HookThe casino does not need luck — it needs E(X)

Put \(£1\) on a single number at a European roulette table and the wheel offers \(37\) pockets — the numbers \(1\) to \(36\), plus the green zero. Hit your number and you are paid \(35\) to \(1\); miss and the pound is gone. Let \(X\) be your profit. Then \(\mathrm{E}(X) = 35 \times \frac{1}{37} + (-1) \times \frac{36}{37} = -\frac{1}{37} \approx -2.7\text{p}\) per pound staked. That number is the casino's entire business model: not a rigged wheel, just a payout ratio set slightly below the true odds, multiplied by millions of spins.

But the expectation alone does not explain why anyone plays. The standard deviation of that same bet is about \(£5.84\) — more than two hundred times the size of the expected loss. On any single night the spread utterly swamps the drift, so winners walk out convinced the game is beatable; over a lifetime of spins the drift grinds every player down to \(-\frac{1}{37}\) per pound while the casino's own nightly total, summed over thousands of bets, becomes almost perfectly predictable. Mean and variance are the two numbers that tell that whole story, and this section is about extracting both — plus mode, median, and the effect of linear transformations — from any discrete random variable AQA can write down, whether as a table, a formula, or the discrete uniform distribution whose mean and variance you must be able to prove from scratch.

ModelWhat a discrete random variable actually is

A discrete random variable takes isolated values \(x_1, x_2, \dots\), each with a probability \(p_i = \mathrm{P}(X = x_i)\), and the whole distribution is pinned down by one constraint: \(\sum p_i = 1\). AQA presents distributions two ways. As a table — values along the top, probabilities beneath — or as a function, such as \(\mathrm{P}(X=x) = kx\) for \(x = 1, 2, 3, 4\). In the function form, the first job is always the same: use \(\sum p_i = 1\) to find the unknown constant.

Once the distribution is known, probabilities of events are sums of the relevant \(p_i\): \(\mathrm{P}(X \ge 3)\) is the sum over the values \(3\) and above, \(\mathrm{P}(2 \le X < 4)\) picks up \(2\) and \(3\). With discrete variables, strict versus non-strict inequalities genuinely matter — \(\mathrm{P}(X > 2)\) and \(\mathrm{P}(X \ge 2)\) differ by the whole of \(\mathrm{P}(X = 2)\), a distinction that evaporates for continuous variables and therefore catches students switching between the two.

Worked example

\(X\) has \(\mathrm{P}(X=x) = kx\) for \(x = 1, 2, 3, 4\). Find \(k\), then \(\mathrm{P}(X \ge 3)\). Total probability: \(k(1 + 2 + 3 + 4) = 10k = 1\), so \(k = 0.1\). The table reads \(0.1, 0.2, 0.3, 0.4\). Then \(\mathrm{P}(X \ge 3) = 0.3 + 0.4 = 0.7\). Two lines, two marks — but only if the constraint line '\(\sum p_i = 1\)' is written down, because that statement is what the first method mark is for.

ModelMean, variance, mode, median — the four summaries

The expectation is the probability-weighted average, \(\mathrm{E}(X) = \sum x_i p_i\) — the long-run mean per trial, which is exactly how the roulette \(-\frac{1}{37}\) was built. For spread, compute \(\mathrm{E}(X^2) = \sum x_i^2 p_i\) first, then use \[\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2,\] with the standard deviation as its square root. The formula's shape — mean of the squares minus square of the mean — is the single most-used identity in the statistics option, and the classic error is squaring the wrong thing: \(\mathrm{E}(X^2)\) squares the values before weighting, never the probabilities.

The other two summaries are read straight off the table. The mode is the value with the largest probability. The median is the smallest value \(m\) for which the cumulative probability \(\mathrm{P}(X \le m)\) reaches at least \(0.5\) — accumulate the probabilities from the left and stop the first time you hit or pass one half. All four summaries can be asked in one question part, so the efficient exam habit is one extended table: values, probabilities, \(x_i p_i\), \(x_i^2 p_i\), cumulative probability.

Worked example

For the distribution above (\(p_i = 0.1, 0.2, 0.3, 0.4\) on \(x = 1, 2, 3, 4\)): \(\mathrm{E}(X) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 3\). \(\mathrm{E}(X^2) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 10\). So \(\mathrm{Var}(X) = 10 - 3^2 = 1\) and the standard deviation is \(1\). Mode: \(4\) (largest probability, \(0.4\)). Median: cumulative probabilities run \(0.1, 0.3, 0.6\) — the first to reach \(0.5\) is at \(x = 3\), so the median is \(3\). Note mode \(\ne\) median \(\ne\) anything special: the three measures agree only for symmetric distributions, and this one is skewed.

MechanismLinear coding — what survives a change of scale

Transform \(X\) linearly into \(Y = aX + b\) and the two summaries respond differently: \[\mathrm{E}(aX + b) = a\,\mathrm{E}(X) + b, \qquad \mathrm{Var}(aX + b) = a^2\,\mathrm{Var}(X).\] The mean is dragged through both the stretch and the shift, but the variance ignores \(b\) completely — adding a constant slides the whole distribution along without changing its spread — and picks up \(a^2\), not \(a\), because variance lives in squared units. A useful corollary: the standard deviation of \(aX+b\) is \(|a|\) times that of \(X\), and it is the modulus, since \(\sqrt{a^2} = |a|\).

This is the machinery behind every 'entry fee' and 'unit conversion' question. A game pays out \(X\) pounds but charges \(£2\) to play: profit is \(X - 2\), so expected profit is \(\mathrm{E}(X) - 2\) while the risk (variance) is untouched. Temperatures recorded in Celsius with known mean and variance convert to Fahrenheit via \(F = 1.8C + 32\): the mean converts like a temperature, the variance multiplies by \(1.8^2 = 3.24\), and the \(32\) vanishes from the spread entirely.

Worked example

With \(\mathrm{E}(X) = 3\) and \(\mathrm{Var}(X) = 1\) from the previous block, let \(Y = 2X - 3\). Then \(\mathrm{E}(Y) = 2(3) - 3 = 3\) and \(\mathrm{Var}(Y) = 2^2 \times 1 = 4\), so the standard deviation doubles from \(1\) to \(2\). The mean happened to stay at \(3\) — coincidence of these numbers, and a good reminder that \(\mathrm{E}(Y) = \mathrm{E}(X)\) does not mean the distributions match: \(Y\) takes values \(-1, 1, 3, 5\), twice as spread out as \(X\).

ModelThe discrete uniform distribution — when every value is equally likely

The discrete uniform distribution on \(\{1, 2, \dots, n\}\) gives every value the same probability \(\frac{1}{n}\). It is the model for a fair die (\(n = 6\)), a fairly drawn raffle ticket numbered \(1\) to \(n\), a random digit generator, a roulette pocket. Its summaries are worth memorising: \[\mathrm{E}(X) = \frac{n+1}{2}, \qquad \mathrm{Var}(X) = \frac{n^2 - 1}{12}.\] For a fair die: mean \(3.5\), variance \(\frac{35}{12} \approx 2.92\).

The modelling judgement is the examinable edge: the uniform model is suitable only when the physical setup genuinely makes each outcome equally likely — a symmetric die, a well-shuffled draw. It fails for the sum of two dice (seven ways to make \(7\), one way to make \(2\)), for the first significant digit of naturally occurring data, for anything with clustering. AQA regularly asks 'state, with a reason, whether a discrete uniform model is appropriate' — the mark is for the reason, which should name the symmetry or its absence.

MechanismProving the uniform mean and variance — pure core meets statistics

AQA requires the proofs, not just the formulas, and they run on the summation results from the pure core (section D): \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) and \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\). Each value has probability \(\frac{1}{n}\), so the expectations are just scaled sums — the proof is three lines of algebra once the right sum is quoted.

This is the cleanest example of how the Further Maths statistics option leans on the pure core, and the factorisation step in the variance proof is where candidates stall: after subtracting, take out the common factor \(\frac{n+1}{12}\) before expanding anything, and the difference-of-squares finish appears by itself.

Worked example

Prove the mean and variance of the discrete uniform distribution on \(\{1, \dots, n\}\). Mean: \[\mathrm{E}(X) = \sum_{r=1}^{n} r \cdot \frac{1}{n} = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \frac{n+1}{2}.\] Second moment: \[\mathrm{E}(X^2) = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}.\] Variance: \[\mathrm{Var}(X) = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 = \frac{n+1}{12}\left[2(2n+1) - 3(n+1)\right] = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}.\] Check against the die: \(n = 6\) gives mean \(\frac{7}{2} = 3.5\) and variance \(\frac{35}{12}\) — matching the direct table calculation, which is precisely the kind of verification worth a closing sentence in a 'show that' answer.

VocabularyKey terms the mark scheme pays for

Discrete random variable

A variable taking isolated values \(x_i\), each with probability \(p_i = \mathrm{P}(X = x_i)\), where \(\sum p_i = 1\). Specified by a table or a probability function.

Expectation E(X)

The probability-weighted mean, \(\mathrm{E}(X) = \sum x_i p_i\) — the long-run average value per trial, not a value the variable need ever actually take.

Variance

\(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\), where \(\mathrm{E}(X^2) = \sum x_i^2 p_i\). Measured in squared units; its square root is the standard deviation.

Mode

The value of the variable with the largest probability. A distribution can be bimodal if two values tie for the largest.

Median of a DRV

The smallest value \(m\) with cumulative probability \(\mathrm{P}(X \le m) \ge 0.5\) — found by accumulating probabilities from the smallest value upwards.

Linear coding

For \(Y = aX + b\): \(\mathrm{E}(Y) = a\mathrm{E}(X) + b\) and \(\mathrm{Var}(Y) = a^2\mathrm{Var}(X)\). Shifts move the mean but never the spread; stretches scale the standard deviation by \(|a|\).

Discrete uniform distribution

Equal probability \(\frac{1}{n}\) on each of \(\{1, 2, \dots, n\}\); mean \(\frac{n+1}{2}\), variance \(\frac{n^2-1}{12}\). Suitable only when symmetry makes all outcomes equally likely.

E(X²)

The second moment, \(\sum x_i^2 p_i\): square the values, then weight by probabilities. Not the same as \([\mathrm{E}(X)]^2\) — the gap between the two is exactly the variance.

TrapsMisconceptions that cost marks

“E(X) is the most likely value of X.”

Actually: That is the mode. \(\mathrm{E}(X)\) is a weighted average and need not be a possible value at all — a fair die has \(\mathrm{E}(X) = 3.5\), which no face shows, and roulette profit has \(\mathrm{E}(X) = -\frac{1}{37}\) pounds, an amount no single spin can lose.

“Var(aX + b) = a·Var(X) + b.”

Actually: Adding \(b\) slides every value equally and changes no gaps, so it cannot affect spread; and the stretch enters squared because variance is in squared units. \(\mathrm{Var}(aX+b) = a^2\mathrm{Var}(X)\) — the \(b\) vanishes and the \(a\) is squared, both for good reasons.

“E(X²) means square E(X).”

Actually: \(\mathrm{E}(X^2) = \sum x_i^2 p_i\): square the values first, then average. For the die, \(\mathrm{E}(X^2) = \frac{91}{6} \approx 15.2\) while \([\mathrm{E}(X)]^2 = 12.25\). If your two numbers come out equal, you have accidentally proved your variable is constant.

“A negative expected value means you will lose every time.”

Actually: It means you lose on average, in the long run. Roulette's \(-2.7\text{p}\) per pound sits inside a standard deviation of about \(£5.84\) per spin — individual sessions are dominated by the spread, which is why the house edge is invisible to a player and unmissable to the house.

ExamWhat examiners want

Build one table and mine it for everything: a row for \(x_i\), for \(p_i\), for \(x_i p_i\), for \(x_i^2 p_i\), and a running cumulative row. That single structure answers the mean, variance, standard deviation, mode and median parts of a question with no re-reading. When the distribution arrives as a function with an unknown constant, write the line '\(\sum p_i = 1\)' explicitly before solving — the statement is a mark, not decoration. Quote the variance identity in symbols before substituting: '\(\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2\)' followed by numbers earns method marks even if arithmetic later slips.

On linear coding, resist recomputing from a new table: the whole point AQA is testing is that \(\mathrm{E}(aX+b)\) and \(\mathrm{Var}(aX+b)\) come from the formulas in one line each. Say what happened to \(b\) in the variance and why — 'adding a constant does not affect spread' is an AO2 mark that costs six words. For the discrete uniform proofs, the examiner is watching for the standard summation formulas quoted from the pure core and for the factorisation \(\frac{n+1}{12}[2(2n+1) - 3(n+1)]\) taken out cleanly; expanding everything into a quartic mess usually still reaches the answer but burns time the paper does not give you. And in any modelling part, tie the uniform distribution's suitability to physical symmetry in the context — 'the die is unbiased so all six outcomes are equally likely' — because contextual justification, not the formula, is what that mark is printed for.