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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,

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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