On Dec 8, 6:28*pm, John Hall wrote:
In article
,
*Teignmouth writes:
Not sure if I'm reading this correct, and it's from Wikipedia, but 3
Standard Deviations = 1 in 370 years, 4 Standard Deviations = 1 in
15,787 years, and 5 Standard Deviations = 1,744,277 years.
http://en.wikipedia.org/wiki/Normal_distribution
I've calculated the December mean from 1971 - 2009 as +5.0c, with a
Standard Deviation of 1.5c, so by my calculations the current CET that
is running at -2.0c is somewhere between 4 & 5 Standard Deviations.
But you aren't comparing like with like. There is going to be much
greater variation in the average for a spell of 7 days (1st to 7th
December) than there is for the whole month. If the average CET for the
whole month turned out to be -2.0C - or even anywhere near that - then
it really would be something.
--
John Hall
* * * * * * * *"I look upon it, that he who does not mind his belly,
* * * * * * * * will hardly mind anything else."
* * * * * * * * * * * * * * * * * * * * * *Dr Samuel Johnson (1709-84)
John, I forgot to say in my analysis that I was stating that if it
continued and ended up at -2c for the whole month it would be between
4 & 5 Standard Deviations, but obviously it probably won't stay that
way.
Prodata, I use an Excel Spreadsheet Function *(STDEV) which estimates
the Standard Deviation based on a sample of the data set 1971-2009 so,
can't say whether they are normally distributed as I don't have a
degree in maths. There is also a SKEW Function, but I'm not sure how
to use that one with the STDEV Function.
If I use STDEVP the Standard Deviation for 1971-2009 is still 1.5c.
*Excel Functions:
STDEV Estimates standard deviation based on a sample
STDEVA Estimates standard deviation based on a sample, including
numbers, text, and logical values
STDEVP Calculates standard deviation based on the entire population
STDEVPA Calculates standard deviation based on the entire population,
including numbers, text, and logical values
The mean December CET for the whole data set from 1659 to 2009 is
+4.1c, with a Standard Deviation of 1.7c, so 3 Standard Deviations =
-1.1c, 4 Standard Deviations = -2.8c & 5 Standard Deviations = -4.5c.
I get the same result for 1659 to 2009 if I use the STDEVP Function.
Hope that helps.