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Let $X$: The waiting time between two earthquakes (in years). We know that $X$ is following a Weibull distribution, i.e. $X \sim W(\gamma=0, \beta, \delta)$. How can I find $\beta$ and $\sigma$ knowing that $P(X>15) = 0.3679$ and $P(X\leq 5) = 0.2835$.

I found the system of equation $-15\beta - \log(0.3679)\delta = 0$ and $-5 \beta - \log(0.7125)\delta = 0$, but it means $\beta = \delta = 0$. The answer is $\beta > 2$ and $\delta < 16$. How is that possible?

EDIT:

The Weibull distribution is defined as

$$F(x) = \begin{cases} 0, & \text{if } x < \gamma \\ 1 - \exp\left[ - \left( \frac{x-\gamma}{\delta} \right)^\beta \right], & \text{if } x \geq \gamma \end{cases}$$

Michael Hardy
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J.Doe
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    $$ \begin{align} \exp\left[ - \left( \frac 5 \delta \right)^\beta \right] & = 1-0.2835 = 0.7165 \ {} \ \exp\left[ - \left( \frac {15}\delta \right)^\beta \right] & = 0.3679 \end{align} $$ $$ \begin{align} & \left( \frac 5 \delta \right)^\beta = - \log 0.7165 \ {} \ & \left( \frac{15} \delta \right)^\beta = -\log 0.3679 \end{align} $$ $$ \begin{align} & \beta\log5-\beta\log\delta = \log(-\log 0.7165) \ {} \ & \beta\log15-\beta\log\delta = \log(-\log 0.3679) \end{align} $$ Subtracting, we get $$ \beta\log3 = \log(-\log 0.3679) - \log(-\log 0.7165) $$ – Michael Hardy Nov 14 '20 at 17:13
  • Then $$ \beta = \frac{\log(-\log 0.3679) - \log(-\log 0.7165)}{\log3} = \cdots $$ – Michael Hardy Nov 14 '20 at 17:13
  • @MichaelHardy You should put it to build a full answer. You put it to build a comment instead. Just saying ;) – J.Doe Nov 14 '20 at 17:26
  • I would have done that if there had not still been some questions in my mind about this. – Michael Hardy Nov 14 '20 at 18:02
  • @MichaelHardy Never too late to change it ;) – J.Doe Nov 14 '20 at 18:05
  • Possibly I'll look at this further later. – Michael Hardy Nov 15 '20 at 00:12

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