On 07/15/2013 12:42 PM, Brian Barker wrote:
don't stop being what they are after some arbitrary number of
significant figures - whether it be one, three, or any other.
At the fourth significant digit, 0 and 9 occur slightly (¿1:10,000?)
more frequently than 1, 2, 3, 4, 5, 6, 7, and 8. For
At 21:03 18/07/2013 +, Toki Jonathan Kantoor wrote:
On 07/15/2013 12:42 PM, Brian Barker wrote:
... don't stop being what they are after some
arbitrary number of significant figures -
whether it be one, three, or any other.
At the fourth significant digit, 0 and 9 occur
slightly
On 07/14/2013 09:54 PM, Dennis E. Hamilton wrote:
However, Benford's law is about the *first* digit of a wide variety of
numbers.
First three digits, not first digit.
The fourth and subsequent digits should be uniformly distributed.
jonathon
--
LibreOffice in a Multi-Lingual Environment.
At 12:26 15/07/2013 +, Toki Jonathan Kantoor wrote:
On 07/14/2013 09:54 PM, Dennis E. Hamilton wrote:
However, Benford's law is about the *first* digit of a wide variety
of numbers.
First three digits, not first digit. The fourth and subsequent
digits should be uniformly distributed.
On 07/13/2013 11:01 PM, Brian Barker wrote:
Unless I misunderstand,the formula =10^RAND() should create random variates in
the range (1,10) following the law.
10^RAND generates a set of random numbers that does _not_ adhere to
Benford's Law. I need a random number generator whose output does
At 21:01 14/07/2013 +, Toki Jonathan Kantoor wrote:
On 07/13/2013 11:01 PM, Brian Barker wrote:
Unless I misunderstand,the formula =10^RAND() should create random
variates in the range (1,10) following the law.
10^RAND generates a set of random numbers that does _not_ adhere to
Benford's
On 2013-07-15 09:01, Toki Kantoor wrote:
On 07/13/2013 11:01 PM, Brian Barker wrote:
Unless I misunderstand,the formula =10^RAND() should create random variates in
the range (1,10) following the law.
10^RAND generates a set of random numbers that does _not_ adhere to
Benford's Law. I need
Kantoor [mailto:toki.kant...@gmail.com]
Sent: Sunday, July 14, 2013 02:01 PM
Cc: users@global.libreoffice.org
Subject: Re: [libreoffice-users] Benford's Law
On 07/13/2013 11:01 PM, Brian Barker wrote:
Unless I misunderstand,the formula =10^RAND() should create random variates in
the range (1,10
At 21:01 14/07/2013 +, Toki Jonathan Kantoor wrote:
10^RAND generates a set of random numbers that does _not_ adhere to
Benford's Law.
Incidentally, if you would like your random numbers in binary instead
of decimal, I can provide an even easier formula for the initial
digit of Benford's
'
Cc: 'users@global.libreoffice.org'
Subject: RE: [libreoffice-users] Benford's Law
Uniform random number generators do not conform to Benford's law.
To get uniform digits in the range 1 to 10, try =FLOOR(10*RAND();1;1)
However, Benford's law is about the *first* digit of a wide variety of numbers
On 07/14/2013 05:01 PM, Toki Kantoor wrote:
On 07/13/2013 11:01 PM, Brian Barker wrote:
Unless I misunderstand,the formula =10^RAND() should create random variates in
the range (1,10) following the law.
10^RAND generates a set of random numbers that does _not_ adhere to
Benford's Law. I
At 18:59 13/07/2013 +, Toki Jonathan Kantoor wrote:
Once upon a time I had an extension that generated random numbers
that adhered to Benford's Law.
Do you need one? Unless I misunderstand, the formula
=10^RAND()
should create random variates in the range (1,10) following the law.
12 matches
Mail list logo
