Let X∼Weibull(k=1,λ=4)X \sim \text{Weibull}(k=1, \lambda=4)X∼Weibull(k=1,λ=4). What is the exact value of the median of XXX?
4ln(2)4 \ln(2)4ln(2)
4ln(2)\frac{4}{\ln(2)}ln(2)4
2ln(4)2 \ln(4)2ln(4)
4/24/24/2