

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





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]






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]





No
If u know the answer plz tell me..





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]





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??





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."





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]





plz implement it





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





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]





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*k1.821255978)*k+1.781477937)*k0.356563782)*k+0.319381530)*k;
y = 1.0  0.398942280401*exp(0.5*x*x)*y;
return (1neg)*y + neg*(1y);
}
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 = 1.0b*n;
if ( x < 0.0 )
n = 1.0  n;
return n;
}
I'm the ocean. I'm a giant undertow.





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]





No, none.
I'm the ocean. I'm a giant undertow.





Thanks my proble is solved






I'm a desperate Geocacher (www.geocaching.com) trying to solve a difficult puzzle.
I see discussions in this forum about the rightmost nonzero digit of 1000000! is 4.
Can anybody figure out the six rightmost nonzero digits of one million factorial?
As an incentive, I'll send $20 (via PayPal) to the first person who posts the correct answer.
Thanks!
Rod





The last six nonzero digits of 1,000,000! are 412544. Please, keep your money.
I wrote this code to calculate your answer:
using System;
class Program
{
const ulong Target = 1000000; const ulong RoundOff = 10000000000;
static void Main(string[] args)
{
ulong factorial = 1;
for (ulong n = 1; n <= Target; n++)
{
factorial *= n;
while (factorial % 10 == 0)
{
factorial /= 10;
}
factorial %= RoundOff;
}
Console.WriteLine("{0}! (truncated) = {1}", Target, factorial);
}
}
Why it works:
When multiplying numbers, the trailing zeros at the end will not change the outcome... so I threw those out.
Also, if you are only interested in the least significant digits, then the uppermost digits will not have an effect on the outcome of the lower digits... so I threw those out.
With all the rounding, I didn't have to worry about overflow so I was able to calculate the "truncated factorials" of very large numbers iteratively.
Enjoy,
Robert C. Cartaino
modified on Wednesday, July 2, 2008 1:26 PM





Way cool! Thank you very much!
Rod





One thing I forgot to mention in the source code; I rounded off to a few more digits than required because the top digit or two might be wrong because of the rounding. In other words, if you need six digits, calculate for eight. The sixdigit answer I gave above is accurate.





Good Day,
It's me again. This question is one of the questions in ACM ICPC 2006  Philippines.
Basically, you need to express a large number as a product of its Primes.
For example:
120 = 2x2x2x3x5 (or 2^3x3^1x5^1)
I'm lost on how to begin to tackle this problem. Please advice.
Thank you.
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.





Think of how you would do it on paper or using a calculator and generalise...
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."





That is my problem, I don't even know how to do it on paper. I have tried, believe me.
Can you just give me some reading material that will at least help me?
Thanks!
It is said that the most complex structures built by mankind are software systems. This is not generally appreciated because most people cannot see them. Maybe that's a good thing because if we saw them as buildings, we'd deem many of them unsafe.





Ian Uy wrote: I'm lost on how to begin to tackle this problem. Please advice.
As a college student, you really should be able to do this. It's basic math.
What you will want to search for is called "prime factorization." The easiesttounderstand technique is called "trial division."
On paper, you start by writing down your large number (let's call it 'n'). Then try to think of another number which divides evenly into n without any remainder. That number is a factor of n. It's easiest to start with small numbers.
In your example, what number divides evenly into 120 ? How about 2 ? Yes, 120/2=60 . So write down 2 and 60 (those are factors of 120 , but not necessarily prime).
Then try to factor each number you wrote down (the 2 and the 60 ). Are there any numbers (other than 1) that divide into 2 evenly? No, so 2 is a "prime factor" of 120 (i.e. 2 cannot be divided any further).
How about the 60 ? 60/2=30 . So, there is another 2 that is a prime factor of 120 . So far we have the factors of 120=2,2,30 . Now factor the 30 and keep going.
Keep going until you run out of numbers that can be divided evenly. Those will be your prime factors.
120 factored = 2,60 (2 is prime)
60 factored = 2,30 (2 is prime)
30 factored = 2,15 (2 is prime)
15 factored = 3,5 (both 3 and 5 are prime)
Done.
The prime factors of 120 are (from the parentheses above) 2,2,2,3,5 .
Now try writing that in code and see what you come up with. You can use this Table of Prime Factors[^] to check your answers.



