A quantile function Q is defined in terms of its distribution function F as:
$Q(p)=inf\{ x\in R:p \le F(x)\},p\in(0,1)$
But i don't understand very well how it works exactly. Suppose we are managing a cdf for Normal Distribution N(0,1)
$F(x) = \frac{1}{2} \left[ 1 + erf \left( \frac{x-\mu}{\sigma \sqrt{2}} \right) \right]$
If i want to know the probability for Q(1) i have to find the minimum $x \in R$ such that $p \le F(x)$ where in this case:
$\text{find the minimum x s.t. } \frac{1}{2} \left[ 1 + erf \left( \frac{x-0}{1 \sqrt{2}} \right) \right] \ge 1$ $\text{for }x = 0.841 \text{ we have } \frac{1}{2} \left[ 1 + erf \left( \frac{0.841}{ \sqrt{2}} \right) \right] = 1$
Hence Q(1) for N(0,1) is 0.841.
Suppose now we want to use the Weibull Distribution with $k = 1$ and $\lambda = 1$, the cdf is $F(x) = 1 - e^{-(x/\lambda)^{k}}$
From an external quantile applet ( Quantile applet ) i know that Q(1) = 0.632
but i'm not able to compute this result:
$\text{find the minimum x s.t. } 1 - e^{-(x/\lambda)^{k}} \ge 1$ $1 - e^{-(0.632/1)^{1}}$ $ = 0.468472 \neq 1$
Where i'm wrong?