以下哪种方法可用于图像的区域分割？A. K-means聚类B. 主成分分析C. 线性回归D. 逻辑回归

基于对海量数据的挖掘与智能分析,许多过去难以把握的复杂关系,现在已经可以解释他们之间的因果关系。A. 正确B. 错误

5判断若两个随机变量的期望和方差都相等,则可以认为他们有相同的分布。A. XB. sqrt

分组频数表销售收入频数频率%累计频数累计频率%80.00 - 89.0025.025.090.00 - 99.0037.5512.5100.00 - 109.00922.51435.0110.00 - 119.001230.02665.0120.00 - 129.00717.53382.5130.00 - 139.00410.03792.5140.00 - 149.0025.03997.5150.00+12.540100.0总和40100.0　　,即(78.648,83.352)。(3)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,。由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为大样本,所以总体均值overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的99%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(77.940,84.096)。7.5(1)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,,。由于总体标准差已知,所以总体均值的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(24.11,25.89)。(2)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,,。由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为大样本,所以总体均值的98%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(113.17,126.03)。(3)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,。由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为大样本,所以总体均值的90%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(3.136,3.702)。7.6(1)已知:总体服从正态分布,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,。由于总体服从正态分布,所以总体均值overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的95%的置信区间为:,即(8646.97,9153.03)。(2)已知:总体不服从正态分布,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。虽然总体不服从正态分布,但由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为大样本,所以总体均值的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(8734.35,9065.65)。(3)已知:总体不服从正态分布,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352未知,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,。虽然总体不服从正态分布,但由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为大样本,所以总体均值overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的90%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(8760.97,9039.03)。(4)已知:总体不服从正态分布,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352未知,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。虽然总体不服从正态分布,但由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为大样本,所以总体均值overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的99%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(8681.95,9118.05)。7.7已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,当overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为0.1、0.05、0.01时,相应的overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352、overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352、overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。根据样本数据计算得:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为大样本,所以平均上网时间的90%的置信区间为:,即(2.88,3.76)。平均上网时间的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(2.79,3.85)。平均上网时间的99%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(2.63,4.01)。7.8已知:总体服从正态分布,但未知,为小样本,,。根据样本数据计算得:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,。总体均值的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(7.11,12.89)。7.9已知:总体服从正态分布,但未知,为小样本,,。根据样本数据计算得:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。从家里到单位平均距离的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(7.18,11.57)。7.10(1)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,。由于为大样本,所以零件平均长度的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(148.87,150.13)。(2)在上面的估计中,使用了统计中的中心极限定理。该定理表明:从均值为overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352、方差为overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的总体中,抽取容量为的随机样本,当充分大时(通常要求),样本均值的抽样分布近似服从均值为、方差为的正态分布。7.11(1)已知:总体服从正态分布,但未知,为大样本,,。根据样本数据计算得:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,。该种食品平均重量的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(100.87,101.77)。(2)根据样本数据可知,样本合格率为overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。该种食品合格率的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(0.82,0.98)。7.12已知:总体服从正态分布,但overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352未知,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352为小样本,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。根据样本数据计算得:,。总体均值overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的99%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(15.64,16.62)。7.13已知:总体服从正态分布,但未知,为小样本,,。根据样本数据计算得:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,。网络公司员工平均每周加班时间的90%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(10.36,16.76)。7.14(1)已知:,,,。总体总比例overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的99%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(0.32,0.70);(2)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,。总体总比例overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的95%的置信区间为:,即(0.78,0.86);(3)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。总体总比例overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的90%的置信区间为:,即(0.46,0.50)。7.15已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,为0.1和0.05时,相应的,。总体总比例的90%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(0.18,0.28)。总体总比例overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(0.17,0.29)。7.16已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,估计误差overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。应抽取的样本量为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。7.17(1)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,。应抽取的样本量为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。(2)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,未知,,。由于overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352未知,可用使用0.5。应抽取的样本量为:。(3)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。应抽取的样本量为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。7.18(1)已知:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,,,。总体中赞成该项改革的户数比例的95%的置信区间为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352,即(0.51,0.77)。(2)已知:,,。应抽取的样本量为:overline (x)pm (z)_(u/2)dfrac (s)(sqrt {n)}=81pm 1.96times dfrac (12)(sqrt {100)}=81pm 2.352。7.20 顾客到银行办理业务时往往需要等待一段时间,而等待时间的长短与许多因素有关,比如,银行业务员办理业务的速度,顾客等待排队的方式等。为此,某银行准备采取两种排队方式进行试验,第一种排队方式是:所有顾客都进入一个等待队列;第二种排队方式是:顾客在三个业务窗口处列队三排等待。为比较哪种排队方式使顾客等待的时间更短,银行各随机抽取10名顾客,他们在办理业务时所等待的时间(单位:分钟)如下:要求:(1)构建第一种排队方式等待时间标准差的95%的置信区间。

样本个数也就是一个样本中所包含的个体数目。()A. 对B. 错

4,当样本容量充分发时,样本统计量(又称样本估计量)充分地靠近总体参数的可能性趋于1,这称为抽样估计的()。A. 有效性B. 无偏性C. 一致性D. 充分性

在其他条件不变的情况下,如果允许误差缩小为原来的1/2,则样本容量( )A. 扩大为原来的2倍B. 扩大为原来的4倍C. 缩小为原来的1/2D. 缩小为原来的1/4

7.设随机变量X和Y相互独立，且X~N(3,4),Y~N(2,9),则Z=3X+Y~( ).A. N(7,21)B. N(7,27)C. N(7,45)D. N(11,45)

统计表内数字必须准确无误，一律用（）表示。A. 罗马数字B. 中文大写数字C. 中文小写数字D. 阿拉伯数字E. 英文数字

样本估计量和总体参数分别是()A. 前者是一个确定值,后者是随机变量B. 两者都是随机变量C. 前者是随机变量,后者是一个确定值D. 两者都是确定值