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Standard Deviation

Posted on 2014-02-11
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Last Modified: 2014-02-11
I am trying to solve an exercise problem from the Practice-It! website.

I am not a student but I have purchased the book(Building Java Programs) from Amazon. I am trying to teach myself how to code. It is not my intention to violate any EE terms of restriction.

Problem link
Exercise 7.6: stdev

 Write a method called stdev that returns the standard deviation of an array of integers. Standard deviation is computed by taking the square root of the sum of the squares of the differences between each element and the mean, divided by one less than the number of elements. (It's just that simple!)

More concisely and mathematically, the standard deviation of an array a is written as follows:
standard deviation

For example, if the array passed contains the values {1, -2, 4, -4, 9, -6, 16, -8, 25, -10}, your method should return approximately 11.237. You may assume that the array passed is non-null and contains at least two values, because the standard deviation is undefined otherwise.

I have written some code but keep that I thought should work but keep failing test cases.
Please review my code an help guide me to a solution.

public static double stdev(int a[]){
		double mean = 0; 
		int sum = 0;
	    //1. Work out the Mean (the simple average of the numbers)
	    for (int i = 0; i < a.length; i++) {
			
			sum += a[i];
			if(i == a.length -1){
				mean = sum/a.length;
			}
		}
	   
	    
	    //2. Then for each number: subtract the Mean and square the result
	    //GET DIFFERENCE
	    double variance = 0;
	    double [] b = new double [a.length];
	    for (int i = 0; i < a.length; i++) {
			//b[i] = Math.sqrt(Math.abs(a[i] - mean));
	    	double diff = Math.abs(a[i] - mean);
	    	b[i] = diff * diff;
		}
	    
	    //3. Then work out the mean of those squared differences.
	   
	    for (int i = 0; i < b.length; i++) {
			variance += b[i]/b.length;
		}
	    
	     
	    //4. Take the square root of that and you are done!
	    double standardDeviation = Math.sqrt(Math.ceil(variance));
	    return  standardDeviation;
	}

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I tried following this algorithm/formula:
http://www.mathsisfun.com/data/standard-deviation.html

I appreciate your help.
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Question by:DOCDGA
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Expert Comment

by:Geisrud
ID: 39850689
I'm no Java expert, but I do have some practice - we can make this a learning exercise for the both of us.  I thought I would toss this first post up as I work through your code.

So, step by step, the first problem I see is your mean result.  You set the mean type to a double (good), but you're finding the result by dividing an int by an int.  That means your result will discard anything after the decimal.  You get a value plus a ".0" in this case, your mean is "2.0".  I always like to toss in a print statement as I code to check that my result is what I expect.

System.out.println("Average = " + mean); //to see value of mean
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LVL 86

Expert Comment

by:CEHJ
ID: 39850696
You're off on the wrong foot. The calculation of the mean is wrong. I haven't looked beyond that.
a. Don't do the division until the loop is finished
b. (and more importantly) 'sum' should be a double. Why?

UPDATE: i see someone's told you why ;)
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Author Comment

by:DOCDGA
ID: 39850754
Okay, I have changed sum to a double and cast my divisor to a double.
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Accepted Solution

by:
Geisrud earned 500 total points
ID: 39850904
In your step 2 code, I would take out the "Math.abs" function.  I get the same result either way, and this may be my rudimentary understanding of standard deviation, but I don't think there's any reference to absolute value in there.

So - I used:
double diff = a[i] - mean;

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Lastly, and most importantly, there are two ways to calculate standard deviation (WTF!?!?!).  I'm guessing your result is ~10.66, but you're "supposed" to be getting 11.237.  This is because one way is figuring for a "sample" population, while the other figures for the entire data set.

This article may help explain that better than I ever could.
http://chemistry.about.com/od/mathsciencefundamentals/a/How-To-Calculate-Standard-Deviation.htm

Suffice it to say, you need to adjust one other line of your code to get the answer they want.  In your step 3, make the change:

variance += b[i]/(b.length-1);

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Author Closing Comment

by:DOCDGA
ID: 39850936
Thanks for your assistance. I had been working on this problem for awhile.
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LVL 14

Expert Comment

by:Geisrud
ID: 39850972
Glad to help - and I learned something along the way too!
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LVL 86

Expert Comment

by:CEHJ
ID: 39851113
and cast my divisor to a double.
Actually that's not necessary
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