編寫標准偏差Java代碼時出現問題

Question

標題很能說明我的問題。

我正在嘗試編寫一個程序，該程序將計算用戶輸入的一組數字的標准偏差。
這比我預期的要困難得多，為此編寫算法很痛苦。
任何幫助將不勝感激！

import java.util.Scanner;

public class StandardDeviation {

    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);

        double num;
        double total;
        int n;

        System.out.print("First Number: ");
        num = input.nextDouble();

        n = 0;
        total = 0;

        while ( num != -1 ) {

            n++;
            total += num;
            System.out.print("Next Number: ");
            num = input.nextDouble();
        }

        double mean;
        double dev;
        double devn;
        double sqrt;

        mean = total/n;
        dev = (total - mean);
        devn = dev/n;
        sqrt = Math.sqrt(devn);

        System.out.println("N= " +n);
        System.out.println("Total= " +total);
        System.out.println("Mean= " +mean);
        System.out.println("Deviation= " +dev);

        System.out.print("The standard deviation is: " +sqrt);
    }
}

Answer 1

那是因為你的算法是錯誤的。 您不能僅通過累積樣本來計算標准差。 您還必須累積其平方。

像這樣：

    n = 0;
    total = 0;
    total_squared = 0;

    while ( num != -1 ) {

        n++;
        total += num;
        total_squared += num*num;
        System.out.print("Next Number: ");
        num = input.nextDouble();
    }

您的解決方案將是：

    mean = total/n;
    stddev = sqrt(total_squared/n - mean*mean);

Answer 2

這是您使用一些樣本數據計算標准差的方法。

double[] a = {10, 20, 30};
double total = 0;

for (double i : a) {
    total += i;
}

double mean = total/a.length;

double sqTot = 0;
for (double i : a) {
    sqTot += Math.pow((i-mean), 2);
}

System.out.println("Standard deviation " + Math.sqrt(sqTot/a.length));

Answer 3

標准偏差的公式是錯誤的。 這樣可以：

n = 0;
total = 0;
total_sq = 0;

while ( num != -1 ) {
    n++;
    total += num;
    total_sq += num*num;
    System.out.print("Next Number: ");
    num = input.nextDouble();
}

對於總體標准偏差：

mean = total/n;
dev_p = sqrt(total_sq/n - mean*mean);

對於偏離標准樣品：

dev_s = dev_p * n/(n-1);