The normal distribution: 68%, 95% and 99.7% measured as areas

Normal distribution & 3σ

Why does the share within one, two or three standard deviations come out the same for every normal curve, whatever μ and σ are?

Slide the mean and the spread and shade the band μ ± kσ. The board integrates the curve numerically over that band: about 0.683, 0.954 and 0.997 for k = 1, 2, 3, with any μ and σ, because changing them only moves and rescales the same picture.

Drag the ringed points on the board.

High school
Normal density
Probability as area
68–95–99.7 rule
Standardisation z = (x − μ)/σ

\mu

0.0

\sigma

1.00

k

1σ

2σ

3σ

2.00

\mu \pm 3\sigma:\ 99.73\%

$P(\mu - k\sigma < X < \mu + k\sigma)$: 0.9545
$\mu - k\sigma$: -2.00
$\mu + k\sigma$: 2.00
$\text{whole area}$: 1.000000
$f(\mu) = \tfrac{1}{\sigma\sqrt{2\pi}}$: 0.3989
$\text{outside, one side}$: 0.02275

Guided tour

1Probability is area
For a continuous quantity the curve is a density: the chance of landing between two values is the area under the curve between them. The whole area is $1$ . Shade $\mu\pm2\sigma$ and the board measures $0.9545$ .
268%, 95%, 99.7%
Within one standard deviation: $0.6827$ . Within two: $0.9545$ . Within three: $0.9973$ , leaving only $0.00135$ in each tail. That is the $3\sigma$ rule: outside $\mu\pm3\sigma$ is about $3$ in $1000$ .
3Change σ, keep the share
Widen to $\sigma=2$ and move to $\mu=1.5$ . The curve is twice as wide and half as tall (peak $0.1995$ ), so the area over $\mu\pm\sigma$ is unchanged: still $0.6827$ .
4Narrow and tall
At $\sigma=0.6$ the peak rises to $0.6649$ , but the total area still reads $1$ : a smaller spread squeezes the same probability into a narrower range.

Explanation

1
The curve
The normal density with mean $\mu$ and standard deviation $\sigma$ is
$f(x)=\frac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}}$
- It is symmetric about $\mu$ and highest there.
- It bends the other way exactly at $\mu\pm\sigma$ (red points): $f''(x)$ has the sign of $(x-\mu)^2-\sigma^2$ .
- The factor $\frac{1}{\sigma\sqrt{2\pi}}$ is there only to make the whole area $1$ ; the board checks this numerically.
2
Why the shares do not depend on μ and σ
Substitute $z=\frac{x-\mu}{\sigma}$ . Then $dx=\sigma\,dz$ , and $x=\mu\pm k\sigma$ becomes $z=\pm k$ :
$\int_{\mu-k\sigma}^{\mu+k\sigma}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx=\int_{-k}^{k}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz$
$\mu$ and $\sigma$ have disappeared. Moving $\mu$ slides the picture; changing $\sigma$ stretches it sideways by $\sigma$ and squashes it by $\sigma$ , and areas are unchanged by that. This is why one table of the standard normal serves every normal distribution.
3
Where 0.6827 comes from
The integral of $e^{-z^2/2}$ has no formula in elementary functions, so the numbers have to be computed. The board does what a table-maker does: it cuts $[\mu-k\sigma,\,\mu+k\sigma]$ into $48$ strips and adds them up with Simpson's rule, which is accurate here to better than $10^{-5}$ . That the whole area is exactly $1$ can be proved (square the integral and switch to polar coordinates), but that proof is beyond this chapter.
4
Why so many things are normal
A quantity made of many small independent contributions, such as a height, a measurement error or the number of heads in many tosses, is close to normal whatever the contributions look like. This is the central limit theorem. You can see the start of it in the binomial chapter: with $p=0.5$ and $n=20$ the bars already trace a bell.

Check yourself

Q01

Heights are normal with $\mu=170$ cm and $\sigma=6$ cm. About what share is between $158$ and $182$ cm?

Q02

If $\sigma$ doubles, what happens to the peak height of the curve?

Q03

A part should measure $10.00$ mm with $\sigma=0.02$ mm. One measures $10.08$ mm. What does the $3\sigma$ rule suggest?

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