Here is a pretty close version I wrote in T/SQL:
create function normdist(@value float,
@mean float,
@sigma float,
@cummulative bit)
returns numeric(12,2)
begin
declare @x float
declare @z float
declare @t float
declare @ans float
declare @returnvalue float
select @x = (@value-@mean)/@sigma
if (@cummulative = 1)
begin
select @z = abs(@x)/sqrt(2.0)
select @t = 1.0/(1.0+0.5*@z)
select @ans = @t*exp(-@z*@z-1.26551223+@t*(1.00002368+@t*(0.37409196+@t*(0.09678418+@t*(-0.18628806+@t*(0.27886807+@t*(-1.13520398+@t*(1.48851587+@t*(-0.82215223+@t*0.17087277)))))))))/2.0
if (@x <= 0)
select @returnvalue = @ans
else
select @returnvalue = 1-@ans
end
else
begin
select @cummulative = exp(-@x*@x/2.0)/sqrt(2.0*3.14159265358979)
end
return cast(@returnvalue * 100 as numeric(12,2))
end
J wrote:
NORMDIST(X, mean, Standard_Dev,True) in SQL 2005?
25-Oct-07
Hi
In Excel I can find the normal distribution with the formula
=NORMDIST(x, mean, standard dev, cumulative
How is this done in SQL 2005? I am having a difficult time finding this on
the web so hopefully someone can assist me with this or at least point me in
the right direction
Thank you in advance for your help
All the best
JC
Previous Posts In This Thread:
On Thursday, October 25, 2007 11:21 AM
J wrote:
NORMDIST(X, mean, Standard_Dev,True) in SQL 2005?
Hi
In Excel I can find the normal distribution with the formula
=NORMDIST(x, mean, standard dev, cumulative
How is this done in SQL 2005? I am having a difficult time finding this on
the web so hopefully someone can assist me with this or at least point me in
the right direction
Thank you in advance for your help
All the best
JC
On Thursday, October 25, 2007 11:46 AM
Russell Fields wrote:
JC,SQL Server implements functions like this through Analysis Services.
JC
SQL Server implements functions like this through Analysis Services. Her
is an Excel / SQL Server page for SQL Server 2000
http://msdn2.microsoft.com/en-us/library/aa178231(SQL.80).asp
RLF
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