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    Re: Lunars using Bennett
    From: Peter Fogg
    Date: 2008 Apr 11, 03:31 +1000

    Frank wrote:
    > Your result, 0.7, is just right ...
    > ... what we need is the standard deviation of a uniform
    > distribution one unit wide ... and
    > that happens to be 0.288. And 0.288*sqrt(N) with N=6 is 0.7 just as you
    > found. It works.
    
    Thanks to Frank (and Bill Noyce and Alex) for throwing much
    appreciated light on this statistical question: the extent of error
    when adding together rounded values, in this case six.
    
    It began when Alex claimed: "the errors will accumulate", which was
    later qualified to: "They will do both things: cancel but still
    accumulate" which seemed difficult but it turns out that this is
    indeed so.  Kind of.
    
    In the example using Bennett tables they mostly cancel out, but
    accumulate to the extent that an error, on average, of  0.7 of a whole
    number can be expected when adding 6 rounded values, compared to
    adding values expressed to the nearest tenth.  Its a fairly modest
    accumulation.
    
    Actually, I prefer to think in terms of the result of my experiment:
    that in 86% of a random sample, the error was within 1 minute of arc,
    which seems quite a usefully practical result for a method that uses
    values rounded to whole minutes of arc, rather than the more commonly
    employed tenths.
    
    Remember that more precise Bennett tables are freely available for use
    with lunars.
    
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