Normal inverse cumulative distribution function

`X = norminv(P,mu,sigma)`

[X,XLO,XUP] = norminv(P,mu,sigma,pcov,alpha)

`X = norminv(P,mu,sigma)`

computes
the inverse of the normal cdf using the corresponding mean `mu`

and
standard deviation `sigma`

at the corresponding probabilities
in `P`

. `P`

, `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, and the values
in `P`

must lie in the interval [0 1].

`[X,XLO,XUP] = norminv(P,mu,sigma,pcov,alpha)`

produces
confidence bounds for `X`

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`

. `XLO`

and `XUP`

are
arrays of the same size as `X`

containing the lower
and upper confidence bounds.

The function `norminv`

computes confidence
bounds for `P`

using a normal approximation to the
distribution of the estimate

$$\widehat{\mu}+\widehat{\sigma}q$$

where *q* is the `P`

th quantile
from a normal distribution with mean 0 and standard deviation 1. 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 may be more accurate.

The normal inverse function is defined in terms of the normal cdf as

$$x={F}^{-1}(p|\mu ,\sigma )=\{x:F(x|\mu ,\sigma )=p\}$$

where

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

The result, *x*, is the solution of the integral
equation above where you supply the desired probability, *p*.

Find an interval that contains 95% of the values from a standard normal distribution.

x = norminv([0.025 0.975],0,1) x = -1.9600 1.9600

Note that the interval `x`

is not the only
such interval, but it is the shortest.

xl = norminv([0.01 0.96],0,1) xl = -2.3263 1.7507

The interval `xl`

also contains 95% of the
probability, but it is longer than `x`

.

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