dqweibull(x, a = .8, qdist = 1.2, b = 1, forceExpectation = F)
pqweibull(q, a = .8, qdist = 1.2, b = 1, forceExpectation = F)
qqweibull(p, a = .8, qdist = 1.2, b = 1, forceExpectation = F)
rqweibull(n = 1, a = .8, qdist = 1.2, b = 1, forceExpectation = F)
qweibullExpectation(a = .8, qdist = 1.2, b = 1)
qweibullHazard(x, a = .8, qdist = 1.2, b = 1, forceExpectation = F)TRUE, the expectation of the distribution is forced to be 1 by letting b be a function of the other parameters.