A bakers oven may be out of use due for two reasons. With probability 0.8 the oven will be damaged from dirt and it will take exactly 5 minutes to repair it. With probability 0.2 the oven will need major repairs and repair time will follow a Weibull distribution with parameters with α = 6 and β = 0.5.

a) If X is the repair time of the next failure, find the cumulative distribution of X. b) Outline an inversion method to generate the failure times.

I am confused with the above problem for a I tried to generate the cdf by using the weibull cdf and adding the probabilities and arrived at:

$F(x)=0$ for $x<1-e^-(\frac{x}{6})^{0.5}$

$F(x)=0.2$ for $1-e^-(\frac{x}{6})^{0.5}<x<5$

$F(x)=1$ for $5<x$

But I am not sure if this is right and then I also am not sure how to use the inversion method on a 3 tier function. Any help would be appreciated!


Correction: the CDF is $F=1-\exp\sqrt{x/6}$ for $x\ge 0$ and $0$ otherwise. (Look up the usual Weibull CDF wherever you prefer, or integrate it's pdf.)I don't know how you got your answer, but any piecewise formula for the CDF add a function of $x$ shouldn't transition at $x$-dependent values. I'll leave you to express $x$ as a function of $F$.

  • $\begingroup$ Should we end up with $F(x)=4x+0.2(1-e^-(\frac{x}{6})^0.5)$ as the cdf for $x>=0$ and $F(x)=0$ otherwise as the cdf then? $\endgroup$ – Mackie Oct 29 '18 at 19:20
  • $\begingroup$ @Mackie The thing you need to multiply by $0.8$ isn't $5x$; it's $1$ for $x\ge 5$, $0$ otherwise. The easiest way to sample is to first randomly pick whether to return $5$ or a Weibull sample. $\endgroup$ – J.G. Oct 29 '18 at 20:06
  • $\begingroup$ How do we arrive at the cdf being $1−exp(-sqrt(x/6))$ from multiplying 1 by 0.8? $\endgroup$ – Mackie Oct 29 '18 at 20:20
  • $\begingroup$ $F(x)=0.8-0.2exp(sqrt(x/6))$? $\endgroup$ – Mackie Oct 29 '18 at 21:13

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