Evaluate the cumulative distribution function of a Weibull distribution
Source:R/Weibull.R
cdf.Weibull.Rd
Evaluate the cumulative distribution function of a Weibull distribution
Usage
# S3 method for class 'Weibull'
cdf(d, x, drop = TRUE, elementwise = NULL, ...)
Arguments
- d
A
Weibull
object created by a call toWeibull()
.- x
A vector of elements whose cumulative probabilities you would like to determine given the distribution
d
.- drop
logical. Should the result be simplified to a vector if possible?
- elementwise
logical. Should each distribution in
d
be evaluated at all elements ofx
(elementwise = FALSE
, yielding a matrix)? Or, ifd
andx
have the same length, should the evaluation be done element by element (elementwise = TRUE
, yielding a vector)? The default ofNULL
means thatelementwise = TRUE
is used if the lengths match and otherwiseelementwise = FALSE
is used.- ...
Arguments to be passed to
pweibull
. Unevaluated arguments will generate a warning to catch mispellings or other possible errors.
Value
In case of a single distribution object, either a numeric
vector of length probs
(if drop = TRUE
, default) or a matrix
with
length(x)
columns (if drop = FALSE
). In case of a vectorized distribution
object, a matrix with length(x)
columns containing all possible combinations.
See also
Other Weibull distribution:
pdf.Weibull()
,
quantile.Weibull()
,
random.Weibull()
Examples
set.seed(27)
X <- Weibull(0.3, 2)
X
#> [1] "Weibull(shape = 0.3, scale = 2)"
random(X, 10)
#> [1] 1.440254e-05 4.128282e+01 2.513340e-03 2.840554e+00 7.792913e+00
#> [6] 1.472187e+00 4.985175e+01 7.900541e+02 1.972819e+01 1.063212e+01
pdf(X, 2)
#> [1] 0.05518192
log_pdf(X, 2)
#> [1] -2.89712
cdf(X, 4)
#> [1] 0.7080417
quantile(X, 0.7)
#> [1] 3.713233