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n?16;x?1950,sx?300;1???95%,t?/2(n?1)?2.1315?x?sn?30016?75;?x?t?/2(n?1)?x?159.8625X?x??x?1950?159.8625:(1792.1375,2111.8625)(2)这批电子管的平均寿命的方差、标准差的置信区间
?1?0.025(16?1)?6.262;??0.025(16?1)?27.488?U222?(n?1)s2?21??/2(n?1)?215586.1;??2L?(n?1)s22??/2(n?1)?49112.34?U??U?464.3125;?2L?2L?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%spn2?p?(1?pnN)?0.0044?0.44%;?p?Z?/2?p?1.32%P?p???2%?1.32%:(0.68%,3.32%)10、解:
(1)已知:n?100, Z?/2?1 p?1?5%,sp?p(1?p)?95%?5%?0.04752?p?spn2?0.02179;?pp?Z?/2*?p?0.02179P?p???95%?2.179%置信区间为(97.179%, 92.821%)。在68.27%概率保证下,废品率的(2) ?(Z?/2)?95.45% Z?/2?2 ?p?Z?/2*?p?4.358%;P?p??p?95%?4.358?.358%, 90.642%)。在95.45%概率保证下,废品率的(3)概率度增大,误差范围置信区间为(也随之而增加。
11、解:(1)计算平均考试成绩的置信区间
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已知: n?100, N?100/1%?10000; ?(Z?/2)?95.45%,Z?/2?2x??xf?fsxn2?76.6;sx2??(x?x)?f2f?129.44?x?(1?nN)?1.132;?x?Z?/2*?x?2.264X?x??x?76.6?2.264在95.45%概率保证下,英语考试(2)考试成绩在80分以上的比重p?48%,sp?p(1?p)?48%?52%?0.2496的平均成绩范围是(78.864, 74.336)分。?p?spn2(1?pnN)?0.0497;?p?Z?/2*?p?0.0994P?p???48%?9.94%的成绩超过80分以上的比重范围是(57.94%, 38.06%)。在95.45%概率保证下,英语考试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*?150g(1)x??xf?fsxn2?150.3;sxnN2??(x?x)?f2f?0.76?x?(1?)?0.0867;?x?Z?/2?x?0.26X?x??x?150.3?0.26在99.73%概率保证下。这批茶叶达到不低于150克的标准要求。(2)p?70100spnp2每包的平均重量范围为(150.56, 150.14),?70%,spnN2?p(1?p)?0.21?p?(1?)?0.0456;?p?Z?/2?p?0.1368P?p???70%?13.68%的合格率范围为(83.68%, 56.32%)。在99.73%概率保证下。这批茶叶
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13、 解:依题意,总体方差未知,且为大样本。
N?2500,n?400.x?3000kg;sx?300;1???95%,Z?/2?1.96(1)X2?x?sxn2(1?nN)?0.7937;?x?Z?/2?x?1.557?1.56X?x??x?3000?1.56:(2998.44kg,3001.56kg) (2)良种率P的置信区间
p?90%,sp?p(1?p)?90%?10%?9%2?p?spn2(1?pnN)?1.374%;?p?Z?/2?p?2.69%P?p???90%?2.69%:(87.31%,92.69%) 14、解:
N?5000;n?200,n1?170,1???95.45%,Z?/2?2p?n1n?85%;sp?p(1?p)?12.75%spnp22?p??2.525%;?p?Z?/2?p?5.05%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?2x?12in?ni?1xi?194;sxnN2?12in?ni?1si?3043.62?x?sxn2(1?)?3.80;?x?Z?/2?x?2?3.8?7.6X?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,
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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?2x??xf?fsxn2?3980;sxnN2??(x?x)?f?12f?1562725.45?x?(1?)?53.037;?x?Z?/2?x?106.074X?x??x?3980?106.074:(3873.926;4086.074)(2)XN?500?10?(3980?106.074):(19369631;20430369)(3)Pp?80500?16%;sp?p(1?p)spn22?p?(1?pnN)?1.6395%;?p?Z?/2?p?3.279%P?p???16%?3.279%:(12.72%,19.279%)18、解:(1)本书稿错字数的置信区间
n?30;N?30?5?150;1???95%,Z?/2?1.96x??xnsxn2?4.733;sx?3.44(N?nN?1)?0.564;?x?Z?/2?x?1.1058?x?X?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??np?0.0034;sp?N?nN?1?(p?n?1p)2?0.00259?x?spn2(p)?0.00042;?p?Z?/2?p?0.00083P?p???0.0034?0.00083:(0.00259,0.00425)
19、解:N=600 M=5;n=30.p=95%,δ整群抽样的抽样误差
2p
=4%;1-α=68.3%,Zα/2=1
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?p??pR?rr(R?12)?40?600?30600?1?3.56%?p?Z?/2?p?1?3.56%?3.56%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p?方差. 采用不重复抽样 。 2r?100,R?1000?pf?fii2i?2% sp??(p?p)?fii2fi?0.45%;?(t)?68.27%, t?1?p?spr(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?Ntsp22222N?p?tsp?7.148?8均误差?p??(3)按重复抽样时,抽样平?p?spr2?0.067!、解: (1)Xx??xri?75;?x?2?(xi?x)2r?1?15.81?x??xr(R?rR?1)?6.77;?x?t?/2(n?1)?x?18.81X?x??x?75?18.81:(56.19,93.81)
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