1、概率论实验概率论实验实验报告2015年6月1 个人中至少有两人生日相同的概率是多少?通过计算机模拟此结果。 程序:n=1,2,10,100;N=10000;M=zeros(1,length(n);for k=1:length(n) for i=1:N X=ceil(365*rand(1,n(k); Y=unique(X); if length(Y)=0&need(i)=0.05&need(i)=0.15&need(i)=0.4&need(i)=0.75&need(i)=0.9&need(i)=x sale(i)=sale(i)+22*x; else sale(i)=sale(i)+22*nee
2、d(i); end end Profit(x+1)=mean(sale); end stem(0:5,Profit) Profitend运行结果:Profit = 0 14.0000 25.8000 31.0000 25.2000 17.2000Profit = 0 12.6800 23.1600 28.5800 24.9800 18.9600Profit = 0 13.0320 23.6440 28.8220 26.3440 20.5880Profit = 0 12.8164 23.5824 28.7978 26.2868 20.5154结论:重复试验次数01百份2百份3百份4百份5百份10
3、次01425.83125.217.2100次012.6823.1628.5824.9818.961000次013.03223.64428.82226.34420.58810000次012.816423.582428.797826.286820.5154理论值012.923.628.826.320.5 随重复试验的次数增多,模拟值逐渐接近理论值。观察发现,当购进量为3百份时,利润期望值最高。4.设总体,是来自总体的一组样本,通过计算机模拟分别画出当时的概率密度曲线,观察当越来越大时的概率密度曲线是否与某正态分布的概率密度曲线接近,以此验证中心极限定理。程序:clearn=2,4,10,20,10
4、0,1000;m=50000;for k=1:length(n) figure temp=rand(n(k),m); One=ones(1,n(k); X=One*temp; p,x=ksdensity(X); mu=0.5*n(k); sigma=sqrt(n(k)/(12); xx=mu-4*sigma:.1:mu+4*sigma; yy=pdf(norm,xx,mu,sigma); hold on plot(x,p,b,xx,yy,r-) legend(n= num2str(n(k),N( num2str(mu) , num2str(sigma) ) title(n= num2str(n
5、(k)end5.就不同的自由度画出分布、分布及F分布的概率密度曲线,每种情况至少画三条曲线,并将分布的概率密度曲线与标准正态分布的概率密度曲线进行比较。程序:figurex=0:.01:5;y=pdf(chi2,x,1);plot(x,y,b)hold ony=pdf(chi2,x,2);plot(x,y,r)y=pdf(chi2,x,3);plot(x,y,g)title(chi2)legend(n=1,n=2,n=3) figurex=-5:.01:5;y=pdf(t,x,1);plot(x,y,b)hold ony=pdf(t,x,2);plot(x,y,c)y=pdf(t,x,10);
6、plot(x,y,g)y=pdf(norm,x,0,1);plot(x,y,r-)title(t)legend(n=1,n=2,n=10,N(0,1) figurex=0:.01:5;y=pdf(f,x,10,10);plot(x,y,b)hold ony=pdf(f,x,10,10);plot(x,y,r)y=pdf(f,x,10,4);plot(x,y,g)title(F)legend(m=10,n=10,m=10,n=10,m=10,n=4)运行结果:n=10时,基本与标准正态重合。6就正态总体的某一个参数,构造置信区间,以检验置信度。即通过随机产生100组数据,构造100个置信区间,观
7、察是否有100(1-)%个区间包含此参数。程序:clearm=100;mu=1.5;sigma=3;alpha=0.1 0.05 0.01;count(3)=0;for k=1:3 for i=1:100 X=mu+sigma*randn(1,m); %XN(mu,sigma) UP=mean(X)+norminv(1-alpha(k)/2,0,1)*sigma/sqrt(m); DOWN=mean(X)-norminv(1-alpha(k)/2,0,1)*sigma/sqrt(m); if DOWNmu count(k)=count(k)+1; end endendp=count/100运行
8、结果:p = 0.9100 0.9600 0.9800p = 0.9200 0.9300 0.9900p = 0.8800 0.9400 1.0000p =0.9100 0.9200 0.9900结论: 对参数mu构造置信区间;当alpha=0.1时,包含mu的区间的概率接近0.9当alpha=0.05时,包含mu的区间的概率接近0.95当alpaca=0.01时,包含mu的区间的概率接近0.997.对于正态总体,当均值已知时,至少用两种方法构造方差的置信度为95%的置信区间,比较两种方法的优劣。验证所构造的置信区间置信度为95%程序:a=1;b=0.5;n=100;alpha=0.05;co
9、unt1=0;count2=0;for i=1:100 X=a+b*randn(1,n); %XN(a,b) UP1=sum(X-a).2)/(chi2inv(alpha/2,n); DOWN1=sum(X-a).2)/(chi2inv(1-alpha/2,n); UP2=n*(mean(X)-a)2/(chi2inv(alpha/2,1); DOWN2=n*(mean(X)-a)2/(chi2inv(1-alpha/2,1); if DOWN1b2&b2UP1 count1=count1+1; end if DOWN2b2&b2UP2 count2=count2+1; endendcount
10、1/100count2/100运行结果:method1 = 0.9300method2 = 0.9500method1 = 0.9500method2 = 0.9200method1 = 0.9600method2 =0.9600所构造的区间满足置信度95%用构造UP1 = 0.3248DOWN1 = 0.1861UP1 = 0.3482DOWN1 = 0.1994UP1 = 0.4066DOWN1 =0.2329用构造UP2 = 222.0616DOWN2 = 0.0434UP2 = 1.9079DOWN2 = 3.7295e-004UP2 = 68.4270DOWN2 =0.0134结论:用构造用构造第一次(0.1861,0.3248)(0.0434,222.0616)第二次(0.1994,0.3482)(3.7295e-004,1.9079)第三次(0.2329,0.4066)(0.0134,68.4270)用构造的区间比用构造的区间长度要长,效果要差。
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