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Six Sigma and miners rescue effort?

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  • #30001

    Jen
    Participant

    Ok, I know this is coming out of left field but I’m just curious … Does anyone know if Six Sigma was used in the rescue efforts of the nine miners in Pennsylvania this past weekend? I’m the communications specialist on Six Sigma within my organization, and I’m always looking for little tidbits I can use to help communicate the value of Six Sigma. I have no reason to think that Six Sigma was used … I just thought that if it was, then it would be a neat story. Thanks for your help. 

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    #77744

    Carl H
    Participant

    I dont know of any 6 sigma used.
    A (bad taste)quiz question related might be:
    1)  If the mining company had 1000 miners working 250 days per year and each day was an opportunity to get trapped what is the sigma level of the copmany if these are the only 9 to be trapped in the 10 year company history?  (About 4.5 long term, 6 sigma short term)
    Carl

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    #77748

    Patrick
    Participant

    Hi carl h,
     
    I was browsing through the forum and saw your problem.  The way I read it, D=9 (miners trapped), OPxU=1000 miners times 10 years times 250 days = 2 500 000.
    This gives a 3.6 DPMO which is roughly 6 sigma LT.
     
    What am I doing wrong?
     
    Patrick

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    #77752

    Srinivas Ranga
    Member

    I do not know about 6Sigma being used in rescue but most of the methods used were Lean techniques. As Lean teaches doing lot in less time with less man power etc,. I have no doubt that Lean techniques were used in the rescue.

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    #77753

    Carl H
    Participant

    Patrick,
    I think you may have looked up 3.6 DPMO on the Zst or short term sigma levle table.  I treated this as long term (10 years) and used a standard normal Z table which gives ~4.5 sigma long term. 
    Carl

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    #77754

    Jen
    Participant

    Appreciate the feedback.

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    #77777

    Patrick
    Participant

    Actually I didn’t. Since we are already talking about a discrete data scenario, I was taught that this is to be considered long term. So now that we have 9 defects divided by the total opportunities (2 500 000) gives you a DPO of 0.0000036 multiplied by 10^6 equals 3.6 DPMO, which is about 6 sigma long term (and considering a standard shift, would be 7.5 sigma short term.Cheers,

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