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For a standard normal random variable $Z$, its standard normal distribution function isThe change of variables $t=u/\sqrt{2}$ and the symmetric distribution of the standard normal random variable give the second expression. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable then givesIn particular, the intervals within one, two, and three standard deviations have calculated probabilities about $0.682689$, $0.954500$, and $0.997300$. These values depend on a normal distribution; they are not universal coverage guarantees for an arbitrary random variable. This is the direct connection between a Gaussian function area and the familiar central intervals.

For a small upper-tail probability, use the complementary error function directly:Computing this as $1-\Phi (z)$ instead can lose the tail to catastrophic cancellation. The Gaussian tail bound gives another useful estimate without requiring an exact error function value.

A Gaussian random variable is a random variable whose probability distribution is a normal distribution. For $\sigma >0$, write $X=\mu +\sigma Z$ with a standard normal random variable $Z$; conversely, standardization of a normal random variable recovers $Z=(X-\mu )/\sigma$. For example, $X\sim N(100,15^2)$ has mean $100$, standard deviation $15$, and variance $225$. The outcome $x=115$ corresponds to standard score $z=1$, one standard deviation above the mean.

Some definitions also allow a Gaussian random variable with $\sigma =0$. This means $X=\mu$ almost surely, with variance zero and probability distribution equal to the Dirac measure ${\delta }_{\mu }$. It has no probability density function with respect to Lebesgue measure: $\mathbb{P}(X=\mu )=1$. The positive-scale density formula and division by $\sigma$ apply only when $\sigma >0$. MIT's constructive definition explicitly includes the constant case. ^[ref].

Let $D=R-L$ be right-arm length minus left-arm length. A linear combination of jointly normal random variables is normal, andwhileThus $D\sim N(2,3)$, and standardization of a normal random variable gives

Set $x_{\ast }=0.5$ in the Gaussian process regression posterior and compute $m_{\ast }$ and $v_{\ast }$ as in part a. Standardization of a normal random variable then giveswhere $\Phi$ is the standard normal cumulative distribution function.

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.

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 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.

A standard normal random variable has the standard normal distribution, mean zero and variance one. If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, the standardization of a normal random variable $(X-\mu )/\sigma$ is a standard normal random variable. Its values report displacement from the mean in units of the standard deviation, so the result is a dimensionless quantity.