Weibull distribution median

Question

I am working with the Weibull model with this pdf:

The standard pdf that the some R functions perform is :

I am required to find the median, which in the standard parameterisation is $\lambda (\ln2)^{\frac{1}{k}}$. I have worked out the median of my given pdf to be $\lambda^{-\alpha}(\ln 2)^{\frac{1}{\alpha}}$.

So here the shape k is $\alpha$; let $\lambda$ be the standard scale, the scale in my pdf can be denoted as $\lambda'$. So $\lambda'= \lambda^{-k}= \lambda^{-\alpha}$. Please check my reparameterisation for me. I have obtained nonsensical results with those $\alpha, \lambda$ values.

Answer 1

Your calculation is wrong. $\lambda'$ is not a scale parameter, so the scale is not $\lambda'$.

The scale is $\lambda$ - i.e. $(\lambda')^{-1/\alpha}$.