SQL Server RANDNORMAL Function

Updated 2023-10-18 16:12:50.517000

Description

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 [westclintech].[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 Type

table

colName	colDatatype	colDesc
Seq	int	A monotonically increasing sequence number
X	float	The random variable

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(*) as [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
-5	1
-4	30
-3	1307
-2	21510
-1	135616
0	683067
1	135698
2	21408
3	1322
4	41

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	99.9969298163252	100
kurt	-0.00220295058182929	0
skew	0.000405814354994069	0
stdev	15.0098425595394	15