Evaluate the probability mass function of a Weibull distribution

Please see the documentation of Weibull() for some properties of the Weibull distribution, as well as extensive examples showing to how calculate p-values and confidence intervals.

Usage

# S3 method for class 'Weibull'
pdf(d, x, drop = TRUE, elementwise = NULL, ...)

# S3 method for class 'Weibull'
log_pdf(d, x, drop = TRUE, elementwise = NULL, ...)

Arguments

d

A Weibull object created by a call to Weibull().

x

A vector of elements whose 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 of x (elementwise = FALSE, yielding a matrix)? Or, if d and x have the same length, should the evaluation be done element by element (elementwise = TRUE, yielding a vector)? The default of NULL means that elementwise = TRUE is used if the lengths match and otherwise elementwise = FALSE is used.

...

Arguments to be passed to dweibull. 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: cdf.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