# Documentation

### This is machine translation

Translated by
Mouse over text to see original. Click the button below to return to the English verison of the page.

# normcdf

Normal cumulative distribution function

## Syntax

`p = normcdf(x)p = normcdf(x,mu,sigma)[p,plo,pup] = normcdf(x,mu,sigma,pcov,alpha)[p,plo,pup] = normcdf(___,'upper')`

## Description

`p = normcdf(x)` returns the standard normal cdf at each value in `x`. The standard normal distribution has parameters `mu = 0` and `sigma = 1`. `x` can be a vector, matrix, or multidimensional array.

`p = normcdf(x,mu,sigma)` returns the normal cdf at each value in `x` using the specified values for the mean `mu` and standard deviation `sigma`. `x`, `mu`, and `sigma` can be vectors, matrices, or multidimensional arrays that all have the same size. A scalar input is expanded to a constant array with the same dimensions as the other inputs. The parameters in `sigma` must be positive.

`[p,plo,pup] = normcdf(x,mu,sigma,pcov,alpha)` returns confidence bounds for `p` when the input parameters `mu` and `sigma` are estimates. `pcov` is the covariance matrix of the estimated parameters. `alpha` specifies 100(1 - `alpha`)% confidence bounds. The default value of `alpha` is 0.05. `plo` and `pup` are arrays of the same size as `p` containing the lower and upper confidence bounds.

`[p,plo,pup] = normcdf(___,'upper')` returns the complement of the normal cdf at each value in `x`, using an algorithm that more accurately computes the extreme upper tail probabilities. You can use `'upper'` with any of the previous syntaxes.

The function `normcdf` computes confidence bounds for `p` using a normal approximation to the distribution of the estimate

`$\frac{X-\stackrel{^}{\mu }}{\stackrel{^}{\sigma }}$`

and then transforming those bounds to the scale of the output `p`. The computed bounds give approximately the desired confidence level when you estimate `mu`, `sigma`, and `pcov` from large samples, but in smaller samples other methods of computing the confidence bounds might be more accurate.

The normal cdf is

`$p=F\left(x|\mu ,\sigma \right)=\frac{1}{\sigma \sqrt{2\pi }}{\int }_{-\infty }^{x}{e}^{\frac{-{\left(t-\mu \right)}^{2}}{2{\sigma }^{2}}}dt$`

The result, p, is the probability that a single observation from a normal distribution with parameters µ and σ will fall in the interval (-∞ x].

The standard normal distribution has µ = 0 and σ = 1.

## Examples

collapse all

What is the probability that an observation from a standard normal distribution will fall on the interval [-1 1]?

```p = normcdf([-1 1]); p(2)-p(1)```
```ans = 0.6827```

More generally, about 68% of the observations from a normal distribution fall within one standard deviation, σ, of the mean, µ.