Plot the Weibull Distribution Hazard Function

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Dear Sirs, I hope you are well. I have set of data in Excel and I want to plot the Weibull distribution function. In the following link, we have an example which is available online. https://www.mathworks.com/examples/statistics/mw/stats-ex39027752-plot-the-weibull-distribution-hazard-function As you can see here the code is: t = 0:0.1:4.5; h1 = exppdf(t,0.8862)./(1-expcdf(t,0.8862)); h2 = wblpdf(t,1,2)./(1-wblcdf(t,1,2)); plot(t,h1,'--',t,h2,'-') My question is about the number 0.8862 in the exponential distribution. How can it be calculated for a set of data to put it in the code? I think this number in h1 is teta, 1 in h2 is alpha and 2 is beta in weibull distribution. I can import the data from excell to matlab instead of t in this code, but I want a help if someone has experience in plotting hazard function for the set of data. Thank you so much for your help and consideration.

Risposte (1)

Star Strider
Star Strider il 9 Mag 2018
According to the documentation for the exppdf (link) function, the second parameter (here 0.8862) is the mean of the data used to calculate it. See the documentation for the Exponential Distribution (link) for details.
  2 Commenti
Hadi Naderiallaf
Hadi Naderiallaf il 9 Mag 2018
Is 0.8862 mean of the data or the mu parameter of exponential distribution?
Star Strider
Star Strider il 9 Mag 2018
As I understand it, both. Note that ‘mu’ is the mean survival time, since that is what the exponential distribution is describing.

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