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Thread: Weibull probability? not sure about how to generate parameters

  1. #1
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    Weibull probability? not sure about how to generate parameters

    So I've got data from a reflux ratio

    0.78, 0.79, 0.87, 0.91, 0.92, 0.92, 0.92, 0.93, 0.94, 0.95, 0.95, 0.95, 0.95, 0.96, 0.96, 0.96, 0.97, 0.97, 0.97, 0.99, 0.99, 0.99, 0.99, 0.99, 0.99


    I've been asked to get the probability less or equal to 0.94. This is fine, take the mean(0.94) and it's SD(0.0557), find the Z value and you get 0.5000 = 50%

    However now it's asking me to find the probability of observing a response of 0.94 or less if it's assumed the data comes from a Weibull distribution.

    I know the CDF is 1-e^-(lambda(x))^beta

    However I'm not really sure how I generate those parameters? or how I go about this?
    Can anyone assist me please?
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  2. #2
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    Re: Weibull probability? not sure about how to generate parameters

    The first P attained (P(X<0.94)=0.5000 is under the assumption of normal distribution.
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  3. #3
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    Re: Weibull probability? not sure about how to generate parameters

    The first thought to come to mind is to find the maximum likelihood estimators for the two Weibull distribution parameters, $\lambda,~k$

    You can find the equations you must numerically solve here, https://en.wikipedia.org/wiki/Weibul...mum_likelihood

    Then construct the estimated CDF using those estimated parameters and solve your problem.

    I'm getting

    $\hat{k} \approx 27.7$

    $\hat{\lambda} \approx 0.962$

    I highly recommend you verify these.
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