Normsdist in VB5
Author Message Normsdist in VB5

I am trying to program a function in VB5 that duplicated the NORMSDIST
function in Excel,
NORMSDIST Returns the standard normal cumulative distribution. The
distribution has a mean of 0 (zero) and a standard deviation of one.

NORMSDIST(Z) where Z   is the value for which you want the distribution.

Excel gives the example NORMSDIST(1.333333) equals 0.908789
_______________________________________________________

PI =  3.14159265358979
The formula I am using  is X =  ( 1/SQR(2 * PI) ) *  EXP(- (Z ^ 2 / 2) )

well it doesn't work, I don't mean I get a syntax error, I just get wrong
numbers returned.

Does anybody know the formula to use ??.

Thank you,

Neil.

Mon, 26 Mar 2001 03:00:00 GMT  Normsdist in VB5

Quote:

>I am trying to program a function in VB5 that duplicated the NORMSDIST
>function in Excel,
>NORMSDIST Returns the standard normal cumulative distribution. The
>distribution has a mean of 0 (zero) and a standard deviation of one.

>NORMSDIST(Z) where Z   is the value for which you want the distribution.

>Excel gives the example NORMSDIST(1.333333) equals 0.908789
>_______________________________________________________

>PI =  3.14159265358979
>The formula I am using  is X =  ( 1/SQR(2 * PI) ) *  EXP(- (Z ^ 2 / 2) )

>well it doesn't work, I don't mean I get a syntax error, I just get wrong
>numbers returned.

The formula you are using is the probability density function, i.e. it
returns the *height* of the curve at point Z, not the area under the curve
to the left of Z. To calculate the area, you need to take the *integral* of
your equation from negative infinity to Z.

I would look around for some statistics packages for VB -- commercial or
otherwise. I have Crescent Software's QuickPak Scientific for DOS Basic,
but its function is accurate to only 4 significant digits. In comparing the
results with those from Excel's function, there are differences in the 6th
decimal place.

Tue, 27 Mar 2001 03:00:00 GMT

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