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强化篇第8章 统计量及其分布

  1. XB(1,12)X \sim B(1,\frac{1}{2})X1,X2,X3X_1,X_2,X_3为来自总体XX的简单随机样本,X\overline{X}为样本均值,则P{X>13}=P\{\overline{X}>\frac{1}{3}\}=( ) A. 38\frac{3}{8} B. 12\frac{1}{2} C. 58\frac{5}{8} D. 78\frac{7}{8}
  2. X1,X2,,XnX_1,X_2,\cdots,X_n是来自总体XB(1,15)X \sim B(1,\frac{1}{5})的简单随机样本,X=1ni=1nXi\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i,若E[(X15)2]<0.01E[(\overline{X}-\frac{1}{5})^2]<0.01,则样本容量nn的最小值为( ) A. 17 B. 18 C. 19 D. 20
  3. X1,,XnX_1,\cdots,X_nY1,,YnY_1,\cdots,Y_n是来自正态总体N(μ,σ2)N(\mu,\sigma^2)的两个相互独立的简单随机样本,X=1ni=1nXi\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_iY=1ni=1nYi\overline{Y}=\frac{1}{n}\sum_{i=1}^{n}Y_i,且满足P{XY>σ}0.05P\{|\overline{X}-\overline{Y}|>\sigma\} \leq 0.05Φ(1.96)=0.975\Phi(1.96)=0.975,则样本容量nn的最小值为( ) A. 7 B. 8 C. 9 D. 10
  4. X1,X2,,XnX_1,X_2,\cdots,X_n是来自总体N(0,1)N(0,1)的简单随机样本,记X=1ni=1nXi\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_iS2=1n1i=1n(XiX)2S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\overline{X})^2T=(X+1)(S2+1)T=(\overline{X}+1)(S^2+1),则E(T)E(T)的值为( ) A. 0 B. 1 C. 2 D. 4
  5. X1,X2,,X10X_1,X_2,\cdots,X_{10}是来自标准正态总体XX的简单随机样本,Y=910(X1019i=19Xi)2Y=\frac{9}{10}(X_{10}-\frac{1}{9}\sum_{i=1}^{9}X_i)^2,则D(Y)=D(Y)=( ) A. 2 B. 1 C. 1100\frac{1}{100} D. 81100\frac{81}{100}
  6. X1,X2,,Xn(n2)X_1,X_2,\cdots,X_n(n \geq 2)为来自正态总体XX的简单随机样本,E(X)=μE(X)=\muD(X)=σ2D(X)=\sigma^2σ>0\sigma>0,记Y=1ni=1nXiμY=\frac{1}{n}\sum_{i=1}^{n}|X_i-\mu|,则D(Y)=D(Y)=( ) A. σ2n(12π)\frac{\sigma^2}{n}(1-\frac{2}{\pi}) B. σ2n(1π2)\frac{\sigma^2}{n}(1-\frac{\pi}{2}) C. σ2n2(12π)\frac{\sigma^2}{n^2}(1-\frac{2}{\pi}) D. σ2n2(1π2)\frac{\sigma^2}{n^2}(1-\frac{\pi}{2})
  7. X1,X2X_1,X_2为来自总体N(0,1)N(0,1)的简单随机样本,记σ^=aX1X2(a>0)\hat{\sigma}=a|X_1-X_2|(a>0),若D(σ^)=1D(\hat{\sigma})=1,则a=a=_____。
  8. 设总体(X,Y)(X,Y)服从N(0,0;1,2;1)N(0,0;1,2;1)(X1,Y1),(X2,Y2)(X_1,Y_1),(X_2,Y_2)是来自总体(X,Y)(X,Y)的简单随机样本,X=X1+X22\overline{X}=\frac{X_1+X_2}{2}Y=Y1+Y22\overline{Y}=\frac{Y_1+Y_2}{2},则E[(XY)2]=E[(\overline{X}-\overline{Y})^2]=( ) A. 32\frac{3}{2} B. 322\frac{3}{2}-\sqrt{2} C. 3222\frac{3}{2}-\frac{\sqrt{2}}{2} D. 32+22\frac{3}{2}+\frac{\sqrt{2}}{2}
  9. 设随机变量XN(0,4)X \sim N(0,4),若X1,X2,,Xn(n>2)X_1,X_2,\cdots,X_n(n>2)是来自总体XX的简单随机样本,则( ) A. 12n(i=1nXi)2χ2(1)\frac{1}{2n}(\sum_{i=1}^{n}X_i)^2 \sim \chi^2(1) B. 116i=1nXi2χ2(n)\frac{1}{16}\sum_{i=1}^{n}X_i^2 \sim \chi^2(n) C. (n1)Xn2i=1n1Xi2t(n1)\sqrt{\frac{(n-1)X_n^2}{\sum_{i=1}^{n-1}X_i^2}} \sim t(n-1) D. (n1)X12i=2nXi2F(1,n1)\frac{(n-1)X_1^2}{\sum_{i=2}^{n}X_i^2} \sim F(1,n-1)
  10. 已知随机变量X,YX,Y(X,Y)(X,Y)的概率密度为f(x,y)=14πex2+(y1)28f(x,y)=\frac{1}{4\pi}e^{-\frac{x^2+(y-1)^2}{8}},则4X2(Y1)2\frac{4X^2}{(Y-1)^2}服从( ) A. χ2(2)\chi^2(2) B. t(1)t(1) C. N(0,22)N(0,2^2) D. F(1,1)F(1,1)
  11. 设随机变量X1,X2,X3,X4X_1,X_2,X_3,X_4相互独立且都服从标准正态分布N(0,1)N(0,1),已知Y=X12+X22X32+X42Y=\frac{X_1^2+X_2^2}{X_3^2+X_4^2},对给定的α(0<α<1)\alpha(0<\alpha<1),数yαy_{\alpha}满足P{Y>yα}=αP\{Y>y_{\alpha}\}=\alpha,则有( ) A. yαy1α=1y_{\alpha}y_{1-\alpha}=1 B. yαyα1α=1y_{\alpha}y_{\frac{\alpha}{1-\alpha}}=1 C. yαy1α=12y_{\alpha}y_{1-\alpha}=\frac{1}{2} D. yαy1α2=12y_{\alpha}y_{1-\frac{\alpha}{2}}=\frac{1}{2}
  12. X1,X2,,X9X_1,X_2,\cdots,X_9是来自正态总体XX的简单随机样本,记Y1=16(X1+X2++X6)Y_1=\frac{1}{6}(X_1+X_2+\cdots+X_6)Y2=13(X7+X8+X9)Y_2=\frac{1}{3}(X_7+X_8+X_9)S2=12i=79(XiY2)2S^2=\frac{1}{2}\sum_{i=7}^{9}(X_i-Y_2)^2Z=2(Y1Y2)SZ=\frac{\sqrt{2}(Y_1-Y_2)}{|S|},证明统计量ZZ服从自由度为2的tt分布。