Standard normal distribution ($N(0,1)$)

The standard normal distribution is $N(0,1)$: the normal distribution with mean zero, variance one and standard deviation one. Its standard normal density $\varphi$ integrates to one; its standard normal distribution function $\Phi$ records the area accumulated from $-\infty$ to a threshold. The distinction matters: $\varphi (0)=1/\sqrt{2\pi }\approx 0.398942$, $\Phi (0)=1/2$, 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,The second area is half the first because the normal distribution is a symmetric probability distribution about its mean. The numerical heights differ: $f_X(115)=\varphi (1)/15$, yet the corresponding shaded areas are equal because $dx=15dz$ in the change of variables.

The standard normal density is

Completing the square in the standard normal density gives $\mathbb{E}[e^{tZ}]=e^{t^2/2}$. Differentiating at zero yields $\mathbb{E}[Z]=0$, $\mathbb{E}[Z^2]=1$, and $\mathbb{E}[Z^4]=3$.

For $Z$ with a standard normal distribution and $a\ge 0$, split the Gaussian integral at zero and complete the square to obtain $\mathbb{E}{e}^{a|Z|}=2e^{a^2/2}\Phi (a)\le 2e^{a^2/2}$, where $\Phi$ is the standard normal distribution function. This converts a bound on an absolute Gaussian random variable into a two-sided Gaussian tail bound using Markov inequality.

The standard normal distribution function is the cumulative distribution function of a standard normal random variable $Z$:Here $\varphi$ is the standard normal density and $\mathrm{erf}$ is the error function. The symmetric probability distribution gives $\Phi (-x)=1-\Phi (x)$ and $\Phi (0)=1/2$, while differentiation gives ${\Phi }^{\prime }(x)=\varphi (x)$. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable gives $\mathbb{P}(X\le x)=\Phi ((x-\mu )/\sigma )$; subtracting two values calculates an interval probability.

The standard normal quantile $z_q$ is the unique real number satisfying $\mathbb{P}(Z\le z_q)=q$ for $Z\sim N(0,1)$ and $0<q<1$.

If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, then $Z=(X-\mu )/\sigma$ has the standard normal distribution. The change-of-variables formula for a probability density gives $f_Z(z)=\sigma f_X(\mu +\sigma z)=\varphi (z)$, the standard normal density. Thus standardization of a normal random variable removes both parameters without changing the probabilities of corresponding events.