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How to find Inverse of NORMSDIST function
ie NORMSINV
It Returns the inverse of the standard normal cumulative distribution. The distribution has a mean of zero and a standard deviation of one.
Syntax
double NORMSINV(double probability)
Probability is a probability corresponding to the normal distribution.
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thanks 4 reply
But I m not getting exact answer when i match my answer from ms excel
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No, that's not what is meant by "inverse". It's the inverse distribution function he's looking for.
I'm the ocean. I'm a giant undertow.
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You should check out http://www.google.com.au/codesearch[^] first when you have a well known function that you want, I had a quick check and there seems to be implementations there.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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I Didn't see any implementations there. If u have seen any then plz provide some link.
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There are multiple implementations there - it is really in your interest to learn how to search for them. I bet you simply typed in "NORMSINV" and just got lots of references to uses of the function, gave up and came back here. You have to be a little creative and try to make the search engine find function definitions. I went to the same page and with my next search came up with several implementations.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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The inverse CDF of the normal distribution doesn't have an analytical form. You can find an approximation here[^].
The other place to look is in the online version of Numerical Recipes[^]. They have an algorithm for the inverse normal CDF.
You can also use the following taken from here[^].
What is a good approximation to the inverse of the
cumulative normal distribution function? Mathematically we can
write this as: Find X such that Q(X) = p for any 0 < p < 1.
(Note: this computes an upper tail probability.) Again, it is
not possible to write this as a closed form expression, so we
resort to approximations. Because of the symmetry of the normal distribution,
we need only consider 0 < p < 0.5. If you
have p > 0.5, then apply the algorithm below to q = 1-p,
and then negate the value for X obtained. (This approximation
is also from Abramowitz and Stegun.)
t = sqrt[ ln(1/p^2) ]
c_0 + c_1*t + c_2*t^2
X = t - ------------------------------
1 + d_1*t + d_2*t^2 + d_3*t^3
c_0 = 2.515517
c_1 = 0.802853
c_2 = 0.010328
d_1 = 1.432788
d_2 = 0.189269
d_3 = 0.001308
See Abramowitz and Stegun; Press, et al.
You should Google a little harder instead of always asking here. These algorithms were worked out a long time ago and have been available for a long time.
I'm the ocean. I'm a giant undertow.
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73Zeppelin wrote: You should Google a little harder instead of always asking here
I think it is more of a case of sheer laziness. He's asking about the FDIST function now, and I pointed him to Wolfram's Mathworld.
"The clue train passed his station without stopping." - John Simmons / outlaw programmer
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
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How to find NORMSDIST function of ms excel.It Returns the standard normal cumulative distribution function. The distribution has a mean of 0 (zero) and a standard deviation of one. Use this function in place of a table of standard normal curve areas.
Syntax
NORMSDIST(z)
Z is the value for which you want the distribution.
Thanks
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drinking beer is a wonderful experience.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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Don't you agree on beer drinking statement?
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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No
If u know the answer plz tell me..
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How can I answer if there is NO question?
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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i want to make function to calculate Normsdist like this:---
Public Function calculateNormsdist(ByVal z As Single) As Single
and it will give the answer
can u implement it??
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sumit7034 wrote: can u implement it??
yes
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
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sumit7034 wrote: can u implement it??
cp9876 wrote: yes
I was pretty sure about.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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plz implement it
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sumit7034 wrote: plz implement it
Why him? Can't you?
"The clue train passed his station without stopping." - John Simmons / outlaw programmer
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
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sumit7034 wrote: i want to make function
sumit7034 wrote: can u implement it?
The two statements above are in contraddiction (unless sumit7034 and cpallini are the same person ).
BTW Now your question is clear, but is too broad and a bit lazy: what have you done so far? What are your troubles about?
I mean, numerically computing the cumulative of the normal distribution isn't such a daunting task, I suppose.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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Implementation 1:
double CNDF(double x)
{
int neg = (x < 0);
if ( neg )
x *= -1;
double k(1/(1+0.2316419*x));
double y=((((1.330274429*k-1.821255978)*k+1.781477937)*k-0.356563782)*k+0.319381530)*k;
y = 1.0 - 0.398942280401*exp(-0.5*x*x)*y;
return (1-neg)*y + neg*(1-y);
}
Implementation 2
double CumNorm(double x)
{
if (x > 6.0)
return 1.0;
if (x < -6.0)
return 0.0;
double b1 = 0.31938153;
double b2 = -0.356563782;
double b3 = 1.781477937;
double b4 = -1.821255978;
double b5 = 1.330274429;
double p = 0.2316419;
double c2 = 0.3989423;
double a = fabs(x);
double t = 1.0 / (1.0 + a * p);
double b = c2*exp((-x)*(x/2.0));
double n = ((((b5*t+b4)*t+b3)*t+b2)*t+b1)*t;
n = 1.0-b*n;
if ( x < 0.0 )
n = 1.0 - n;
return n;
}
I'm the ocean. I'm a giant undertow.
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Wow, no iterations?
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
[My articles]
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No, none.
I'm the ocean. I'm a giant undertow.
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Thanks my proble is solved
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