• 2022-07-25
    假设正态总体[tex=6.0x1.286]/ZR0dAzaI7eKAw6bIvA7MwpDqCSuYOEmaTtlrlZ7K9Dq3LIAqxndmg7Srqig3x0U[/tex],[tex=5.786x1.286]mSwy1LlzIpZh/7u+rnVzC9EnwTiH9euTn7eiWJ2j04+x6Zff+rqFbpuPMeWBrGJS[/tex],且[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]相互独立;[tex=7.357x1.286]4bGv4GNhfHifuCST4hq27TUnKcULSEGkpmlzOaOCxYpERoxgf5hWAMaHNGJT6FVL[/tex]和[tex=6.429x1.286]ZBO5sFjiB9zbl+iHfnFejzSRQdr0ET1eL+e09RRBOQJV24fE+eVUcuyoI9CqVZ+w[/tex]是分别来自总体[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]的简单随机样本,[tex=0.929x1.286]ZAhNd0JrcSurz1OlXw327Q==[/tex]和[tex=1.071x1.286]YdE70j7tnA4A/fKDJXvcfw==[/tex]为总体[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]样本均值和方差;[tex=0.857x1.286]V77uFwc0bxKcDA/4/HJMVw==[/tex]和[tex=1.071x1.357]rPCKOY0U/0WlI7UHVA0hAg==[/tex]为总体[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]的样本均值和方差,而[tex=12.643x2.286]0qTzrdRCC9Cei4Mn5aXoBwCWsqWewXLISOPmCyqv+jofBdZ1o9F9erlq4DDxFcxCyjXHEeVSNIGvXsS1bTJWow==[/tex]是总体[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]的联合样本方差. 证明:[tex=1.5x1.357]d9Q2+RNeOnxlKW/Htzx/FSAH7Ic/xsz4+oVUujOterw=[/tex]是[tex=1.0x1.286]51n47HV7nln8qIGpThl1pg==[/tex]的无偏估计量.
  • 证:由于对任意总体,[tex=1.071x1.286]YdE70j7tnA4A/fKDJXvcfw==[/tex]和[tex=1.071x1.357]rPCKOY0U/0WlI7UHVA0hAg==[/tex]都是[tex=1.0x1.286]51n47HV7nln8qIGpThl1pg==[/tex]的无偏估计量,[tex=14.929x2.286]92iwahuv52hAl5vC9AcqMPx8lKiA+gSskN94182caVyAVT9M4lUHJxbeNANQSYkrDz79sjBM05eqGJo272OrIPJNg9jLc0oVJeS58zaf28o=[/tex][tex=13.071x2.214]xKFzmaTWZNO2inj3DypeI1MufnbRHmmQ7jTpcsIhi+qUFV6iz7X4zasbpks/hXfQEk5Gg/ALIVStOFXnj8urOPV6EAzDvCuvlC7ZKXCo4lc=[/tex],故[tex=1.5x1.357]d9Q2+RNeOnxlKW/Htzx/FSAH7Ic/xsz4+oVUujOterw=[/tex]也是[tex=1.0x1.286]51n47HV7nln8qIGpThl1pg==[/tex]的无偏估计量.

    举一反三

    内容

    • 0

      若随机向量[tex=2.786x1.286]d8ZGztHRaPoTHI8v2JwIGQ==[/tex]服从二维正态分布,则①[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex],[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]一定相互独立;②若[tex=3.5x1.286]sKaD0gq7ZfmqhDuxwY0565jK5tQQMeY1a44eA15r+0I=[/tex],则[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex],[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]一定相互独立;③[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]都服从一维正态分布;④若[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex],[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]相互独立, 则[tex=6.143x1.286]1FUpcitV2qNzFaSobaOhNfKUbfF8QOwkW6yD2rc0W2g=[/tex],几种说法中正确的是 A: ①②③④ B: ②③④ C: ①③ D: ①②④

    • 1

      袋中有5个号码1,2,3,4,5,从中任取3个,记这3个号码中最小的的号码为[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex],最大的号码为[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex] .(1)求[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]的联合分布律;(2)[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]与[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]是否相互独立 .

    • 2

      某厂销售收入[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]与利润[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]的统计资料如表所示。[img=631x178]1790c8ce45dac14.png[/img]若[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]与[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]有线性关系,试求出[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]关于[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]的线性回归方程。

    • 3

      设随机变量[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]的数学期望[tex=1.714x1.286]tnqXjXNHESmtAydX2nd1FQ==[/tex]和[tex=1.571x1.286]9HHQOQ6kFW8m23SI56qi0g==[/tex]存在,证明:假如[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]独立,则[tex=7.0x1.286]Fsc4c/MsrMbL1SEpyKHrDmKSWwNmUF4ydiRy0R1FUw0=[/tex].

    • 4

      假设随机变量[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]的联合概率密度为[tex=17.571x3.143]EPaISH7F+7OFqeEao9lVbRHesk4tplA2VrcCvwQ3rO0t9Qq8Iw/niDFSpYDusNul2n6lMAa/nNo6fxngQQtlYClfavo3+nsShxM9BlAXlm07xYNG1+7omwt7s4WdO9vNijRJOmbFVFR9SeYuI5TFgQ==[/tex].证明随机变量[tex=0.929x1.286]uswT/CEcOIwMpCvTz/zeaA==[/tex]和[tex=0.857x1.286]h9C4nePGcGllh55hxKIsUw==[/tex]不独立,但是[tex=1.286x1.286]ZIiW0MT/rNSURu/rNXyUxw==[/tex]和[tex=1.214x1.286]gnrbKJKP0x+Xz9YnDSiKgQ==[/tex]独立.