How to generate numbers with a normal distribution

I'm trying to seed some data, is there anyway to generate numbers in SQL Server that follow a normal distribution curve?

Like: I will specify the mean, the standard deviation and the count and I get a list of numbers back?

回答1:

quoting from http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=135026 you might try with

CREATE FUNCTION [dbo].[udf_NORMSDIST]
(
            @x FLOAT
)
RETURNS FLOAT
AS
/****************************************************************************************
NAME: udf_NORMSDIST
WRITTEN BY: rajdaksha
http://www.sqlteam.com/forums/topic.asp?TOPIC_ID=135026
DATE: 2009/10/29 
PURPOSE: Mimics Excel's Function NORMSDIST

Usage: SELECT dbo.udf_NORMSDIST(.5)


REVISION HISTORY

Date Developer Details
2010/08/11 LC Posted Function


*****************************************************************************************/

BEGIN 
DECLARE @result FLOAT
DECLARE @L FLOAT
DECLARE @K FLOAT
DECLARE @dCND FLOAT
DECLARE @pi FLOAT
DECLARE @a1 FLOAT
DECLARE @a2 FLOAT
DECLARE @a3 FLOAT
DECLARE @a4 FLOAT
DECLARE @a5 FLOAT

--SELECT @L = 0.0
SELECT @K = 0.0
SELECT @dCND = 0.0

SELECT @a1 = 0.31938153
SELECT @a2 = -0.356563782
SELECT @a3 = 1.781477937
SELECT @a4 = -1.821255978
SELECT @a5 = 1.330274429
SELECT @pi = 3.1415926535897932384626433832795

SELECT @L = Abs(@x)

IF @L >= 30
BEGIN
IF sign(@x) = 1
SELECT @result = 1
ELSE
SELECT @result = 0
END
ELSE
BEGIN
-- perform calculation
SELECT @K = 1.0 / (1.0 + 0.2316419 * @L)
SELECT @dCND = 1.0 - 1.0 / Sqrt(2 * @pi) * Exp(-@L * @L / 2.0) *
(@a1 * @K + @a2 * @K * @K + @a3 * POWER(@K, 3.0) + @a4 * POWER(@K, 4.0) + @a5 * POWER (@K, 5.0))
IF (@x < 0)
SELECT @result = 1.0 - @dCND
ELSE
SELECT @result = @dCND

END

RETURN @result
END

You can obviously offset the results to change the mean and wrap this in a function to iterate as needed to return the desired number of sample points. Let me know if you need help for this.

回答2:

random normal distribution

UPDATE TABLE 
SET COLUMN = CAST(SQRT(-2*LOG(RAND()))*SIN(2*PI()*RAND(CHECKSUM(NEWID())))as decimal(5,3))    
from TABLE

回答3:

For instance (stupid SO body parser): http://www.sqlservercentral.com/articles/SQL+Uniform+Random+Numbers/91103