QCE Mathematical Methods - Unit 4 - Continuous random variables and the normal distribution
Continuous Random Variables | QCE Mathematical Methods
Understand QCE Mathematical Methods continuous random variables, density functions, cumulative distribution functions, expected value and variance.
Updated 2026-05-18 - 4 min read
QCAA official coverage - Mathematical Methods 2025 v1.3
Exact syllabus points covered
- Use relative frequencies and histograms obtained from data to estimate probabilities associated with a continuous random variable.
- Understand probability density functions, cumulative distribution functions, and probabilities associated with a continuous random variable given by integrals.
- Calculate the expected value, $E(X)=\mu=\int_{-\infty}^{\infty}xp(x)\,dx$, of a continuous random variable where $p(x)$ is the probability density function.
- Calculate the variance, $\operatorname{Var}(X)=\sigma^2=\int_{-\infty}^{\infty}(x-\mu)^2p(x)\,dx$, and standard deviation $\sigma$, of a continuous random variable.
- Understand standardised normal variables ($z$-values, $z$-scores) and use these to compare samples.
A continuous random variable can take any value in an interval. Time, height, mass, distance and temperature are common examples. Unlike a discrete random variable, a continuous variable is not handled by adding probabilities at individual points.
For a continuous random variable:
$ P(X=a)=0 $
That does not mean values cannot occur. It means probability is measured across intervals, not exact points.
Original Sylligence diagram for continuous density cdf.
Probability density functions
A probability density function, usually written $p(x)$ or $f(x)$, gives a curve whose area represents probability. The total area under the curve must be:
$ \int_{-\infty}^{\infty}p(x)\,dx=1 $
For an interval:
$ P(a\le X\le b)=\int_a^b p(x)\,dx $
A function is a valid probability density function when:
- $p(x)\ge 0$ over its domain
- the total area under the curve is $1$
- probabilities are found by integrating over intervals
If the density is defined only on a finite interval, it is treated as $0$ outside that interval. For example, if $p(x)=2x$ for $0\le x\le 1$, then $p(x)=0$ for $x<0$ and $x>1$.
From histograms to density
Real continuous data often starts as a relative-frequency histogram. In a density histogram, each bar area represents relative frequency:
$ \text{bar area}=\text{relative frequency} $
Because:
$ \text{bar area}=\text{width}\times\text{height} $
the density height is:
$ \text{density}=\frac{\text{relative frequency}}{\text{class width}} $
This is why density can be greater than $1$ for a narrow interval. The height is not the probability; the area is.
Cumulative distribution functions
The cumulative distribution function, $F(x)$, gives the probability that the random variable is less than or equal to a value:
$ F(x)=P(X\le x) $
In integral form:
$ F(x)=\int_{-\infty}^{x}p(t)\,dt $
For a continuous distribution:
$ P(a\le X\le b)=F(b)-F(a) $
This is the statistics version of the definite-integral idea: accumulated probability up to $b$ minus accumulated probability up to $a$.
Because $P(X=a)=0$, the endpoint style does not change the answer:
$ P(a<X<b)=P(a\le X\le b)=P(a<X\le b) $
That is different from discrete random variables, where including or excluding an endpoint can matter.
Expected value and variance
For a continuous random variable:
$ E(X)=\mu=\int_{-\infty}^{\infty}xp(x)\,dx $
Variance is:
$ \operatorname{Var}(X)=\sigma^2=\int_{-\infty}^{\infty}(x-\mu)^2p(x)\,dx $
Standard deviation is $\sigma$.
The shortcut formula also works:
$ \operatorname{Var}(X)=E(X^2)-[E(X)]^2 $
where:
$ E(X^2)=\int_{-\infty}^{\infty}x^2p(x)\,dx $
For functions of a continuous random variable, the expected value is found by integrating the function times the density:
$ E(g(X))=\int_{-\infty}^{\infty}g(x)p(x)\,dx $
Worked example
Worked validation example
Histograms and relative frequency
Histograms from real data can estimate the shape of a continuous distribution. A large sample with sensible class widths can make the distribution easier to see. A small sample or poor bin choice can hide the shape.
For interval questions involving infinity, use the same area idea:
$ P(X>a)=1-F(a) $
and:
$ P(X\le a)=F(a) $
If a density is given in pieces, integrate over the piece or pieces that overlap the interval. Sketching the support first helps you avoid integrating over a region where the density is actually $0$.