如何使用JAVA来计算标准偏差 [英] How to calculate standard deviation using JAVA

问题描述

我很新来的，
此刻我想计算的Java标准偏差（我GOOGLE了它哈哈）
但我有很多问题，在得到它的工作。

我有一个由用户inputed十个值，然后我需要计算的标准偏差
我的理解到目前为止感谢的人谁也回答
是
我发现阵列的平均
然后完成计算

 双倍二级=总[2];
                双三=总[3];
                双四=总[3];
                双五=总[4];
                双六=总[6];
                七夕=总[7];
                双八=总[8];
                双九=总[9];
                双十=总[10];                双十一= average_total;
平均=一个+两台+三位四个都+ 5 +六秒以上7 + 8 + 9 +10 + 11;平均=平均/ 11;
// 1 =一个 - 均值;
//System.out.println(\"I真的希望这种打印出的值：+一个）;
* /
 // 11 = average_total  - 均值;
 // 11 = Math.pow（average_total，average_total）;
 // stand_dev =（一个+两台+三位四个都+ 5 +六秒以上7 + 8 + 9 +10 +十一）;
 // stand_dev = stand_dev  - 均值;
// stand_dev =（stand_dev  - 平均值）*（stand_dev  - 平均值）;
// stand_dev =（stand_dev / 11）;
// stand_dev =的Math.sqrt（stand_dev）;
 

我已经有存储在10个值的数组我的数据，但我也不太清楚
    如何打印出数据的数组，然后做了计算，不必在这里存储输入code
    数据的一些别的地方，我操纵

感谢您的时间，很多AP preciated：）

解决方案

 计算平均阵列。  通过循环值       数组值=（索引值 - 平均值）^ 2  计算新阵列的总和。  由数组长度划分总和  平方根它
 

编辑：

我会告诉你如何通过数组循环，一切都是pretty多少只是一个不同的计算同样的步骤。

  //计算平均值。INT总= 0;的for（int i = 0; I＆LT; array.length，我++）{
   总+ =阵列[我] //这是总结的所有值的计算
}双平均=总/ array.length;
 

EDIT2：

阅读你的code之后，你正在做的不对的地方在于你是不是通过值循环和平均正确减去它。

又名这一部分。

  

十一= average_total - 均值;结果
  11 = Math.pow（average_total，average_total）;

您需要做到这一点。

 的for（int i = 0; I＆LT; array.length，我++）{
   数组[我] = Math.pow（（数组[我] -mean），2）
}
 

基本上你需要NEWVALUE = OLDVALUE数组中改变每个值 - 平均值（平均）

然后计算总和......然后开方的。

I'm very new here, at the moment I am trying to calculate standard deviation with Java (I have googled it haha) but I am having a lot of issues on getting it working

I have ten values that are inputed by a user which I then have to calculate the standard deviation of my understanding so far thanks to people who have replied is I find the mean of the array then complete the calculations

                double two = total[2];
                double three = total[3];
                double four = total[3];
                double five = total[4];
                double six = total[6];
                double seven = total[7];
                double eight = total[8];
                double nine = total[9];
                double ten = total[10];

                double eleven = average_total;


mean = one + two + three + four + five + six + seven + eight + nine + ten + eleven;

mean = mean/11;
//one = one - mean;
//System.out.println("I really hope this prints out a value:" +one);
*/
 //eleven = average_total - mean;
 //eleven = Math.pow(average_total,average_total);
 //stand_dev = (one + two + three + four + five + six + seven + eight + nine + ten + eleven);
 //stand_dev = stand_dev - mean;
// stand_dev = (stand_dev - mean) * (stand_dev - mean);
// stand_dev = (stand_dev/11);
// stand_dev = Math.sqrt(stand_dev);

I already have my data that is stored in an array of 10 values but I am not too sure how to print the data out of the array then do the calculations with out having to store the enter code here data some where else that I have manipulated

Thank you for your time, much appreciated :)

解决方案

  calculate mean of array.

  loop through values

       array value = (indexed value - mean)^2    

  calculate sum of the new array.

  divide the sum by the array length

  square root it 

edited:

I'll show you how to loop through the array and everything is pretty much this same step just with a different calculation.

// calculating mean.

int total = 0;

for(int i = 0; i < array.length; i++){
   total += array[i]; // this is the calculation for summing up all the values
}

double mean = total / array.length;

edit2:

After reading your code, the part you are doing wrong is that you are not looping through the values and subtracting it with average correctly.

aka this part.

eleven = average_total - mean;
eleven = Math.pow(average_total,average_total);

you need to do this.

for(int i = 0; i < array.length; i++){
   array[i] = Math.pow((array[i]-mean),2)
}

essentially you need to change every value in the array with newvalue = oldvalue - mean(average).

then calculate the sum... then square root that.

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