Difference between two formulas of process sigma

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    Logan Luo

    Does anybody test the difference between following formulas of process sigma?

    process Sigma = NORMSINV(1-((Total Defects) / (Total Opportunities))) + 1.5
    process Sigma = 0.8406 + SQRT(29.37 – 2.221 * (ln(DPMO)))
    The second one can not handle DPMO which is greater than 560000.
    Can anyone explain the calculation behind NORMSINV(P)?


    Dr. Steve W.

    it is the inverse funciton of cumulative normal CDF: it maps the (0, 1)
    interval into (-inf, +inf) and maps .5 to 0, .68 to 1, etc…


    Satya Gopal Kalluri

    Both the formulae attempt to do the same thing.  They calculate that point on the x-axis of a standard normal distribution*, to the right of which the area under the said standard normal distribution = (Total Defects) / (Total Opportunities).

    The value 1.5 takes into account the long term mean shift of the process.
    * Standard Normal Distribution = Normal Distribution with Mean = 0 and Standard Deviation = 1

    The first formula is calculated by a method used by MS Excel, the details of which I do not know.

    I also do not know how the second formula is derived.  Maybe it comes from a method that approximates the normal distribution.  In fact, I would like to know further details.

    But, as far as the second formula is concerned, it does not work above a DPMO value of 560000 because of the error values generated from the SQRT() function.
    Error values from the SQRT() function would be generated when the result from the calculation:
    [29.37 – 2.221 x ln(DPMO)] gives a negative value.  This then depends on the result of ln(DPMO) or the figure of DPMO.
    The DPMO figure for which the above difference turns negative is not 560000, but it is:
    = EXP(29.37/2.221) = 553364.9868568

    The difference between the results given by the two formulae is <= 0.01 when 0.9141494 <= DPMO <= 271009
    Satya Gopal

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