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Normal inverse cumulative distribution function

`X = norminv(P)`

X = norminv(P,mu,sigma)

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

`X = norminv(P)`

computes the inverse of the standard normal cdf. The
standard normal distribution has parameters `mu = 0`

and ```
sigma =
1
```

. `P`

can be a vector, matrix, or multidimensional array, and
the values in `P`

must lie in the interval [0 1].

`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]) 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]) xl = -2.3263 1.7507

The interval `xl`

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

.

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