统计学原理复习思考题与练习题参考答案 下载本文

内容发布更新时间 : 2024/12/26 1:44:45星期一 下面是文章的全部内容请认真阅读。

n?16;x?1950,sx?300;1???95%,t?/2(n?1)?2.1315?x?sn16X?x??x?1950?159.8625:(1792.1375,2111.8625)?300?75;?x?t?/2(n?1)?x?159.8625(2)这批电子管的平均寿命的方差、标准差的置信区间

2?1?0.0252(16?1)?6.262;??0.025(16?1)?27.488?U2(n?1)s2(n?1)s22?2?215586.1;?L?2?49112.34?1??/2(n?1)??/2(n?1)22?U??U?464.3125;?L??L?221.613平均寿命的方差的置信区间为(49112.34,215586.1);标准差的置信区间为(221.613, 464.3125)。

9、解:依题意,此为不重复抽样,且为大样本。

N?1,000,000;n?1,000.p?2%.Z?/2?3s2p?p(1?p)?0.0196?1.96%n)?0.0044?0.44%;?p?Z?/2?p?1.32%nNP?p??p?2%?1.32%:(0.68%,3.32%)(1?10、解:

?p?s2p(1)已知:n?100, Z?/2?1 p?1?5%,sp?p(1?p)?95%?5%?0.047522?p??0.02179;?p?Z?/2*?p?0.02179nP?p??p?95%?2.179%sp在68.27%概率保证下,废品率的置信区间为(97.179%, 92.821%)。(2) ?(Z?/2)?95.45% Z?/2?2 ?p?Z?/2*?p?4.358%;P?p??p?95%?4.358%在95.45%概率保证下,废品率的置信区间为(99.358%, 90.642%)。(3)概率度增大,误差范围也随之而增加。

11、解:(1)计算平均考试成绩的置信区间

已知: n?100,  N?100/1%?10000;    ?(Z?/2)?95.45%,Z?/2?2x??xf?f2?76.6;sx?22(x?x)f??f?129.44sn?x?x(1?)?1.132;?x?Z?/2*?x?2.264nNX?x??x?76.6?2.264在95.45%概率保证下,英语考试的平均成绩范围是(78.864, 74.336)分。(2)考试成绩在80分以上的比重p?48%,sp?p(1?p)?48%?52%?0.2496?p?n)?0.0497;?p?Z?/2*?p?0.0994nNP?p??p?48%?9.94%(1?sp2在95.45%概率保证下,英语考试的成绩超过80分以上的比重范围是(57.94%, 38.06%)。12、解:依题意,此为总体方差未知;不重复抽样,为大样本。计算样本指标如下表所示。

n?100,N?100/1%?10000,?(Z?/2)?99.73%,Z?/2?3X*?150gxf?(1)x??f2?150.3;sx2(x?x)???f2f?0.76sn?x?x(1?)?0.0867;?x?Z?/2?x?0.26nNX?x??x?150.3?0.26在99.73%概率保证下。这批茶叶每包的平均重量范围为(150.56, 150.14),达到不低于150克的标准要求。(2)p?702?70%,sp?p(1?p)?0.211002?p?n)?0.0456;?p?Z?/2?p?0.1368nNP?p??p?70%?13.68%(1?sp在99.73%概率保证下。这批茶叶的合格率范围为(83.68%, 56.32%)。

13、 解:依题意,总体方差未知,且为大样本。

2N?2500,n?400.x?3000kg;sx?300;1???95%,Z?/2?1.96(1)X2sxn?x?(1?)?0.7937;?x?Z?/2?x?1.557?1.56nNX?x??x?3000?1.56:(2998.44kg,3001.56kg) (2)良种率P的置信区间

p?90%,s2p?p(1?p)?90%?10%?9%?p?n)?1.374%;?p?Z?/2?p?2.69%nNP?p??p?90%?2.69%:(87.31%,92.69%)(1?N?5000;n?200,n1?170,1???95.45%,Z?/2?2p?n1?85%;s2p?p(1?p)?12.75%ns2ps2p 14、解:

?p??2.525%;?p?Z?/2?p?5.05%nP?p??p?85%?5.05%:(79.95%,90.05%)NL?N?P?3997.5?3998;NU?N?P?4502.5?4503根据计算,在95.45%置信度下,该批树苗的成活率的置信区间为79.95%~90.05%之间。成活总数的置信区间为3998~4503株之间。

15、解:根据题意,等比例类型抽样

N?4000;n?200;ni/n?Ni/N;1???95.45%,Z?/2?212122x??nixi?194;sx??nisi2?3043.6ni?1ni?12sxn?x?(1?)?3.80;?x?Z?/2?x?2?3.8?7.6nNX?x??x?194?7.6;(186.4,201.6)XN?4000?(194?7.6);(745600,806400)16、解:(1)随机起始点。d=300/15=20。

3,23,43,63,83,103,123,143,163,183,203,223,243,263,283。 (2)半距起点时,抽中学生的编号为

10,30,50,70,90、110、130、150、170、190、210、230、250、270、290。 (3)采取对称取点时,

3, 37;43; 77; 83; 117; 123; 157; 163; 197; 203;237;243; 277; 283。

17、解:依题意,此为无关标志排队的等距抽样。

(1)Xn?500;N?500?10;1???95.45%,Z?/2?2xf?x??f?3980;sx2(x?x)f???f?12?1562725.452sxn?x?(1?)?53.037;?x?Z?/2?x?106.074nNX?x??x?3980?106.074:(3873.926;4086.074)(2)XN?500?10?(3980?106.074):(19369631;20430369)(3)P80p??16%;s2p?p(1?p)500?p?n)?1.6395%;?p?Z?/2?p?3.279%nNP?p??p?16%?3.279%:(12.72%,19.279%)(1?n?30;N?30?5?150;1???95%,Z?/2?1.96x?s2p18、解:(1)本书稿错字数的置信区间

?x?4.733;snx?3.442sxN?n?x?()?0.564;?x?Z?/2?x?1.1058nN?1X?x??x?4.733?1.1058:(3.628,5.839)XN?(4.733?1.1058)?150:(544.13,875.87)(2)本书平均每页错字率的置信区间

?p?0.0034;sp?np??(p?p)n?12?0.00259s2N?np?x?()?0.00042;?p?Z?/2?p?0.00083nN?1P?p??p?0.0034?0.00083:(0.00259,0.00425)

19、解:N=600 M=5;n=30.p=95%,δp=4%;1-α=68.3%,Zα/2=1 整群抽样的抽样误差

2

4`0?30??3.56%rR?130600?1?p?Z?/2?p?1?3.56%?3.56%?p?2?pR?r()?P?p??p?95%?3.56%:(91.44%,98.56%)在68.3%的置信度下,这批商品的合格率的置信区间为(91.44%,98.56%)。 20、解:

(1)样本平均废品率及其方差. 采用不重复抽样 。 r?100,R?1000?pfp??fiii?2%  sp2?(p?p)??fii2fi?0.45%;?(t)?68.27%, t?1s2rp?p?(1?)?0.064%;?p??p*t?0.064%rRP?p??p?2%?0.064%以 概率保证程度68.27%,估计这批零件的废品率区间为(2.064%, 1.936%)(2) ?(t)?95.45%, t?2; P?2.5%, r??r?Nt2sp2222N?p?tsps2pr?7.148?8(3)按重复抽样时,抽样平均误差?p???p??0.067!、解:

(1)Xx?x?ri?75;?x?(?(xi?x)2r?1?15.81?x??x2R?rrR?1X?x??x?75?18.81:(56.19,93.81))?6.77;?x?t?/2(n?1)?x?18.81