Standard Deviation. Explain.

[Aktungbby]

^ I see it so clearly now; Jeff! You've got it in a nutshell!I didn't know air guns were so complicated.

[ikalugin]

Err you don't really need to create an object (class) for this, a simple function should do just fine.

[Jeff-Groves]

Yep. A new function is plenty good enuff. I don't like adding some monster package I'll never use.

[Eichhörnchen]

Nothing worse than having a monster package and never using it.

[Pisces]

I have never understood why it has to divide by N-1. Can anyone explain why it is less 1?

I mean, simple averages are simply the sum divided by N. That makes sense to spread the (sum of) differences accross all samples.

[Tango589]

Originally Posted by Eichhornchen

Nothing worse than having a monster package and never using it.

Fnaar, fnaar.

[vienna]

You know, when I saw the title of this thread, I was a bit confused. I live in Hollywood, California; here, a standard deviation is anything that does not require above normal medical attention, a lengthy session in a confessional, or an excessive bail...


<O>

[Chad]

I am a bit rusty on my C++, but maybe we can get you going in the right direction with some psuedo-code!

We know we will have a list of items with numeric values, for simplicity sake, let's assume these are doubles.

So, let's pass that to stubbed method:

public void CalculateStandardDev(double listOfDoubles)
{

}

Now, we know the length of our array using the following:

listOfDoubles.size();

This will be our 'n' variable

public void CalculateStandardDev(double listOfDoubles)
{
     int sizeOfList = listOfDoubles.size();
}

Now we can find the mean by adding an additional local parameter called temporarySum and adding to it while in scope. This may be where my C++ rusty-ness shows

public void CalculateStandardDev(double listOfDoubles)
{
     int sizeOfList = listOfDoubles.size();
     double temporarySum = 0;

     for(int i=0; i<sizeOfList; i++)
     {
         temporarySum += listOfDoubles[i];
     }
}