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# expinv

Exponential inverse cumulative distribution function

## Syntax

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

## Description

X = expinv(P,mu) computes the inverse of the exponential cdf with parameters specified by mean parameter mu for the corresponding probabilities in P. P and mu 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 input. The parameters in mu must be positive and the values in P must lie on the interval [0 1].

[X,XLO,XUP] = expinv(X,mu,pcov,alpha) produces confidence bounds for X when the input mean parameter mu is an estimate. pcov is the variance of the estimated mu. 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 bounds are based on a normal approximation for the distribution of the log of the estimate of mu. If you estimate mu from a set of data, you can get a more accurate set of bounds by applying expfit to the data to get a confidence interval for mu, and then evaluating expinv at the lower and upper end points of that interval.

The inverse of the exponential cdf is

$x={F}^{-1}\left(p|\mu \right)=-\mu \mathrm{ln}\left(1-p\right)$

The result, x, is the value such that an observation from an exponential distribution with parameter µ will fall in the range [0 x] with probability p.

## Examples

Let the lifetime of light bulbs be exponentially distributed with µ = 700 hours. What is the median lifetime of a bulb?

```expinv(0.50,700)
ans =
485.2030
```

Suppose you buy a box of "700 hour" light bulbs. If 700 hours is the mean life of the bulbs, half of them will burn out in less than 500 hours.