HookWhy CERN waited for five sigma
On 4 July 2012, physicists at CERN's Large Hadron Collider announced they had found the Higgs boson. What made it a discovery rather than a rumour was a single phrase: the signal was 'five sigma'. In the language of the Normal distribution, five sigma means the bump in their data sat five standard deviations away from what you would expect if the particle did not exist — and the chance of a fluctuation that large occurring purely by luck is about 1 in 3.5 million. Particle physics has a hard convention: nothing below five sigma is allowed to count as a discovery. That threshold is not really physics at all. It is a statement about the tail of a bell-shaped curve, and reading it correctly is the whole of this section.
Statistical distributions are the models that turn 'how likely is this?' into a number you can defend. In S4 you will meet the binomial distribution, which counts successes in a fixed number of independent trials, and learn to compute its probabilities exactly; the Normal distribution, the bell curve that describes heights, measurement errors and — through five sigma — particle physics, and how to find probabilities from it by standardising; and the deep link between the two, the reason a lumpy binomial histogram smooths into a Normal bell as the number of trials grows. Finally you will learn to choose: to look at a situation and say which model fits it, and, just as importantly, when neither one does.
ModelDiscrete distributions and the binomial
A discrete random variable \(X\) takes separate, countable values, each with a probability, and those probabilities must sum to one: \(\sum P(X=x)=1\). The simplest distributions are just listed in a table — the score on a spinner, the number of faulty items in a box — and the sum-to-one rule is your first check and often a way to find a missing value.
The headline discrete model is the binomial distribution, and it applies only when four conditions all hold: a fixed number \(n\) of trials; each trial has just two outcomes, success or failure; the probability of success \(p\) is constant from trial to trial; and the trials are independent. When they do, we write \(X\sim B(n,p)\), and the probability of exactly \(r\) successes is \[P(X=r)=\binom{n}{r}p^r(1-p)^{n-r},\] where \(\binom{n}{r}\) counts the ways to place the \(r\) successes among the \(n\) trials. The mean number of successes is \(E(X)=np\). In practice you will use a calculator's cumulative binomial function for 'at least' and 'at most' questions, but writing the model and the event out first is what earns the marks — and forces you to confirm the four conditions actually hold.
A seed supplier claims a 90% germination rate. A gardener sows 10 seeds; let \(X\) be the number that germinate, modelled as \(X\sim B(10,\,0.9)\). Find \(P(X=8)\) and \(P(X\ge 8)\).
Exactly eight: \[P(X=8)=\binom{10}{8}(0.9)^8(0.1)^2=45\times 0.43046721\times 0.01=0.1937\ (4\text{ d.p.}).\] For 'at least eight', add the three top cases. \(P(X=9)=\binom{10}{9}(0.9)^9(0.1)^1=10\times 0.0387420489=0.3874\), and \(P(X=10)=(0.9)^{10}=0.3487\). So \[P(X\ge 8)=0.1937+0.3874+0.3487=0.9298.\] The mean is \(np=10\times 0.9=9\) seeds, which sits comfortably inside that 93% chance of eight or more — a quick sanity check that the arithmetic is pointing the right way.
ModelThe Normal distribution and standardising
The Normal distribution models a continuous quantity — a length, a mass, a time — that clusters symmetrically around a central value. We write \(X\sim N(\mu,\sigma^2)\), where \(\mu\) is the mean (the centre of the bell) and \(\sigma^2\) is the variance (\(\sigma\) sets the spread). The curve is symmetric about \(\mu\), its points of inflection — where it changes from bending down to bending up — sit exactly at \(\mu\pm\sigma\), and the total area beneath it is one. Because probability is area, the probability of any single exact value is zero: you always ask for an interval, and \(P(X\le a)=P(X\lt a)\) for a continuous variable. Roughly 68% of the distribution lies within one standard deviation of the mean, 95% within two, and 99.7% within three — which is why 'five sigma' is so extreme.
To find a probability you standardise: convert your value to the number of standard deviations it lies from the mean, \[Z=\frac{X-\mu}{\sigma},\] turning any Normal variable into the standard Normal \(Z\sim N(0,1)\). Tables or a calculator then give \(\Phi(z)=P(Z\le z)\), the area to the left. A quick sketch with the required region shaded is the habit that prevents the most common Normal error — forgetting to subtract from one when you want the upper tail.
A machine fills bags with a mean of 500 g and standard deviation 4 g, modelled as \(X\sim N(500,\,4^2)\). Find the probability a bag is underweight at below 495 g, and the probability it lies between 495 g and 505 g.
Standardise 495: \(Z=\dfrac{495-500}{4}=-1.25\). From tables \(\Phi(1.25)=0.8944\), so \[P(X\lt 495)=P(Z\lt -1.25)=1-0.8944=0.1056,\] about 11%. For the central band, 505 standardises to \(Z=+1.25\), so by symmetry \[P(495\lt X\lt 505)=\Phi(1.25)-\Phi(-1.25)=0.8944-0.1056=0.7888.\] Nearly 79% of bags fall in that window. The points of inflection sit at \(500\pm 4\), i.e. 496 g and 504 g — the shoulders of the bell, a full standard deviation either side of the mean.
MechanismThe link between binomial and Normal
The two models are not strangers. Draw the probability histogram of a binomial with bars of width 1 and height \(P(X=r)\), so — following the histogram rule from S2 — the area of each bar is its probability. For small \(n\) the picture is lopsided, but as \(n\) grows the bars grow symmetric and pile into a smooth bell that is well approximated by a Normal with the same mean \(np\) and variance \(np(1-p)\). This is why the Normal turns up everywhere: whenever an outcome is the sum of many small, independent, similar effects, the total is approximately Normal.
Francis Galton built a machine to show it. His 'quincunx', devised in the 1870s, drops balls through a triangular array of pins; at each pin a ball goes left or right — an independent Bernoulli trial — and the balls collect at the bottom in a bell-shaped heap. Each ball's final position is a binomial count, and the heap is the Normal curve. The five-sigma rule at CERN rides on the same bridge: the raw event counts are discrete, but because they aggregate many independent collisions, physicists judge them on the Normal scale, where 'five standard deviations out' has a precise, tiny probability.
Consider tossing a fair coin 100 times, \(X\sim B(100,\,0.5)\). Its mean is \(np=50\) and its standard deviation is \(\sqrt{np(1-p)}=\sqrt{100\times 0.5\times 0.5}=\sqrt{25}=5\).
Because \(n\) is large and \(p=0.5\) makes the distribution symmetric, the binomial histogram is almost perfectly bell-shaped, close to \(N(50,\,5^2)\). The '99.7% within three standard deviations' rule then predicts that nearly all outcomes lie in \(50\pm 3\times 5\), i.e. between 35 and 65 heads. Getting 30 heads would be four standard deviations from the mean — the kind of departure that, in a physics experiment, would start people talking about a discovery.
CaseChoosing the model — and when neither fits
Selecting a distribution is an examined skill in its own right, and the marks are as much in rejecting the wrong model as in fitting the right one. Reach for the binomial when you are counting successes in a fixed number of independent trials with a constant probability — the number of sixes in twelve rolls, the number of left-handers in a class. Reach for the Normal when you are measuring a continuous quantity that clusters symmetrically around a mean — heights, times, manufacturing tolerances.
The model fails when its conditions fail. A binomial dies the moment the trials stop being independent or \(p\) stops being constant: drawing cards without replacement changes the probability at every draw, so 'number of hearts in five cards dealt' is not binomial. A Normal is wrong for strongly skewed data such as incomes, and it is a poor fit whenever the spread is large relative to the mean, because the bell then assigns a non-negligible — and impossible — probability to values below zero for a quantity like height or waiting time. The discipline is a checklist: count or measurement? a fixed number of independent trials? constant probability? symmetric spread? Answer those honestly and the right model, or the honest admission that neither fits, falls out.
Classify three situations. The number of sixes in 12 rolls of a fair die: a fixed 12 trials, two outcomes, constant \(p=\tfrac16\), independent — so \(X\sim B(12,\,\tfrac16)\), a clean binomial. The time an athlete takes to run 100 m across many races: a continuous measurement clustering around a personal best — reasonably Normal, \(T\sim N(\mu,\sigma^2)\). But the number of hearts when 5 cards are dealt from one shuffled pack: the probability of a heart starts at \(\tfrac{13}{52}\) and changes with every card removed, so the trials are not independent and \(p\) is not constant — not binomial, and choosing \(B(5,\,\tfrac14)\) here would be exactly the modelling error the spec is testing.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Write the distribution in full before you compute anything — \(X\sim B(n,p)\) or \(X\sim N(\mu,\sigma^2)\). It is worth a mark, and it forces you to confirm the model's conditions actually hold. For binomial probabilities, put the event in symbols — \(P(X\ge 8)\), not just a calculator keystroke — so the examiner can see whether you wanted 'at least', 'more than' or 'exactly'; the boundary is where marks leak away.
For Normal questions, always draw a small bell and shade the region you want before standardising with \(Z=\dfrac{X-\mu}{\sigma}\); the sketch is what catches the 'one minus' error on upper tails. Remember that \(\Phi(z)\) is the area to the left, so an upper tail is \(1-\Phi(z)\) and a symmetric lower tail is \(\Phi(-z)=1-\Phi(z)\). When a question asks you to select or criticise a model, list the conditions and check each against the context out loud — 'independent? constant probability? continuous and symmetric?' — because the examiner is marking your judgement, not just your final number.