece over its own sub-interval and add. The cdf is then also piecewise, continuous at the joins.
• The mode at an endpoint. When $f$ is monotone on the support there is no interior critical point, so the mode is the endpoint where $f$ is largest. Do not chase a derivative that never vanishes.
• A density that exceeds $1$. This is fine: $f$ is a density, not a probability. On a narrow support the height can be well above $1$ while the area stays $1$.
• Spotting symmetry. If $f$ is symmetric about a centre $c$, then $E(X) =$ median $= c$ immediately, and $P(X > c) = \frac12$, saving you an integral.