15?0.06?6%0p(1?p)0.06?0.94?p???0.015?1.5%n250由F(t)?68.27% t?1P? ?p?t??p?1?1.5%?1.5% p??p?P?p??p 6%-1.5%?P?6%?1.5% 4.5%?P?7.5% (2)
由F(t)?95.45% t?2 ?p?t?μp?2?1.5%?3% p?Δp?P?p?Δp 6%-3%?P?6%?3% 3%?P?9%
4. (1)
t?1 ??0.015 p(1- p)?2%?(1-2%)?0.0196 1?98%?2%n??88(块)20.01598%?2%?22(块) (2) n?20.03
5.
?Xf32?600?36?300X???33.33f600?300?222?σifi20?600?30?300????566.7 600?300?fi2
566.7?0.794 Δ?tμx?2?0.79?41.588n90031.7?42X?34.918μx???2
6. p
N?1000,000? n
μp?p(1?p)n(1?)?nN3.32?00 ? p2?%?(11000?2%2% t32%)(1?1000)?0.00442 1000,0002%?0.004?42?3?P0.68%??P7.
0.0?04423已知： x?170 σ?12 n?10 160.5?x?179.5σ2122 μx???3.7947n10 x ? t μx?16.05 17-t0?3.794?716.05 t?2.50349查表得： F (t?)98.83%8.
?xf32000x1???800元 σ1?141.42 σ2?2000040?f?xf39500x2???658.33元 σ2?123.88 σ2?15347.2260?f
20000?40?15347.22?60σ2??17208.3340?60Nt2?σ2100,000?22?17208.33n???304.99?305(人)22222N?Δ?t?σ100,000?15?2?17208.33(1) 随机抽样：
?xf?71500?715x??f100 σ2?220252Nt2?σ2100,000?2?22025n???390.02?391(人)2N?Δ2?t2?σ2100,000?15?2?222025(2)
σ222025Δ?tμ?t??2??15.01n391
Δ15.01t???1.9999?2μx7.50539.
18?23?33?40?45?31.8m2r5 (x1?18, x2?23, x3?33, x4?40, x5?45)
x?i?1?xir?64?9?25?(x?x1)σ???32.67 n3212225?1?16?(x?x2) σ???14n3264?4?36?(x?x3)2 σ3???34.67n3264?4?100?(x?x4)2σ4???56n32?(x?x5)144?1?1692σ5???104.67n3各群群间方差的平均数：1r212 ????σi??(32.67?14?34.67?56?104.67)?48.4ri?15各群群间方差：22r(18?31.8)2?(23?31.8)2?(33?31.8)2?(40?31.8)2?(45?31.8)2 ??102.165两个阶段抽样的抽样平均误差为：δ2R?r?2N?nμx????rR?1nN?1 ?102.1610?548.4100?3????3.81510?115100?1Δ?tμx?2?3.81?7.62
δ2?i?1?(xi?x)r224.18?X?39.42 10.
平均袋重(千克) xi 48.5 49 49.5 50 50.5 50.6 51 51.5 52 合计 批数r 1 2 5 7 3 1 3 1 1 24 xr 48.5 98.0 247.5 350.0 151.5 50.6 153.0 51.5 52.0 1202.6 xi?x (xi?x)2r -1.61 -1.11 -0.61 -0.11 0.39 0.49 0.89 1.39 1.89 - 2.5921 2.4642 1.8605 0.0847 0.4563 0.2401 2.3763 1.9321 3.5721 15.5784 x?.6?xr1202??50.11(千克)24?ri?12?(xi?x)rrδx2??r?15.5784?0.6491(千克)242δxR?r0.64911440?24μx?()???0.02705?0.9840?0.163rR?1241440?1Δx?tμx?2?0.163?0.326
? 50.11-0.326?X?50.11?0.326即 49.78(千克)?X?50.44(千克)各批一等品包装质量比重(%)pi 95 96 97 98 98.5 99 合计 批数r 3 1 5 9 1 5 24 pir 2.85 0.96 4.85 8.82 0.985 4.95 23.415 pi?p ?pi?p?2r -0.0256 0.001966 -0.0156 0.0002433 -0.0056 0.0001565 0.0044 0.0144 - 0.0001737 0.001037 0.003677 0.01 0.0001 p??pir23.415??97.56%r24?i?12δp??(pi?p)r?rr2?0.003677?0.0153$
0.000153μp??0.9840?0.247$Δp?tμp?2?0.247%?0.494%? 97.56%-0.494%?P?97.56%?0.494%即 97.07%?P?98.05%
11. (1)
已知：x?1800小时 S?6小时 n?100个σ2S26计算：μx????0.6(小时)nn10 F(t)?68.27% t?1Δx?t?μx?1?0.6?0.6(小时)极限误差为0.6小时。(2)
已知：Δx?0.4小时 S?6小时 t?1t2σ212?6236计算：n?2???225(只) 2Δx0.40.16应抽取225只灯泡进行测试。(3)
已知：Δx?0.4小时 F(t)?95.45% t?2 S?6小时 t2σ222?62计算：n?2??900(只)2Δx0.4应抽取900只灯泡进行测试。(4)
已知：Δx?0.6小时 t?2 S?6小时
t2σ222?62计算：n?2??400(只)Δx0.62
应抽取400只灯泡进行测试。(5)
通过以上计算可以看到，抽样单位数和概率之间是正比关系，即当概率提高时，抽样单位数也会增加；抽样单位数和允许误差（极限误差）之间是反比关系，即当极限误差范围扩大时，相应的抽样单位数就会减少。
12. 已知：σ?12 n?400 x?21根据题意假设H：0 : X?X0?20 H1 : X?X0?20用Z统计量代入上述数据：
x?X021?20??1.67?/n12/400由α?0.05所对应的临界Z0值.05?1.64Z?原假设，则可以说体明的总平均值会超20。过
因Z?Zα为拒绝域，题Z中?1.67，Zα＝1.64，故拒绝
13.
提出假设： H0:??50 H1:??50因总体方差未知，宜采用统计量t: t?根据资料计算：n ?50.20(克) S?x?x??0S/n?x?49.8?51?50.5?49.5?49.2?50.2?51.2?50.3?49.7?50.610?(x?x)2n?1x??050.20?50 t???0.97S/n0.65/10由??0.1 查??0.1双侧，自由度为10?1?9得?0.65
拒绝域为t?t?，题中t?t?，故接受原假设，袋即重每量符合要求。14.
(9) t0.1?1.83