Random Variables

Random variables was worth 14 marks on NESA's sample paper for the 2024 syllabus, the second-biggest focus area after Applications of calculus. School trial papers often give it less time than the HSC does, so make sure you practise all three parts.

• Discrete random variables: probability distributions as tables and graphs, the uniform distribution, expected value $E(X)$, variance $\text{Var}(X)$ and standard deviation.
• Continuous random variables: probability density functions (PDFs), cumulative distribution functions (CDFs), the continuous uniform distribution, and the mean and variance by integration.
• The normal distribution: its shape and features, the empirical rule (68–95–99.7), $z$-scores, standard normal tables, and finding $\mu$ or $\sigma$ from a probability.

Probability basics from Year 11 (MA11-PD) are assumed.

A random variable $X$ assigns a number to each outcome of a random process. It is discrete if it takes separate values (counts, scores on a die). Its probability distribution lists each value $x$ with $P(X=x)$, often written $P(x)$. The probabilities are all between 0 and 1 and sum to 1. They can be estimated from data using relative frequencies, which also sum to 1.

In a uniform discrete distribution all values are equally likely, like the score on a fair die.

Mean and variance

where $E(X^2)=\sum x^{2}P(x)$ and the standard deviation is $\sigma =\sqrt{\text{Var}(X)}$. The expected value is the long-run average, not necessarily a value $X$ can take. The second variance formula is usually quicker.

Worked example 1

The number of goals $X$ a junior soccer team scores in a game has this distribution.

The probabilities sum to 1, so this is a valid distribution.

So $\sigma =0.9$ goals. To draw the distribution, use a column graph with $x$ on the horizontal axis and $P(x)$ as the column heights.

A continuous random variable takes any value in an interval (heights, times, masses). Probabilities belong to intervals, not single values, so

and it doesn't matter whether interval endpoints are included: $P(a<X<b)=P(a\le X\le b)$.

The PDF and the CDF

The probability density function $f(x)$ is a curve where area means probability:

A function is a valid PDF on its domain if:

• $f(x)\ge 0$ for every $x$ in the domain, and
• the total area under it is 1: $\int f(x)dx=1$ over the whole domain.

The cumulative distribution function gives the probability up to $x$:

for a PDF on $[a,b]$. The CDF is non-decreasing, starts at 0 and finishes at 1. It links to the PDF both ways:

Mode, median and quantiles

The mode is the $x$-value where $f(x)$ is largest (use calculus or the graph). The median $m$ solves $F(m)=0.5$; other quantiles work the same way, e.g. the 90th percentile solves $F(x)=0.9$.

Mean and variance

These are the continuous versions of the discrete formulas, with the sum replaced by an integral. If the integral is outside the course, the syllabus expects you to use technology.

The continuous uniform distribution

If $X$ is equally likely to be anywhere in $[a,b]$, its PDF is a horizontal line:

Probabilities are rectangle areas, and the mean is the midpoint $\frac{a+b}{2}$.

Worked example 2

A continuous random variable has PDF $f(x)=k(3-x)$ for $0\le x\le 3$, and 0 otherwise.

(a) Find $k$. The total area must be 1:

so $k=\frac{2}{9}$. (Check: $f(x)\ge 0$ on $[0,3]$.) The graph is a triangle, and $\frac{1}{2}\times 3\times 3k=1$ gives the same answer.

(b) Find the CDF and $P(X<1)$.

So $P(X<1)=F(1)=\frac{5}{9}$. Check: $F(3)=1$.

(c) Find $E(X)$ and $\text{Var}(X)$.

The mean is below the midpoint 1.5 because more of the area is on the left.

(d) Find the median and the mode. The median solves $F(m)=0.5$:

Only $m=3-\frac{3}{\sqrt{2}}\approx 0.879$ lies in $[0,3]$. The PDF is a decreasing straight line ($f^{\prime }(x)=-\frac{2}{9}$), so it is largest at the left end: the mode is $x=0$.

Many naturally occurring measurements (heights, masses, measurement errors, test scores) are approximately normally distributed. We write $X\sim N(\mu ,{\sigma }^2)$: mean $\mu$, standard deviation $\sigma$ (so variance ${\sigma }^2$). Its PDF is

Features of the curve:

• bell-shaped and symmetric about $x=\mu$, with total area 1;
• mean, median and mode are all equal to $\mu$;
• points of inflection at $x=\mu \pm \sigma$;
• the tails approach the $x$-axis as $x\to \pm \infty$ but never touch it.

The empirical rule

For any normal distribution, approximately:

• 68% of values lie within 1 standard deviation of the mean;
• 95% within 2 standard deviations;
• 99.7% within 3 standard deviations.

Use symmetry to split these: for example 34% lie between $\mu$ and $\mu +\sigma$, and 2.5% lie above $\mu +2\sigma$. A quick sketch with $\mu$ and $\mu \pm \sigma ,\pm 2\sigma ,\pm 3\sigma$ marked makes these questions much easier.

z-scores

A $z$-score says how many standard deviations a value is above (positive) or below (negative) the mean:

If $X\sim N(\mu ,{\sigma }^2)$ then $Z\sim N(0,1)$, the standard normal distribution. Use $z$-scores to compare results from different groups (the higher $z$-score is the better relative performance) and, with a table of $P(Z\le z)$, to find probabilities. Useful facts from symmetry:

Worked example 3

The battery life of brand A torches is normally distributed with mean 42 hours and standard deviation 3 hours.

(a) Use the empirical rule to estimate the percentage lasting between 39 and 48 hours.

39 is $\mu -\sigma$ and 48 is $\mu +2\sigma$. So the percentage is $34\%+47.5\%=81.5\%$.

(b) Find $P(X<44.1)$, given $P(Z<0.7)=0.7580$.

so $P(X<44.1)=0.7580$.

(c) Brand B torches have a mean battery life of 40 hours, and 10% of them last longer than 48 hours. Given $P(Z<1.2816)=0.9$, find the standard deviation for brand B.

The top 10% starts at $z=1.2816$, so

Finding $\mu$ works the same way: set up $z=\frac{x-\mu }{\sigma }$ with the known $z$ and solve.

• Forgetting to check that probabilities sum to 1 (discrete) or the area is 1 (continuous) before using a distribution.
• Using $E(X^2)-\mu$ instead of $E(X^2)-{\mu }^2$ for the variance.
• Giving the standard deviation when the variance is asked for, or the other way round. In $N(\mu ,{\sigma }^2)$ the second number is the variance.
• Integrating over the wrong interval for a CDF, or forgetting the CDF is 0 to the left of the domain and 1 to the right.
• Thinking $P(X=a)$ is $f(a)$ for a continuous variable. It is 0; $f(a)$ is a density, not a probability.
• Using the empirical rule halves wrongly: 95% within 2 standard deviations means 2.5% in each tail, not 5%.
• Sign errors with negative $z$-scores; a sketch of the shaded area prevents most of these.

• Show that $f$ is a PDF: state both conditions ($f(x)\ge 0$ and total area 1) and show the integral.
• Find a probability from a PDF: write the integral with limits before evaluating. For a CDF, write $F(b)-F(a)$.
• Normal questions: sketch the bell curve, mark the mean and the value(s), shade the required area, then compute $z$. Markers reward correct $z$-scores even if the table lookup goes wrong.
• Justify / compare: in comparison questions, compute both $z$-scores and state which is higher and what that means in context.
• Read the question for which tool it wants: "use the empirical rule" means 68–95–99.7, not the table.
• Give probabilities as decimals or fractions, and percentages only when asked.

• MA11-PD Probability and data: probability rules, relative frequency and data displays are assumed.
• MA12-IC Integral calculus: every continuous probability, CDF, mean and variance is a definite integral.
• MA12-DC Differential calculus: a PDF is the derivative of its CDF.
• MA12-AC Applications of calculus: finding where a PDF is largest, or its points of inflection, uses stationary point tests.