Hexa Scholar
MathProbability

The normal distribution and its tail areas

Three bell-shaped curves have different centers and widths. Is the probability of landing more than two standard deviations from the center the same for all three, or larger for the wider curves?

Stage 1 asks for your estimates before the page computes any area. The later stages explain the answer and use it in problems about exam scores, machine parts and daily stock returns.

Before you start · 3 quick checks on a CDF, the SD of aX + b and the function e^(−z²/2), about 4 min
Setup

Normal distributions. Fix a number \(\mu\) and a number \(\sigma \gt 0\). A random variable \(X\) is normal with mean \(\mu\) and standard deviation \(\sigma\) when it has the density

\[f(t) = \frac{1}{\sigma\sqrt{2\pi}}\,e^{-(t - \mu)^{2}/(2\sigma^{2})}\]

for every real \(t\). This is written \(X \sim N(\mu, \sigma^{2})\); the second number is the variance \(\sigma^{2}\), not \(\sigma\). The density is a bell-shaped curve, symmetric about \(\mu\), highest at \(t = \mu\), where its height is \(1/(\sigma\sqrt{2\pi})\), and positive at every \(t\). Its total area is 1, \(E[X] = \mu\) and \(\operatorname{SD}(X) = \sigma\); stage 3 shows why. The lesson on PMF, PDF and CDF introduced densities and CDFs, and the lesson on expectation and variance defined \(E[X]\) and \(\operatorname{SD}(X)\).

The standard normal. With \(\mu = 0\) and \(\sigma = 1\) the density is

\[\varphi(u) = \frac{1}{\sqrt{2\pi}}\,e^{-u^{2}/2}.\]

A random variable \(Z\) with this density is called standard normal, and its CDF is

\[\Phi(z) = P(Z \le z) = \int_{-\infty}^{z} \varphi(u)\,du.\]

\(\Phi\) has no formula in terms of elementary functions; its values come from numerical methods, printed tables or software. The Black–Scholes formula writes \(N(d)\) for \(\Phi(d)\).

Fine print: tables of Φ and negative z

A printed table usually lists \(\Phi(z)\) for \(z \ge 0\) in steps of 0.01, rounded to four decimal places. Values for negative \(z\) come from the symmetry of \(\varphi\) about 0: the area below \(-z\) equals the area above \(z\), so \(\Phi(-z) = 1 - \Phi(z)\) for every \(z\).

1 See

How much area lies in the tails?

How the estimates work

Click one range for each curve. A choice is locked once clicked, and nothing about any curve is shown until all three are in. Then the page marks each choice, explains it, adds a second scale under each panel and lets you redraw the panels on their own scales. Answer again clears all of it.

2 Try

Areas under any normal curve

Two predictions first. The explorer below them opens once both are answered.

The explorer opens when you have answered the two questions above.

3 Prove

Every normal probability is a standard normal probability

A schematic bean machine: gold balls fall from a funnel through a triangle of dark pegs into a row of narrow bins. The balls in the bins form a symmetric pile, tallest in the middle bin and lower toward both sides, with the outermost bins empty.
An idealized bean machine. At every row of pegs, each ball bounces left or right with equal chances, independently of its other bounces. The bin it lands in then has a binomial distribution, and with many rows that distribution is close to a normal curve. The pile drawn here is a schematic of that shape; real piles of finitely many balls vary.
Theorem

Assumptions. Let \(\mu\) be any real number and \(\sigma \gt 0\), and let \(X\) be normal with mean \(\mu\) and SD \(\sigma\), with density \(f(t) = \frac{1}{\sigma\sqrt{2\pi}}e^{-(t - \mu)^{2}/(2\sigma^{2})}\). Let \(\varphi\) and \(\Phi\) be the standard normal density and CDF of the setup. The case \(\mu = 0\), \(\sigma = 1\) is included.

Conclusion. For every real \(x\) and every \(k \gt 0\),

\[\begin{aligned} P(X \le x) &= \Phi\left(\frac{x - \mu}{\sigma}\right), \\ P(|X - \mu| \gt k\sigma) &= 2(1 - \Phi(k)). \end{aligned}\]

Facts the proof uses.

  • for a random variable with a density, \(P(X \le x)\) is the area under the density to the left of \(x\), and a single point has probability 0;
  • the substitution rule for integrals, also over infinite intervals;
  • the symmetry \(\varphi(-u) = \varphi(u)\), which, with the total area 1, gives \(\Phi(-z) = 1 - \Phi(z)\) for every \(z\).

Each option is one sentence about the \(X\), \(\mu\) and \(\sigma\) of the theorem. Pick the one that is true in every case the theorem covers, \(\mu = 0\) and \(\sigma = 1\) included.

What else follows
  • z-scores. \(Z = (X - \mu)/\sigma\) is standard normal: for every real \(z\), \(P(Z \le z) = P(X \le \mu + \sigma z) = \Phi(z)\). Conversely, if \(Z\) is standard normal, then \(X = \mu + \sigma Z\) is normal with mean \(\mu\) and SD \(\sigma\). The number \((x - \mu)/\sigma\) is called the z-score of \(x\).
  • Any interval. For \(a \le b\), with or without the ends, \[\begin{aligned} P(a \le X \le b) &= \Phi\left(\frac{b - \mu}{\sigma}\right) \\ &\quad - \Phi\left(\frac{a - \mu}{\sigma}\right). \end{aligned}\]
  • Quantiles. For \(0 \lt p \lt 1\), the value \(c\) with \(P(X \le c) = p\) is \(c = \mu + \sigma z_p\), where \(z_p\) is the number with \(\Phi(z_p) = p\).
  • Option prices. In the Black–Scholes formula, \(N(d_1)\) and \(N(d_2)\) are values of \(\Phi\). There the log of the stock price at maturity is normal under the pricing measure, and \(N(d_2)\) is the probability, under that measure, that the call ends in the money, found by the same change of variable.
Optional: why the density has total area 1, mean μ and SD σ

Letting \(x\) grow in line 3 of the proof shows that the total area under \(f\) equals the total area under \(\varphi\). For \(\varphi\), write \(I = \int_{-\infty}^{\infty} e^{-u^{2}/2}\,du\). Then \(I^{2}\) is the integral of \(e^{-(u^{2} + v^{2})/2}\) over the whole plane, and in polar coordinates

\[\begin{aligned} I^{2} &= \int_{0}^{2\pi}\int_{0}^{\infty} e^{-r^{2}/2}\,r\,dr\,d\theta \\ &= 2\pi. \end{aligned}\]

So \(I = \sqrt{2\pi}\), and \(\varphi\) has total area 1.

With \(X = \mu + \sigma Z\) from the box above, the rules for \(aZ + b\) in the lesson on expectation and variance give \(E[X] = \mu + \sigma E[Z]\) and \(\operatorname{SD}(X) = \sigma\operatorname{SD}(Z)\). \(E[Z] = \int_{-\infty}^{\infty} u\varphi(u)\,du = 0\), because \(u\varphi(u)\) is an odd function whose integral converges. Integrating by parts with \(\varphi'(u) = -u\varphi(u)\),

\[\begin{aligned} &E[Z^{2}] \\ &= \int_{-\infty}^{\infty} u \cdot u\varphi(u)\,du \\ &= \Big[-u\varphi(u)\Big]_{-\infty}^{\infty} \\ &\quad + \int_{-\infty}^{\infty} \varphi(u)\,du \\ &= 0 + 1, \end{aligned}\]

so \(\operatorname{Var}(Z) = 1\), \(\operatorname{SD}(Z) = 1\), \(E[X] = \mu\) and \(\operatorname{SD}(X) = \sigma\).

4 Apply

Problems

Each problem states its model, and has hints and a full solution. Answers are decimals; a printed table of \(\Phi\) or software both work.

Key point To find a normal probability, turn each endpoint into \(z = (x - \mu)/\sigma\) and use \(\Phi\), with \(\Phi(-z) = 1 - \Phi(z)\). To find a cutoff with a given probability, find \(z\) from \(\Phi\) and convert back with \(x = \mu + \sigma z\). The results describe the model: data with heavier tails put more probability far from the mean than a normal model does.
Level Intro Prerequisites densities and CDFs (the lesson on PMF, PDF and CDF), expectation, variance and SD (the lesson on expectation and variance) Time ~35 min

Further reading