2658 - 概率论与数理统计 练习题
时间限制 : 1 秒
内存限制 : 128 MB
1、将一枚均匀的硬币连掷三次，其中有两次都出现正面的概率为
A.0.125 B. 0.25 C. 0.375 D. 0.5
2、设随机变量X的概率密度为
，则c值为
<span>A.</span>－<span>0.5 B. 0 C. 0.5 D. 1</span>
<span>3</span>、设随机变量<img src="http://tk.hustoj.com:80/attached/image/20140511/20140511104042_84548.png" alt="" />，则<span>E[X<sup>2</sup>]</span> 值为<span></span>
<span>A. 0.25 B. 0.5 C. 1 D. 1.25</span>
<span>4</span>、设X的分布函数为F(x)，则Y=3X+1的分布函数G(y)为<span></span>
<span>A. F(y/3-1/3) B. F(3y+1) C. 3F(y)+1 D. F(y)/3-1/3</span>
<span>5</span>、如果X,Y,满足D(X+Y)=D（X-Y），则必有<span></span>
<span>A. X</span>与<span>Y</span>独立 <span>B. X</span>与<span>Y</span>不相关<span> </span><span>C. D(X)=0 </span> <span> D. D(Y)=0</span>
<br />
6、设随机变量<span>X</span>服从（–<span>a</span>，<span>2a</span>）上的均匀分布，<span>E[X]=3</span>，则<span>a</span>值为<span></span>
<span>A. 0 B. 3 C. 6 D. 12</span>
7、设标准正态分布的上<span>0.05</span>分位点为<i><span>z</span></i>=1.645,则分布函数在该点的函数值为<span></span>
<span>A. 0.05 B. 1.645 C. 0.95 D. 1</span>
8、设总体期望为<span>0</span>.<span>50</span>，方差为<span>0</span>.<span>0001</span>，容量为<span>16</span>的样本均值为<span>0</span>.<span>54</span>，则<span>U</span>统计量的值为<span></span>
<span>A. 0.0001 B. 0.50 C. 0.54 D. 16</span>
9、已知事件A，B相互独立且互不相容，P(AB)=<u> </u><u></u>
<u><br />
10、<span style="line-height:1.5;">甲、乙两人独立的对同一目标射击一次，其命中率分别为0.6和0.5，现已知目标被命中，则它是甲射中的概率为（三位小数）</span><u> </u><span style="line-height:1.5;">。</span>
<span></span>
<br />
<span style="line-height:1.5;"><br />
<span style="line-height:1.5;">11、</span><span style="line-height:1.5;">设</span><span style="line-height:1.5;"></span><span style="line-height:1.5;">是来自正态总体</span><span style="line-height:1.5;"><img src="http://tk.hustoj.com:80/attached/image/20140511/20140511122127_31218.png" alt="" /></span><span style="line-height:1.5;">的样本，令</span><span style="line-height:1.5;"><img src="http://tk.hustoj.com:80/attached/image/20140511/20140511122137_96077.png" alt="" /></span><span style="line-height:1.5;"><br />
则当C= 时，CY～
。
<span style="line-height:1.5;"><br />
<span style="line-height:1.5;">12、设<span>X</span>、<span>Y</span>相互独立，<span>E[X]=</span><i>a</i>，<span>E[Y] =</span><i>b</i>，E[X<sup>2</sup>]=<i>c</i>，<span>E[Y<sup>2</sup>]=d</span>，则<span>E[</span><i>b</i>X–<i>a</i>Y]为多少？</span>
<span style="line-height:1.5;"><br />
<span style="line-height:1.5;">13、设总体<span>X</span>服从<span>(0, </span><i>a</i>)上的均匀分布，其中<i>a</i>未知。今对<span>X</span>进行<span>10</span>次简单抽样，样本<span></span>的观测值为<span>0.60</span>、<span>0.58</span>、<span>0.61</span>、<span>0.55</span>、<span>0.59</span>、<span>0.63</span>、<span>0.52</span>、<span>0.67</span>、<span>0.49</span>、<span>0.62</span>。试用矩估计法由一阶矩估计<i>a</i>的值<span>（三位小数）</span>。</span>
<span style="line-height:1.5;"><br />
<span style="line-height:1.5;">14、<span>设总体</span><span>X</span><span>服从</span><span>(0, </span><i>a</i><span>)上的均匀分布，其中</span><i>a</i><span>未知。今对</span><span>X</span><span>进行</span><span>10</span><span>次简单抽样，样本</span><span></span><span>的观测值为</span><span>0.60</span><span>、</span><span>0.58</span><span>、</span><span>0.61</span><span>、</span><span>0.55</span><span>、</span><span>0.59</span><span>、</span><span>0.63</span><span>、</span><span>0.52</span><span>、</span><span>0.67</span><span>、</span><span>0.49</span><span>、</span><span>0.62</span><span>。试用矩估计法由二阶矩估计</span><i>a</i><span>的值<span>（三位小数）</span>。</span><span></span></span>
<span style="line-height:1.5;"><br />
<span style="line-height:1.5;">15、设一机电系统由四种零件构成，它们分别占系统的<span>94%</span>、<span>3%</span>、<span>2%</span>、<span>1%</span>，四种零件的故障率依次为<span>0.02</span>，<span>0.05</span>，<span>0.10</span>，<span>0.15</span>。若任意零件出故障，则整个系统不能正常工作。求该机电系统能正常工作的概率<span>（三位小数）</span>？</span>
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C语言解答
#include "stdio.h" int main() { int a; float b; char c; while(scanf("%d",&a) != EOF) { switch(a){ case 0: printf("1\n");break; case 1: c='C';break; case 2: c='C';break; case 3: c='D';break; case 4: c='A';break; case 5: c='B';break; case 6: c='C';break; case 7: c='C';break; case 8: c='D';break; case 9: printf("0\n");break; case 10: b=0.75;break; case 11: printf("8\n");break; case 12: printf("0\n");break; case 13: b=0.586*2;break; case 14: b=1.018;break; case 15: b=0.976;//b=0.94*0.02+0.03*0.05+0.02*0.10+0.01*0.15;break; } if(a==1||a==2||a==3||a==4||a==4||a==5|a==6||a==7||a==8) printf("%c\n",c); else if(a==10||a==13||a==14||a==15) printf("%.3f\n",b); } return 0; }
C++解答
#include <stdio.h> int main() { int a; while(scanf("%d",&a) != EOF) { if (a==1) { printf("C\n"); } else if (a==2) { printf("C\n"); } else if (a==3) { printf("D\n"); } else if (a==4) { printf("A\n"); } else if (a==5) { printf("B\n"); } else if (a==6) { printf("C\n"); } else if (a==7) { printf("C\n"); } else if (a==8) { printf("D\n"); } else if (a==9) { printf("0\n"); } else if (a==10) { printf("0.750\n"); } else if (a==11) { printf("8\n"); } else if (a==12) { printf("0\n"); } else if (a==13) { printf("1.172\n"); } else if (a==14) { printf("1.019\n"); } else if (a==15) { printf("0.976\n"); } else { printf("1"); } } return 0; }
Java解答
import java.util.*; public class Main { public static void main(String args[]) { Scanner cin = new Scanner(System.in); int a; double b; while (cin.hasNext()) { a = cin.nextInt(); if (a==1){ System.out.printf("C\n"); } else if (a==2){ System.out.printf("C\n");} else if (a==3){ System.out.printf("D\n");} else if (a==4){ System.out.printf("A\n");} else if (a==5){ System.out.printf("B\n");} else if (a==6){ System.out.printf("C\n");} else if (a==7){ System.out.printf("C\n");} else if (a==8){ System.out.printf("D\n");} else if (a==9){ System.out.printf("0\n");} else if (a==10){ System.out.printf("0.750\n");} else if (a==11){ System.out.printf("8\n");} else if (a==12){ System.out.printf("0\n");} else if (a==13){ System.out.printf("1.172\n");} else if (a==14){ System.out.printf("1.019\n");} else if (a==15){ System.out.printf("0.976\n");} else if (a==0){ System.out.printf("1\n");} } } }