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1,分布计算36除以3与3的积商是多少这道题怎么做分布计算

36/3*3=36/9=4
(880-36x15)÷17=20

分布计算36除以3与3的积商是多少这道题怎么做分布计算

2,正态分布常用公式

你这里a>0,所以P因为P{|x|≥a}=P{x≥a}+P{x≤-a},标准正态分布P{x≥a}=P{x≤-a},所以P{x≥a}=P{|x|≥a}/2
自然对数的底,是(1+1/n)^n在n趋向正无穷时的极限,其值约等于2.718281828。计算的时候直接代入就行了。一般数学软件里面都可以直接用。 至于为什么正态分布公式里会有e存在,我想是在计算分布的时候总体越大越精确,所以会取一个n趋向正无穷的数,那么出现了上面的极限,就用e来代替了。

正态分布常用公式

3,怎么用科学计算器 算正态分布值

计算方法:(1) P(Z≥0.3)=0.3821 a 进入单一变量统计模b 进入正态分布模式 shift 1 7 3 0.3 ) =(2) p(-0.3≤Z≤0.3)=0.2358 a 进入单一变量统计模式 Mode 3 1 AC b 进入正态分布模式 shift 1 7 3 0.3 ) - shift 1 7 3 (-)0.3 )=
计算方法:(1)p(z≥0.3)a进入单一变量统计模式mode31acb进入正态分布模式shift1730.3)=(2)p(-0.3≤z≤0.3)=0.2358a进入单一变量统计模式mode31acb进入正态分布模式shift1730.3)-shift173(-)0.3)=

怎么用科学计算器 算正态分布值

4,如何使用韦伯分布函数

weibull(x,alpha,beta,cumulative)x 参数值。alpha 分布参数。beta 分布参数。cumulative 指明函数的形式。说明如果x、alpha 或 beta 为非数值型,函数 weibull 返回错误值 #value!。 如果x < 0,函数 weibull 返回错误值 #num!。 如果alpha ≤ 0 或 beta ≤ 0,函数 weibull 返回错误值 #num!。 韦伯累积分布函数的计算公式如下: 韦伯概率密度函数的计算公式如下: 当alpha = 1,函数 weibull 返回指数分布: 示例如果您将示例复制到空白工作表中,可能会更易于理解该示例。操作方法创建空白工作簿或工作表。 请在“帮助”主题中选取示例。不要选取行或列标题。 从帮助中选取示例。 按ctrl+c。 在工作表中,选中单元格 a1,再按 ctrl+v。 若要在查看结果和查看返回结果的公式之间切换,请按 ctrl+`(重音符),或在“工具”菜单上,指向“公式审核”,再单击“公式审核模式”。 1234ab数据说明105计算函数的数值20α 分布参数100β 分布参数公式说明(结果)=weibull(a2,a3,a4,true)在上述条件下使用韦伯累积分布函数的结果 (0.929581)=weibull(a2,a3,a4,false)在上述条件下使用韦伯概率密度函数的结果 (0.035589)
密度函数: x≤0时,p(x)=0; x﹥0时,p(x)=aλx^(a-1)exp(-λx^a). 累计分布函数: x≤0时,F(x)=0; x>0时,F(x)=∫aλt^(a-1)exp(-λt^a)dt 积分(0,x) =-∫exp(-λt^a)d(-λt^a) =- exp(-λt^a) t从0到x =1- exp(-λx^a) 结论:x≤0时,F(x)=0; x>0时,F(x)=1- exp(-λx^a)

5,如何求概率分布

求出概率分布函数然后在整个区间上积分
好,我试着描述了一下从1 -5 的分布几率 以及变化:in 1d6 case, the result is as following:the probability of getting a 1 is 0.17the probability of getting a 2 is 0.17the probability of getting a 3 is 0.17the probability of getting a 4 is 0.17the probability of getting a 5 is 0.17the probability of getting a 6 is 0.17in 2d6 case, the result is as following:the probability of getting a 2 is 0.03the probability of getting a 3 is 0.06the probability of getting a 4 is 0.08the probability of getting a 5 is 0.11the probability of getting a 6 is 0.14the probability of getting a 7 is 0.17the probability of getting a 8 is 0.14the probability of getting a 9 is 0.11the probability of getting a 10 is 0.08the probability of getting a 11 is 0.06the probability of getting a 12 is 0.03in 3d6 case, the result is as following:the probability of getting a 3 is 0.00the probability of getting a 4 is 0.01the probability of getting a 5 is 0.03the probability of getting a 6 is 0.05the probability of getting a 7 is 0.07the probability of getting a 8 is 0.10the probability of getting a 9 is 0.12the probability of getting a 10 is 0.12the probability of getting a 11 is 0.13the probability of getting a 12 is 0.12the probability of getting a 13 is 0.10the probability of getting a 14 is 0.07the probability of getting a 15 is 0.05the probability of getting a 16 is 0.03the probability of getting a 17 is 0.01the probability of getting a 18 is 0.00in 4d6 case, the result is as following:the probability of getting a 4 is 0.00the probability of getting a 5 is 0.00the probability of getting a 6 is 0.01the probability of getting a 7 is 0.02the probability of getting a 8 is 0.03the probability of getting a 9 is 0.04the probability of getting a 10 is 0.06the probability of getting a 11 is 0.08the probability of getting a 12 is 0.10the probability of getting a 13 is 0.11the probability of getting a 14 is 0.11the probability of getting a 15 is 0.11the probability of getting a 16 is 0.10the probability of getting a 17 is 0.08the probability of getting a 18 is 0.06the probability of getting a 19 is 0.04the probability of getting a 20 is 0.03the probability of getting a 21 is 0.02the probability of getting a 22 is 0.01the probability of getting a 23 is 0.00the probability of getting a 24 is 0.00in 5d6 case, the result is as following:the probability of getting a 5 is 0.00the probability of getting a 6 is 0.00the probability of getting a 7 is 0.00the probability of getting a 8 is 0.00the probability of getting a 9 is 0.01the probability of getting a 10 is 0.02the probability of getting a 11 is 0.03the probability of getting a 12 is 0.04the probability of getting a 13 is 0.05the probability of getting a 14 is 0.07the probability of getting a 15 is 0.08the probability of getting a 16 is 0.09the probability of getting a 17 is 0.10the probability of getting a 18 is 0.10the probability of getting a 19 is 0.09the probability of getting a 20 is 0.08the probability of getting a 21 is 0.07the probability of getting a 22 is 0.05the probability of getting a 23 is 0.04the probability of getting a 24 is 0.03the probability of getting a 25 is 0.02the probability of getting a 26 is 0.01the probability of getting a 27 is 0.00the probability of getting a 28 is 0.00the probability of getting a 29 is 0.00the probability of getting a 30 is 0.006d6上面的东西都和武器无缘了。。。不是长期拥有的东西也没必要去在意它到底出多少点吧? 大概规律就是中间多两边少,成钟型分布。下面是程序:/* * to change this template, choose tools | templates * and open the template in the editor. */package diceandprobability;import java.text.decimalformat;import java.text.numberformat;import java.util.arraylist;import java.util.iterator;import java.util.list;import java.util.random;/** * * @author yichuan */public class main /** * @param args the command line arguments */ public static void main(string[] args) final int total_roll = 1000000; final int num_dice = 2; final string prompt = "in " + num_dice + "d6 case, the result is as following:\n"; random ran = new random(); list result = new arraylist(); int num = 0; int count = 0; for(int i = 0; i < total_roll; i++){ for(int j = 0; j < num_dice; j ++){ num += ran.nextint(6)+1; } result.add(num); num = 0; } system.out.println(prompt); for(int i = num_dice; i <= num_dice * 6; i ++){ iterator iter = result.iterator(); while(iter.hasnext()){ if (iter.next() == i){ count ++; } } double amount = count / (total_roll + 0.0); numberformat format = new decimalformat("#0.00000"); system.out.println("the probability of getting a " + i + " is " + format.format(amount)); count = 0; } } }

6,求概率分布

题很简单,描述很麻烦。直接给结果自己看看。概率分布:pEX=3x1/20+4x3/20+5x6/20+6x10/20=21/4DX=(3-ex)^2+(4-ex)^2...=嘿嘿,自己算了哈!哥也要考概率了,加油哈!
好,我试着描述了一下从1 -5 的分布几率 以及变化:in 1d6 case, the result is as following:the probability of getting a 1 is 0.17the probability of getting a 2 is 0.17the probability of getting a 3 is 0.17the probability of getting a 4 is 0.17the probability of getting a 5 is 0.17the probability of getting a 6 is 0.17in 2d6 case, the result is as following:the probability of getting a 2 is 0.03the probability of getting a 3 is 0.06the probability of getting a 4 is 0.08the probability of getting a 5 is 0.11the probability of getting a 6 is 0.14the probability of getting a 7 is 0.17the probability of getting a 8 is 0.14the probability of getting a 9 is 0.11the probability of getting a 10 is 0.08the probability of getting a 11 is 0.06the probability of getting a 12 is 0.03in 3d6 case, the result is as following:the probability of getting a 3 is 0.00the probability of getting a 4 is 0.01the probability of getting a 5 is 0.03the probability of getting a 6 is 0.05the probability of getting a 7 is 0.07the probability of getting a 8 is 0.10the probability of getting a 9 is 0.12the probability of getting a 10 is 0.12the probability of getting a 11 is 0.13the probability of getting a 12 is 0.12the probability of getting a 13 is 0.10the probability of getting a 14 is 0.07the probability of getting a 15 is 0.05the probability of getting a 16 is 0.03the probability of getting a 17 is 0.01the probability of getting a 18 is 0.00in 4d6 case, the result is as following:the probability of getting a 4 is 0.00the probability of getting a 5 is 0.00the probability of getting a 6 is 0.01the probability of getting a 7 is 0.02the probability of getting a 8 is 0.03the probability of getting a 9 is 0.04the probability of getting a 10 is 0.06the probability of getting a 11 is 0.08the probability of getting a 12 is 0.10the probability of getting a 13 is 0.11the probability of getting a 14 is 0.11the probability of getting a 15 is 0.11the probability of getting a 16 is 0.10the probability of getting a 17 is 0.08the probability of getting a 18 is 0.06the probability of getting a 19 is 0.04the probability of getting a 20 is 0.03the probability of getting a 21 is 0.02the probability of getting a 22 is 0.01the probability of getting a 23 is 0.00the probability of getting a 24 is 0.00in 5d6 case, the result is as following:the probability of getting a 5 is 0.00the probability of getting a 6 is 0.00the probability of getting a 7 is 0.00the probability of getting a 8 is 0.00the probability of getting a 9 is 0.01the probability of getting a 10 is 0.02the probability of getting a 11 is 0.03the probability of getting a 12 is 0.04the probability of getting a 13 is 0.05the probability of getting a 14 is 0.07the probability of getting a 15 is 0.08the probability of getting a 16 is 0.09the probability of getting a 17 is 0.10the probability of getting a 18 is 0.10the probability of getting a 19 is 0.09the probability of getting a 20 is 0.08the probability of getting a 21 is 0.07the probability of getting a 22 is 0.05the probability of getting a 23 is 0.04the probability of getting a 24 is 0.03the probability of getting a 25 is 0.02the probability of getting a 26 is 0.01the probability of getting a 27 is 0.00the probability of getting a 28 is 0.00the probability of getting a 29 is 0.00the probability of getting a 30 is 0.006d6上面的东西都和武器无缘了。。。不是长期拥有的东西也没必要去在意它到底出多少点吧? 大概规律就是中间多两边少,成钟型分布。下面是程序:/* * to change this template, choose tools | templates * and open the template in the editor. */package diceandprobability;import java.text.decimalformat;import java.text.numberformat;import java.util.arraylist;import java.util.iterator;import java.util.list;import java.util.random;/** * * @author yichuan */public class main /** * @param args the command line arguments */ public static void main(string[] args) final int total_roll = 1000000; final int num_dice = 2; final string prompt = "in " + num_dice + "d6 case, the result is as following:\n"; random ran = new random(); list result = new arraylist(); int num = 0; int count = 0; for(int i = 0; i < total_roll; i++){ for(int j = 0; j < num_dice; j ++){ num += ran.nextint(6)+1; } result.add(num); num = 0; } system.out.println(prompt); for(int i = num_dice; i <= num_dice * 6; i ++){ iterator iter = result.iterator(); while(iter.hasnext()){ if (iter.next() == i){ count ++; } } double amount = count / (total_roll + 0.0); numberformat format = new decimalformat("#0.00000"); system.out.println("the probability of getting a " + i + " is " + format.format(amount)); count = 0; } } }

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