Can anyone see whats up with this progam PLEASSSE!

Question

public class Mean
     public static void main(String args[])
          int counter = 0;
          int sum = 0;
          float mean = 0;
          float dev = 0;
          float sum2 = 0;
          float total = 0;
          for(int x =1;x<=5;x++)
               counter = counter+1;
               sum = sum+x;
          mean = sum/counter;
          System.out.println("The mean is "+mean);
          for(float y=1; y<=5;y++)
                    sum2 = (Math.sqrt(y-mean));
                    total = total+sum2;
                    dev = (Math.sqrt(total/mean));
               System.out.println("The deviant is "+dev);
I THINK THE SQUARE ROOT IS WRONG IM NOT SURE HOW TO DO THEM

Answer 1

Ok for debuggin i putputted your parameters before sqrt with the following code
public class Mean
public static void main(String args[])
int counter = 0;
int sum = 0;
double mean = 0d;
double dev = 0d;
double sum2 = 0d;
double total = 0d;
for(int x =1;x<=5;x++)
counter = counter+1;
sum = sum+x;
mean = sum/counter;
System.out.println("The mean is "+mean);
for(double y=1d; y<=5;y++)
sum2 = (Math.sqrt(y-mean));
total = total+sum2;
dev = (Math.sqrt(total/mean));
System.out.println("Before Deviant: \n Total is:"+total+"\n Mean is: " +mean+ "\n Deviant is: " +dev+ "\n");
System.out.println("The deviant is "+dev);
}and got the following results.
The mean is 3.0
Before Deviant:
Total is:NaN
Mean is: 3.0
Deviant is: NaN
Before Deviant:
Total is:NaN
Mean is: 3.0
Deviant is: NaN
Before Deviant:
Total is:NaN
Mean is: 3.0
Deviant is: NaN
Before Deviant:
Total is:NaN
Mean is: 3.0
Deviant is: NaN
Before Deviant:
Total is:NaN
Mean is: 3.0
Deviant is: NaN
The deviant is NaN
Total does not seem to be right. This is because in the statement
sum2 = (Math.sqrt(y-mean));
y is less than mean and you cannot take a sqrt of a negative number. Thus thiserror propagates to give a total as undefined too. Now please figure out your logic to be right. I dont remember high school equations of deviants and stuff but I do remeber there is a very easy equaiton to calculate deviant independent of the mean.