设X ∼ B ( 1 , 1 2 ) X \sim B(1,\frac{1}{2}) X ∼ B ( 1 , 2 1 ) ,X 1 , X 2 , X 3 X_1,X_2,X_3 X 1 , X 2 , X 3 为来自总体X X X 的简单随机样本,X ‾ \overline{X} X 为样本均值,则P { X ‾ > 1 3 } = P\{\overline{X}>\frac{1}{3}\}= P { X > 3 1 } = ( )
A. 3 8 \frac{3}{8} 8 3
B. 1 2 \frac{1}{2} 2 1
C. 5 8 \frac{5}{8} 8 5
D. 7 8 \frac{7}{8} 8 7
设X 1 , X 2 , ⋯ , X n X_1,X_2,\cdots,X_n X 1 , X 2 , ⋯ , X n 是来自总体X ∼ B ( 1 , 1 5 ) X \sim B(1,\frac{1}{5}) X ∼ B ( 1 , 5 1 ) 的简单随机样本,X ‾ = 1 n ∑ i = 1 n X i \overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i X = n 1 ∑ i = 1 n X i ,若E [ ( X ‾ − 1 5 ) 2 ] < 0.01 E[(\overline{X}-\frac{1}{5})^2]<0.01 E [( X − 5 1 ) 2 ] < 0.01 ,则样本容量n n n 的最小值为( )
A. 17
B. 18
C. 19
D. 20
设X 1 , ⋯ , X n X_1,\cdots,X_n X 1 , ⋯ , X n 与Y 1 , ⋯ , Y n Y_1,\cdots,Y_n Y 1 , ⋯ , Y n 是来自正态总体N ( μ , σ 2 ) N(\mu,\sigma^2) N ( μ , σ 2 ) 的两个相互独立的简单随机样本,X ‾ = 1 n ∑ i = 1 n X i \overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i X = n 1 ∑ i = 1 n X i ,Y ‾ = 1 n ∑ i = 1 n Y i \overline{Y}=\frac{1}{n}\sum_{i=1}^{n}Y_i Y = n 1 ∑ i = 1 n Y i ,且满足P { ∣ X ‾ − Y ‾ ∣ > σ } ≤ 0.05 P\{|\overline{X}-\overline{Y}|>\sigma\} \leq 0.05 P { ∣ X − Y ∣ > σ } ≤ 0.05 ,Φ ( 1.96 ) = 0.975 \Phi(1.96)=0.975 Φ ( 1.96 ) = 0.975 ,则样本容量n n n 的最小值为( )
A. 7
B. 8
C. 9
D. 10
设X 1 , X 2 , ⋯ , X n X_1,X_2,\cdots,X_n X 1 , X 2 , ⋯ , X n 是来自总体N ( 0 , 1 ) N(0,1) N ( 0 , 1 ) 的简单随机样本,记X ‾ = 1 n ∑ i = 1 n X i \overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i X = n 1 ∑ i = 1 n X i ,S 2 = 1 n − 1 ∑ i = 1 n ( X i − X ‾ ) 2 S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\overline{X})^2 S 2 = n − 1 1 ∑ i = 1 n ( X i − X ) 2 ,T = ( X ‾ + 1 ) ( S 2 + 1 ) T=(\overline{X}+1)(S^2+1) T = ( X + 1 ) ( S 2 + 1 ) ,则E ( T ) E(T) E ( T ) 的值为( )
A. 0
B. 1
C. 2
D. 4
设X 1 , X 2 , ⋯ , X 10 X_1,X_2,\cdots,X_{10} X 1 , X 2 , ⋯ , X 10 是来自标准正态总体X X X 的简单随机样本,Y = 9 10 ( X 10 − 1 9 ∑ i = 1 9 X i ) 2 Y=\frac{9}{10}(X_{10}-\frac{1}{9}\sum_{i=1}^{9}X_i)^2 Y = 10 9 ( X 10 − 9 1 ∑ i = 1 9 X i ) 2 ,则D ( Y ) = D(Y)= D ( Y ) = ( )
A. 2
B. 1
C. 1 100 \frac{1}{100} 100 1
D. 81 100 \frac{81}{100} 100 81
设X 1 , X 2 , ⋯ , X n ( n ≥ 2 ) X_1,X_2,\cdots,X_n(n \geq 2) X 1 , X 2 , ⋯ , X n ( n ≥ 2 ) 为来自正态总体X X X 的简单随机样本,E ( X ) = μ E(X)=\mu E ( X ) = μ ,D ( X ) = σ 2 D(X)=\sigma^2 D ( X ) = σ 2 ,σ > 0 \sigma>0 σ > 0 ,记Y = 1 n ∑ i = 1 n ∣ X i − μ ∣ Y=\frac{1}{n}\sum_{i=1}^{n}|X_i-\mu| Y = n 1 ∑ i = 1 n ∣ X i − μ ∣ ,则D ( Y ) = D(Y)= D ( Y ) = ( )
A. σ 2 n ( 1 − 2 π ) \frac{\sigma^2}{n}(1-\frac{2}{\pi}) n σ 2 ( 1 − π 2 )
B. σ 2 n ( 1 − π 2 ) \frac{\sigma^2}{n}(1-\frac{\pi}{2}) n σ 2 ( 1 − 2 π )
C. σ 2 n 2 ( 1 − 2 π ) \frac{\sigma^2}{n^2}(1-\frac{2}{\pi}) n 2 σ 2 ( 1 − π 2 )
D. σ 2 n 2 ( 1 − π 2 ) \frac{\sigma^2}{n^2}(1-\frac{\pi}{2}) n 2 σ 2 ( 1 − 2 π )
设X 1 , X 2 X_1,X_2 X 1 , X 2 为来自总体N ( 0 , 1 ) N(0,1) N ( 0 , 1 ) 的简单随机样本,记σ ^ = a ∣ X 1 − X 2 ∣ ( a > 0 ) \hat{\sigma}=a|X_1-X_2|(a>0) σ ^ = a ∣ X 1 − X 2 ∣ ( a > 0 ) ,若D ( σ ^ ) = 1 D(\hat{\sigma})=1 D ( σ ^ ) = 1 ,则a = a= a = _____。
设总体( X , Y ) (X,Y) ( X , Y ) 服从N ( 0 , 0 ; 1 , 2 ; 1 ) N(0,0;1,2;1) N ( 0 , 0 ; 1 , 2 ; 1 ) ,( X 1 , Y 1 ) , ( X 2 , Y 2 ) (X_1,Y_1),(X_2,Y_2) ( X 1 , Y 1 ) , ( X 2 , Y 2 ) 是来自总体( X , Y ) (X,Y) ( X , Y ) 的简单随机样本,X ‾ = X 1 + X 2 2 \overline{X}=\frac{X_1+X_2}{2} X = 2 X 1 + X 2 ,Y ‾ = Y 1 + Y 2 2 \overline{Y}=\frac{Y_1+Y_2}{2} Y = 2 Y 1 + Y 2 ,则E [ ( X ‾ − Y ‾ ) 2 ] = E[(\overline{X}-\overline{Y})^2]= E [( X − Y ) 2 ] = ( )
A. 3 2 \frac{3}{2} 2 3
B. 3 2 − 2 \frac{3}{2}-\sqrt{2} 2 3 − 2
C. 3 2 − 2 2 \frac{3}{2}-\frac{\sqrt{2}}{2} 2 3 − 2 2
D. 3 2 + 2 2 \frac{3}{2}+\frac{\sqrt{2}}{2} 2 3 + 2 2
设随机变量X ∼ N ( 0 , 4 ) X \sim N(0,4) X ∼ N ( 0 , 4 ) ,若X 1 , X 2 , ⋯ , X n ( n > 2 ) X_1,X_2,\cdots,X_n(n>2) X 1 , X 2 , ⋯ , X n ( n > 2 ) 是来自总体X X X 的简单随机样本,则( )
A. 1 2 n ( ∑ i = 1 n X i ) 2 ∼ χ 2 ( 1 ) \frac{1}{2n}(\sum_{i=1}^{n}X_i)^2 \sim \chi^2(1) 2 n 1 ( ∑ i = 1 n X i ) 2 ∼ χ 2 ( 1 )
B. 1 16 ∑ i = 1 n X i 2 ∼ χ 2 ( n ) \frac{1}{16}\sum_{i=1}^{n}X_i^2 \sim \chi^2(n) 16 1 ∑ i = 1 n X i 2 ∼ χ 2 ( n )
C. ( n − 1 ) X n 2 ∑ i = 1 n − 1 X i 2 ∼ t ( n − 1 ) \sqrt{\frac{(n-1)X_n^2}{\sum_{i=1}^{n-1}X_i^2}} \sim t(n-1) ∑ i = 1 n − 1 X i 2 ( n − 1 ) X n 2 ∼ t ( n − 1 )
D. ( n − 1 ) X 1 2 ∑ i = 2 n X i 2 ∼ F ( 1 , n − 1 ) \frac{(n-1)X_1^2}{\sum_{i=2}^{n}X_i^2} \sim F(1,n-1) ∑ i = 2 n X i 2 ( n − 1 ) X 1 2 ∼ F ( 1 , n − 1 )
已知随机变量X , Y X,Y X , Y 且( X , Y ) (X,Y) ( X , Y ) 的概率密度为f ( x , y ) = 1 4 π e − x 2 + ( y − 1 ) 2 8 f(x,y)=\frac{1}{4\pi}e^{-\frac{x^2+(y-1)^2}{8}} f ( x , y ) = 4 π 1 e − 8 x 2 + ( y − 1 ) 2 ,则4 X 2 ( Y − 1 ) 2 \frac{4X^2}{(Y-1)^2} ( Y − 1 ) 2 4 X 2 服从( )
A. χ 2 ( 2 ) \chi^2(2) χ 2 ( 2 )
B. t ( 1 ) t(1) t ( 1 )
C. N ( 0 , 2 2 ) N(0,2^2) N ( 0 , 2 2 )
D. F ( 1 , 1 ) F(1,1) F ( 1 , 1 )
设随机变量X 1 , X 2 , X 3 , X 4 X_1,X_2,X_3,X_4 X 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) N ( 0 , 1 ) ,已知Y = X 1 2 + X 2 2 X 3 2 + X 4 2 Y=\frac{X_1^2+X_2^2}{X_3^2+X_4^2} Y = X 3 2 + X 4 2 X 1 2 + X 2 2 ,对给定的α ( 0 < α < 1 ) \alpha(0<\alpha<1) α ( 0 < α < 1 ) ,数y α y_{\alpha} y α 满足P { Y > y α } = α P\{Y>y_{\alpha}\}=\alpha P { Y > y α } = α ,则有( )
A. y α y 1 − α = 1 y_{\alpha}y_{1-\alpha}=1 y α y 1 − α = 1
B. y α y α 1 − α = 1 y_{\alpha}y_{\frac{\alpha}{1-\alpha}}=1 y α y 1 − α α = 1
C. y α y 1 − α = 1 2 y_{\alpha}y_{1-\alpha}=\frac{1}{2} y α y 1 − α = 2 1
D. y α y 1 − α 2 = 1 2 y_{\alpha}y_{1-\frac{\alpha}{2}}=\frac{1}{2} y α y 1 − 2 α = 2 1
设X 1 , X 2 , ⋯ , X 9 X_1,X_2,\cdots,X_9 X 1 , X 2 , ⋯ , X 9 是来自正态总体X X X 的简单随机样本,记Y 1 = 1 6 ( X 1 + X 2 + ⋯ + X 6 ) Y_1=\frac{1}{6}(X_1+X_2+\cdots+X_6) Y 1 = 6 1 ( X 1 + X 2 + ⋯ + X 6 ) ,Y 2 = 1 3 ( X 7 + X 8 + X 9 ) Y_2=\frac{1}{3}(X_7+X_8+X_9) Y 2 = 3 1 ( X 7 + X 8 + X 9 ) ,S 2 = 1 2 ∑ i = 7 9 ( X i − Y 2 ) 2 S^2=\frac{1}{2}\sum_{i=7}^{9}(X_i-Y_2)^2 S 2 = 2 1 ∑ i = 7 9 ( X i − Y 2 ) 2 ,Z = 2 ( Y 1 − Y 2 ) ∣ S ∣ Z=\frac{\sqrt{2}(Y_1-Y_2)}{|S|} Z = ∣ S ∣ 2 ( Y 1 − Y 2 ) ,证明统计量Z Z Z 服从自由度为2的t t t 分布。