我尝试在 C 中计算小于 100 个数字的标准偏差

Question

#include<stdio.h>
#include<math.h>
int main()
{        
    int n;
    scanf("%d",&n);
    int i=0,sum=0,average=0;
    int num[100]={0};
    double dvalue[100]={0.0};
    double variance=0.0,s=0.0;
    for(i=0;i<n;i++){//I just know this method to get the number array;
        scanf("%d",&num[i]);
        sum+=num[i];
    }
    average=sum/n;
    for(i=0;i<n;i++){
        dvalue[i]=num[i]-average;
        dvalue[i]=pow(dvalue[i],2.0);//to get the (x-average）……2
        variance+=dvalue[i];
    }
    s=sqrt(variance);
    return 0;
}

这是我的代码，但无法正确执行 answer.I 我是新手，这是我第一次通过 Whosebug 寻求帮助，我无法获得输出，为什么？谢谢大家，我一开始其实是debug的，但是不知道怎么修改，再次感谢

Answer 1

有两点需要更正：
1. average 的类型应该是 double
2. 你错误地计算了方差。取 (x- 平均值) 的总和后，你应该将它除以 n

#include<stdio.h>
#include<math.h>
int main()
{
    int n;
    scanf("%d", &n);
    int i = 0, sum = 0;
    double average = 0;
    int num[100] = { 0 };
    double dvalue[100] = { 0.0 };
    double variance = 0.0, s = 0.0;
    for (i = 0; i < n; i++) { //I just know this method to get the number array;
        scanf("%d", &num[i]);
        sum += num[i];
    }
    printf("sum %d\n", sum);
    average = sum;   
    average = average / n;
    printf("average %lf\n", average);
    for (i = 0; i < n; i++) {
        dvalue[i] = num[i] - average;
        dvalue[i] = pow(dvalue[i], 2.0); //to get the (x-average)……2
        variance += dvalue[i];
    }
    variance = variance / n; // population variance
    s = sqrt(variance);
    printf("varience %lf\n", variance);
    printf("Population Standard deviation %lf", s);

    return 0;
}

根据@chux 的建议编辑了答案

我尝试在 C 中计算小于 100 个数字的标准偏差

I try to calculate a Standard deviation of less than 100 numbers in C

c

standard-deviation