= Standard normal distribution
{title2=$N(0,1)$}
{wiki}
= Standard normal
{synonym}
The <standard normal distribution> is $N(0,1)$: the <normal distribution> with <mean> zero, <variance> one and <standard deviation> one. Its <standard normal density> $\phi$ integrates to one; its <standard normal distribution function> $\Phi$ records the area accumulated from $-\infty$ to a threshold. The distinction matters: $\phi(0)=1/\sqrt{2\pi}\approx0.398942$, $\Phi(0)=1/2$, and $\mathbb P(Z=0)=0$ for a <standard normal random variable> $Z$.
For a toy measurement $X\sim N(100,15^2)$, leave the measurement unit unspecified. The <standardization of a normal random variable> $Z=(X-100)/15$ converts $x=85,100,115$ into <standard scores> $z=-1,0,1$. Using the <standard normal distribution function>,
$$
\begin{aligned}
\mathbb P(85\leq X\leq115)&=\Phi(1)-\Phi(-1)\approx0.6826894921,\\
\mathbb P(100\leq X\leq115)&=\Phi(1)-\Phi(0)\approx0.3413447461,\\
\mathbb P(X=100)&=0.
\end{aligned}
$$
The second area is half the first because the <normal distribution> is a <symmetric probability distribution> about its <mean>. The numerical heights differ: $f_X(115)=\phi(1)/15$, yet the corresponding shaded areas are equal because $dx=15\,dz$ in the <change of variables>.
\Image[/standard-normal-area-and-z-score.png]
{title=Standardization changes density heights and coordinates but preserves probabilities}
{description=The same toy <normal distribution> is shown before and after <standardization of a normal random variable>. The blue and orange areas together have <probability> approximately $0.6826894921$; the orange area alone has <probability> approximately $0.3413447461$. The threshold $115$ becomes <standard score> $1$. Each <probability density function> has total area one over the whole <real line>; a single outcome has zero <probability>.}
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