RANDNORMAL
Updated: 31 March 2014 
Use the table-valued function RANDNORMAL to generate a sequence of random numbers from the normal distribution with mean @mu and standard deviation @sigma.
Syntax
SELECT * FROM [wctMath].[wct].[RANDNORMAL](
  <@Rows, int,>
 ,<@mu, float,>
 ,<@sigma, float,>)
Arguments
@Rows 
the number of rows to generate. @Rows must be of the type int or of a type that implicitly converts to int.
@mu
the mean of the distribution. @mu must be of the type float or of a type that implicitly converts to float.
@sigma
the standard deviation of the distribution. @sigma must be of the type float or of a type that implicitly converts to float.
Return Types
RETURNS TABLE (
      [Seq] [int] NULL,
      [X] [float] NULL
)
Remarks
·         @sigma must be greater than zero.
·         If @mu is NULL then @mu is set to zero.
·         If @sigma is NULL then @sigma is set to 1.
·         If @Rows is less than 1 then no rows are returned.
Examples
In this example we create a sequence 1,000,000 truncated random numbers from a standard normal distribution and COUNT the results. Elementary statistics leads us to expect the results to be distributed approximately like this:
    
        
            | X | COUNT | 
        
            | -4 | 31 | 
        
            | -3 | 1318 | 
        
            | -2 | 21400 | 
        
            | -1 | 135905 | 
        
            | 0 | 682689 | 
        
            | 1 | 135905 | 
        
            | 2 | 21400 | 
        
            | 3 | 1318 | 
        
            | 4 | 31 | 
    
SELECT
   X,
   COUNT(*)
FROM (
   SELECT 
      wct.TRUNC(x,0) as x
   FROM
      wct.RANDNORMAL(
         1000000, --@Rows
         NULL,    --@mu
         NULL     --@sigma
         )
   )n
GROUP BY
   X
ORDER BY
   1
This produces the following result. Your results will be different.
    
        
            | X | COUNT | 
        
            | -4 | 40 | 
        
            | -3 | 1315 | 
        
            | -2 | 21185 | 
        
            | -1 | 136077 | 
        
            | 0 | 682787 | 
        
            | 1 | 135712 | 
        
            | 2 | 21556 | 
        
            | 3 | 1289 | 
        
            | 4 | 39 | 
    
In this example we generate 1,000,000 random numbers from a normal distribution with a mean of 100 and a standard deviation of 15. We calculate the mean, standard deviation, skewness, and excess kurtosis from the resultant table and compare those values to the expected values for the distribution.
DECLARE @size as int = 1000000
DECLARE @mu as float = 100
DECLARE @sigma as float = 15
DECLARE @mean as float = @mu
DECLARE @var as float = POWER(@mu, 2)
DECLARE @stdev as float = @sigma
DECLARE @skew as float = 0
DECLARE @kurt as float = 0
 
SELECT
   stat,
   [RANDNORMAL],
   [EXPECTED]
FROM (
   SELECT
      x.*
   FROM (
      SELECT
         AVG(x) as mean_NORMAL,
         STDEVP(x) as stdev_NORMAL,
         wct.SKEWNESS_P(x) as skew_NORMAL,
         wct.KURTOSIS_P(x) as kurt_NORMAL
      FROM 
         wct.RANDNORMAL(@size,@mu,@sigma)
      )n
   CROSS APPLY(
      VALUES
         ('RANDNORMAL','avg', mean_NORMAL),
         ('RANDNORMAL','stdev', stdev_NORMAL),
         ('RANDNORMAL','skew', skew_NORMAL),
         ('RANDNORMAL','kurt', kurt_NORMAL),
         ('EXPECTED','avg',@mean),
         ('EXPECTED','stdev',@stdev),
         ('EXPECTED','skew',@skew),
         ('EXPECTED','kurt',@kurt)
      )x(fn_name,stat,val_stat)     
   )d
PIVOT(sum(val_stat) FOR fn_name in([RANDNORMAL],[EXPECTED])) P
This produces the following result (your result will be different).
    
        
            | stat | RANDNORMAL | EXPECTED | 
        
            | avg | 100.0010563 | 100 | 
        
            | kurt | -0.001653991 | 0 | 
        
            | skew | 0.004052033 | 0 | 
        
            | stdev | 15.00480078 | 15 | 
    
 
See Also