The Normal (or Gaussian) distribution describes a value that clusters around a center $\mu$ and spreads out symmetrically with typical deviation $\sigma$. Heights, measurement errors, and averages of many repetitions all follow this shape closely.
Probability density function
Cumulative distribution function
The integral of $f$ has no closed formula. It is written with the symbol $\Phi$ after standardizing, which is what Section 02 is about.
Parameters and support
Where these come from
Unlike other distributions, nothing needs to be computed: the Normal is defined so that its two parameters are directly its mean and its variance. Writing $X \sim N(\mu, \sigma^2)$ already tells you both. Watch the convention: the second parameter is the variance $\sigma^2$, not the standard deviation.
Adult heights in a country are well modeled by $X \sim N(170, 8^2)$, in centimeters. What fraction of people are at most $180$ cm tall?
A printed z table lists $\Phi(z)$ for $z$ rounded to two decimals. The row picks the first decimal of $z$, the column picks the second. For $z = 1.25$: row $1.2$, column $.05$.
This kind of table is normally used in exams and textbooks, where a calculator with a built-in $\Phi$ function is not available and the probability has to be looked up by hand instead.
| z | .00 | .01 | .02 | .03 | .04 | .05 |
|---|---|---|---|---|---|---|
| 1.0 | 0.8413 | 0.8438 | 0.8461 | 0.8485 | 0.8508 | 0.8531 |
| 1.1 | 0.8643 | 0.8665 | 0.8686 | 0.8708 | 0.8729 | 0.8749 |
| 1.2 | 0.8849 | 0.8869 | 0.8888 | 0.8907 | 0.8925 | 0.8944 |
| 1.3 | 0.9032 | 0.9049 | 0.9066 | 0.9082 | 0.9099 | 0.9115 |
Move $\mu$ to slide the bell along the axis and $\sigma$ to stretch or squeeze it. The total area stays $1$, so a wider bell must be a shorter one. Move $t$ to shade $P(X \le t)$.
The 68-95-99.7 rule
About $68\%$ of the probability lies within $1\sigma$ of the mean, $95\%$ within $2\sigma$, and $99.7\%$ within $3\sigma$. Values beyond $3\sigma$ are rare, which is why "how many sigmas away" is a natural way to describe how unusual an observation is.
Linear transformations stay Normal
Shifting and scaling a Normal variable produces another Normal variable. This is what makes standardization possible.
Sums of independent Normals
Adding independent Normal variables gives a Normal again. Means add, and variances add.
| Symbol | Meaning |
|---|---|
| $X \sim N(\mu, \sigma^2)$ | a Normal variable with mean $\mu$ and variance $\sigma^2$ |
| $\mu$ | mean, the center of the bell |
| $\sigma,\ \sigma^2$ | standard deviation and variance, the width of the bell |
| $\Phi$ | CDF of the standard Normal, defined in Section 02 |
Since the Normal CDF has no formula, its values are tabulated. Tabulating a table for every $(\mu, \sigma)$ pair would be hopeless, so every Normal is converted to the single reference case $Z \sim N(0,1)$, called the standard Normal.
The z-score
Subtracting $\mu$ centers the variable at $0$ and dividing by $\sigma$ rescales its spread to $1$. The value $z$ answers "how many standard deviations above or below the mean is this observation."
Any probability becomes a Φ lookup
Symmetry shortcut
Tables usually list only $z \ge 0$. Negative values follow directly from the symmetry of the bell.
Scores on an exam follow $X \sim N(70, 10^2)$. A student scores $85$. How unusual is that?
Same recipe every time
Every Normal probability question follows these three moves: standardize, look up $\Phi$, and use symmetry or complements to get the final probability. The interactive below replaces the printed table.
Slide $z$ along the standard Normal. The shaded area is $\Phi(z) = P(Z \le z)$, what a printed z table lists. Check the symmetry: the readout at $z$ and at $-z$ always adds up to $1$.
Take independent copies $X_1, X_2, \ldots$ of a variable with mean $\mu$. The Law of Large Numbers says the running average approaches $\mu$ as $n$ grows.
Why it works
The average has mean $\mu$ and variance $\sigma^2/n$, which shrinks to $0$ as $n$ grows, so the average stays close to $\mu$.
Not a promise about individual outcomes
A streak of low rolls does not make the next roll "due" to be high. That reasoning is the gambler's fallacy.
On a European roulette wheel, a $1$€ bet on red pays out with probability $18/37$. The casino's profit per bet is a random variable with mean $\mu = 1/37 \approx 0.027$€.
Connection to simulation
Every "resample" button on these visual guides relies on the LLN: estimating a probability by simulating many trials works because the observed frequency converges to the true probability.
Roll a fair die over and over and track the running average, which should approach $\mu = 3.5$. Early on the path swings wildly, then the swings shrink. Resample to see that every path is different but they all converge to the same value.
The LLN says $\bar{X}_n$ is close to $\mu$. The CLT adds that its fluctuations around $\mu$ follow a bell curve for large $n$, whatever distribution the $X_i$ have.
Equivalent forms
The sum $S_n = X_1 + \cdots + X_n$ follows the same rule.
The √n scale
Uncertainty shrinks like $\sigma/\sqrt{n}$: halving it takes four times as much data. That is why polls of a few thousand people can describe millions.
Serving one customer takes an Exponential time with mean $2$ minutes, so each service time has $\mu = 2$ and $\sigma^2 = 4$. What is the chance that serving $36$ customers takes more than $80$ minutes in total?
The slider $n$ sets how many dice get rolled and averaged together each time. Click roll and watch each new average appear as a bar. At $n=1$ every value from $1$ to $6$ is equally likely, a flat spread. Raise $n$ to average more dice per roll: the bars concentrate near $3.5$ and begin to match the bell curve on top: $N\!\left(3.5, \frac{35/12}{n}\right)$, the Normal predicted by the CLT.
| Object | Statement | Use it for |
|---|---|---|
| $N(\mu, \sigma^2)$ | $f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-(x-\mu)^2 / 2\sigma^2}$, mean $\mu$, variance $\sigma^2$ | values built from many small independent effects |
| z-score | $Z = \frac{X-\mu}{\sigma} \sim N(0,1)$, so $P(X \le x) = \Phi\big(\frac{x-\mu}{\sigma}\big)$ and $\Phi(-z) = 1 - \Phi(z)$ | turning any Normal question into a $\Phi$ lookup |
| LLN | $\bar{X}_n \to \mu$ as $n \to \infty$, since $\text{Var}(\bar{X}_n) = \sigma^2/n \to 0$ | the long-run value of an average |
| CLT | $\bar{X}_n \approx N\big(\mu, \frac{\sigma^2}{n}\big)$ and $S_n \approx N(n\mu, n\sigma^2)$ for large $n$ | probabilities about sums and averages, regardless of the underlying distribution |