Standard Normal Probability Function

A Standard Normal Probability Function is a Gaussian function with parameters [math]\displaystyle{ \mu=0, \sigma=1 }[/math].

AKA: [math]\displaystyle{ \mathcal{N}(0,1) }[/math], z-Distribution.
Context:
- It can represented/calculated by equation [math]\displaystyle{ f(z) = \frac{1}{2 \pi} e^{\frac{-z^2}{2}} }[/math].
- …
Counter-Example(s):
- General Normal Distribution.
See: Gaussian Distribution, Probability Density Function, Gaussian Integral, Inflection Point.

References

(Wikipedia, 2016) ⇒ https://en.wikipedia.org/wiki/normal_distribution#Standard_normal_distribution Retrieved:2016-10-4.
- The simplest case of a normal distribution is known as the standard normal distribution. This is a special case when μ=0 and σ=1, and it is described by this probability density function: : [math]\displaystyle{ \phi(x) = \frac{e^{- \frac{\scriptscriptstyle 1}{\scriptscriptstyle 2} x^2}}{\sqrt{2\pi}}\, }[/math] The factor [math]\displaystyle{ 1/\sqrt{2\pi} }[/math] in this expression ensures that the total area under the curve [math]\displaystyle{ \phi(x) }[/math] is equal to one. ^[1] The ½ in the exponent ensures that the distribution has unit variance (and therefore also unit standard deviation). This function is symmetric around x=0, where it attains its maximum value [math]\displaystyle{ 1/\sqrt{2\pi} }[/math] ; and has inflection points at +1 and −1. Authors may differ also on which normal distribution should be called the "standard" one. Gauss defined the standard normal as having variance [math]\displaystyle{ \sigma^2 = \frac{1}{2} }[/math], that is : [math]\displaystyle{ \phi(x) = \frac{e^{-x^2}}{\sqrt\pi}\, }[/math] Stigler goes even further, defining the standard normal with variance [math]\displaystyle{ \sigma^2 = \frac{1}{(2\pi)} }[/math] : : [math]\displaystyle{ \phi(x) = e^{-\pi x^2} }[/math]

http://math2.org/math/stat/distributions/z-dist.htm
- QUOTE: The z- is a N(0, 1) distribution, given by the equation: [math]\displaystyle{ f(z) = \frac{1}{2 \pi} e^{\frac{-z^2}{2}} }[/math]
  The area within an interval [math]\displaystyle{ (a,b) = \operatorname{normalcdf}(a,b) = \int_a^b e^2, dz }[/math] (It is not integrable algebraically.)
  The Taylor expansion of the above assists in speeding up the calculation: [math]\displaystyle{ \operatorname{normalcdf}(-\infty,z) = \frac{1}{2} + \frac{1}{\sqrt{2 \pi}} \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)2^k k!} }[/math]