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The normal distribution

Every bell curve is one bell stretched and shifted, so one set of areas serves them all, as long as the data really are bell-shaped.

Helps to know first
This topic

First, the picture

Bells of every centre and width slide and squeeze onto one bell. Watch the shaded bands fill to the same shares.

Most values hide near the middle

Bells of any centre and width, standardised onto one bell, and the area near its middle.

Three bells: adult heights (centre 170 cm, standard deviation 8), exam scores (500 and 100) and radio noise (0 mV and 0.5), each on its own axis.

band edge k
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area inside
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0.00 / 12.60 s
Describe this picture

Three small bells, for adult heights (centre 170 cm, standard deviation 8), exam scores (500 and 100) and radio noise (0 mV and 0.5), each on its own axis in its own units, and one large standard bell beside or below them. In the clip each small bell slides and squeezes as its axis turns into zz from −4 to 4, its own numbers fading out and the zz numbers fading in, until all three are the standard bell. Then bands within one, two and three standard deviations of the centre fill in turn on the large bell, listed with their areas: 68.27 %, 95.45 % and 99.73 %. Readouts give the band edge kk and the area inside. After the clip a slider, “Band edge, k standard deviations”, sets kk from 0 to 3.5; at k=1.96k=1.96 the area reads 95.00 %.

One bell, stretched and shifted

Measure the heights of many adults, the scores on a large exam or the noise on a radio, and the histograms often take the same shape: a hump in the middle, falling away evenly on both sides. That shape is the normal distribution, or bell curve. Its density, for a centre μ\mu and a width σ\sigma, is

fX(x)=1σ2π e−(x−μ)2/(2σ2).f_X(x)=\frac{1}{\sigma\sqrt{2\pi}}\,e^{-(x-\mu)^2/(2\sigma^2)}.

I write it N(μ,σ2)\mathcal{N}(\mu,\sigma^2). A variable XX that follows it has mean μX=μ\mu_X=\mu and standard deviation σX=σ\sigma_X=\sigma. Here a bare σ\sigma is the bell’s standard deviation; signal processing pages use the same letter for something else.

The clip showed the one fact that makes the bell easy to use: every bell is the same bell, moved and stretched. Subtract the centre and divide by the width:

Z=X−μXσX.Z=\frac{X-\mu_X}{\sigma_X}.

This is standardising. ZZ measures how many standard deviations XX lies from its mean, and it follows the standard normal law N(0,1)\mathcal{N}(0,1), with mean 0 and standard deviation 1. A value of ZZ is called a zz-score.

Areas under the one bell

Since every bell is the standard bell in other units, one set of areas serves them all. The distribution function of ZZ has its own letter:

Φ(z)=Pr⁡{Z≤z}.\Phi(z)=\Pr\{Z\le z\}.

Φ\Phi has no formula in everyday functions. Software computes it: norm.cdf in SciPy, or the normal cumulative function of any spreadsheet. Its curve reads like this:

00.510.8413−3−2−10123z
Fig. The standard normal distribution function Φ(z). Read it like a table: at z = 1 the curve stands at 0.8413, so 84.13 % of standard normal values lie below 1.

The bell is symmetric, so Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z): the area below −1-1 is 1−0.8413=0.15871-0.8413=0.1587. The area within kk standard deviations of the mean is

Pr⁡{∣Z∣<k}=Φ(k)−Φ(−k)=2Φ(k)−1.\Pr\{\lvert Z\rvert<k\}=\Phi(k)-\Phi(-k)=2\Phi(k)-1.

Three values are worth knowing by heart:

WithinAreaOutside
1 standard deviation0.6827about 1 in 3
2 standard deviations0.9545about 1 in 22
3 standard deviations0.9973about 1 in 370

And the edge that holds 95 % is 1.96 standard deviations, to two decimals: 2Φ(1.96)−1=0.952\Phi(1.96)-1=0.95.

After the clip, drag the band’s edge and watch the area. Most of it is gathered quickly: one standard deviation already holds two thirds, and past three there is almost nothing left.

Not every shape is a bell

These areas belong to the bell. Other shapes give other areas, so check the shape before you use them.

Take the exponential waiting time Exp(1)\mathrm{Exp}(1) of Continuous random variables, with mean 1 and standard deviation 1. Within one standard deviation of the mean, from 0 to 2, lies 1−e−2=0.86471-e^{-2}=0.8647 of it, not 0.6827. Within two, from −1-1 to 3, lies 1−e−3=0.95021-e^{-3}=0.9502, not 0.9545.

What holds for every shape is much weaker. Whatever the law, at least 1−1/k21-1/k^2 of it lies within kk standard deviations of the mean: at least 75 % within two. The bell gives 95.45 %. This guarantee, Chebyshev’s inequality, is proved in The law of large numbers.

Watch out

The 68, 95 and 99.7 % areas are facts about the bell, not about all data. Waiting times, incomes and file sizes are lopsided, and for them a zz-score does not translate into these shares.

Sums of bells

Add two independent bell-shaped variables and the sum is a bell too. Their means add, and their variances add:

X1+X2∼N(μ1+μ2, σ12+σ22),X_1+X_2\sim\mathcal{N}\big(\mu_1+\mu_2,\ \sigma_1^2+\sigma_2^2\big),

where X1∼N(μ1,σ12)X_1\sim\mathcal{N}(\mu_1,\sigma_1^2) and X2∼N(μ2,σ22)X_2\sim\mathcal{N}(\mu_2,\sigma_2^2), and ∼\sim reads “follows”. The standard deviations do not add: they combine like the sides of a right triangle, σ12+σ22\sqrt{\sigma_1^2+\sigma_2^2}. Why variances add for independent variables, and why the sum stays a bell, is the subject of Sums of random variables.

Worked example

1. Heights. Adult heights in a population follow N(170,82)\mathcal{N}(170,8^2) cm. What share is taller than 186 cm? Standardise: z=(186−170)/8=2z=(186-170)/8=2. So

Pr⁡{X>186}=1−Φ(2)=0.0228,\Pr\{X>186\}=1-\Phi(2)=0.0228,

about 2 people in 100.

What share is between 160 and 178 cm? The edges standardise to (160−170)/8=−1.25(160-170)/8=-1.25 and (178−170)/8=1(178-170)/8=1:

Φ(1)−Φ(−1.25)=0.8413−0.1056=0.7357.\Phi(1)-\Phi(-1.25)=0.8413-0.1056=0.7357.

2. A journey. Your drive takes N(30,42)\mathcal{N}(30,4^2) minutes and the walk after it N(20,32)\mathcal{N}(20,3^2) minutes, independently. The whole journey is N(50, 16+9)=N(50,52)\mathcal{N}(50,\,16+9)=\mathcal{N}(50,5^2): a standard deviation of 5 minutes, not 4+3=74+3=7. The chance it takes more than an hour is Pr⁡{Z>(60−50)/5}=1−Φ(2)=0.0228\Pr\{Z>(60-50)/5\}=1-\Phi(2)=0.0228.

And the chance it takes between 45 and 55 minutes, one standard deviation either side, is 0.6827.

Where you’ll meet this

Measurement error and noise. Errors that come from many small, independent causes added together pile into a bell, the reason given by The central limit theorem. Thermal noise in electronics is the classic case, as in Random variables for signals in the signal processing course.

Test scores and heights. Large exams are often scaled so that scores form a bell; heights of adults of one sex in one population are close to one. “In the top 2.3 %” is a statement about 1−Φ(2)1-\Phi(2).

Tolerances. A factory that keeps a part’s size within three standard deviations of the target expects about 3 parts in 1000 outside, if the sizes are bell-shaped.

In signal processing · a bell keeps its shape

The bell is its own Fourier transform: a Gaussian pulse has a Gaussian spectrum, as Transforms of common signals in the signal processing course shows. That is one reason it appears in filter design and in the spread of pulses in time and frequency.

Reference card

QuantityFormulaNotes
Density N(μ,σ2)\mathcal{N}(\mu,\sigma^2)1σ2πe−(x−μ)2/(2σ2)\frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/(2\sigma^2)}mean μ\mu, standard deviation σ\sigma
StandardiseZ=(X−μX)/σXZ=(X-\mu_X)/\sigma_XZ∼N(0,1)Z\sim\mathcal{N}(0,1)
Distribution functionΦ(z)=Pr⁡{Z≤z}\Phi(z)=\Pr\{Z\le z\}software, not a table; Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z)
Band of kk SD2Φ(k)−12\Phi(k)-10.6827, 0.9545, 0.9973
95 % bandk=1.96k=1.96
Sum of independent bellsmeans add, variances addSDs combine as σ12+σ22\sqrt{\sigma_1^2+\sigma_2^2}
Any shapeat least 1−1/k21-1/k^2 within kk SDChebyshev

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