standard deviation and mean for an array

[peterdfl]

hey everyone, i just started doing arrays and the program i'm trying to create is supose to take 8 random numbers from the user then calculates the standard deviation, and the mean and provide the standard scores, right now i'm have 6 errors because of the array. The error are are saying ; epected at the end of the method, but i did that and it doesn work. if you can help if would be greatly appreciated!!! :)
here's my code do far

import java.util.Scanner;
public class Statistics
{
	public static void main(String[] args)
	{

		Scanner keyboard= new Scanner(System.in);
		int Mean = 0;
		double SD= 0.0;
		int Count= 0;
		System.out.print("How many numbers will you be entering? ");
		int NumAmount = keyboard.nextInt();


		System.out.println("Thank you. Please enter the " + NumAmount + " values below:");

		GetNumbers(NumAmount);
		int ListNum= GetNumbers(NumAmount);
		double [] values = new double [NumAmount];

		StanDev(values, NumAmount);
		System.out.println("Statistics for the given numbers:");
}

	static int GetNumbers(int NumAmount)
	{
		for (int n =0; n<=NumAmount;n++)
		{
		int [] values = new int [NumAmount];
		Scanner keyboard= new Scanner(System.in);
		int numbers = keyboard.nextInt();
		System.out.println(numbers);
		values [n]=numbers;
	}

	static double StanDev(double values,int NumAmount)

	{
		for(int i = 0; i < NumAmount; i++)
		{
		sum += values[i];
	    double mean = sum / NumAmount;
	    double sq_diff_sum = 0;
		}

	   	for(int i = 0; i <NumAmount ; ++i)
		{
	       double diff = values[i] - mean;
       	sq_diff_sum += diff * diff;

		}
		double deviance= sq_diff_sum / n;
    	return sqrt(deviance);

		}
	}
}

[sukatoa]

Please check those braces that were not paired....

specially at GetNumbers() and StanDev() method....

[sukatoa]

Java doesn't allow also to compile a method inside a method.

[Eranga]

And also you have used some members without declaring. Careful about adding parenthesis in the code. That's the major issue you have comes with.