For each element of x, compute the cumulative distribution function (CDF) of the exponential distribution with mean parameter mu. The size of p is the common size of x and mu. A scalar input functions as a constant matrix of the same size as the other inputs.

Default value is mu = 1.

A common alternative parameterization of the exponential distribution is to use the parameter $\lambda$ defined as the mean number of events in an interval as opposed to the parameter $\mu$, which is the mean wait time for an event to occur. $\lambda$ and $\mu$ are reciprocals, i.e. $\mu \; =\; 1\; /\; \lambda$.

When called with three output arguments, i.e. [p, plo, pup], expcdf computes the confidence bounds for p when the input parameter mu is an estimate. In such case, pcov, a scalar value with the variance of the estimated parameter mu, is necessary. Optionally, alpha, which has a default value of 0.05, specifies the 100 * (1 - alpha) percent confidence bounds. plo and pup are arrays of the same size as p containing the lower and upper confidence bounds.

[…] = expcdf (…, "upper") computes the upper tail probability of the exponential distribution with parameter mu, at the values in x.

Further information about the exponential distribution can be found at https://en.wikipedia.org/wiki/Exponential_distribution

See also: expinv, exppdf, exprnd, expfit, explike, expstat

Source Code: expcdf