文章目录
选择题
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X₁,X₂,X₃,X₄独立同分布,其密度函数均为
f ( x ) = { 2 x , 0 < x < 1 , 0 , o t h e r s . f(x)= \begin{cases} 2x,&0<x<1, \\ 0,&others. \end{cases} f(x)={ 2x,0,0<x<1,others.
则max(X₁,X₂,X₃,X₄)的密度函数是()。
A. f max ( x ) = { 0 , x < 0 , x 2 , 0 ≤ x < 1 , 1 , x ≥ 1. f_{ \max }(x)= \begin{cases} 0,&x<0, \\ x^{2},&0 \le x<1, \\ 1,&x \ge 1. \end{cases} fmax(x)=⎩⎪⎨⎪⎧0,x2,1,x<0,0≤x<1,x≥1.
B. f max ( x ) = { 0 , x < 0 , x 8 , 0 ≤ x < 1 , 1 , x ≥ 1. f_{ \max }(x)= \begin{cases} 0,&x<0, \\ x^{8},&0 \le x<1, \\ 1,&x \ge 1. \end{cases} fmax(x)=⎩⎪⎨⎪⎧0,x8,1,x<0,0≤x<1,x≥1.
C. f ( x ) = { 8 x 7 , 0 < x < 1 , 0 , o t h e r s . f(x)= \begin{cases} 8x^{7},&0<x<1, \\ 0,& \ others. \end{cases} f(x)={ 8x7,0,0<x<1, others.
D. f ( x ) = { 8 x 8 , 0 < x < 1 , 0 , o t h e r s . f(x)= \begin{cases} 8x^{8},&0<x<1, \\ 0,&others. \end{cases} f(x)={ 8x8,0,0<x<1,others.
【正确答案:C】 -
设随机变量X,Y独立同分布,且 P { X = − 1 2 } = P { X = 1 2 } = 1 2 P \{ X=- \frac {1}{2} \} =P \{ X= \frac {1}{2} \} = \frac {1}{2} P{ X=−21}=P{ X=21}=21,Y服从标准正态分布,则有()。
A. P { X + Y ≥ 0 } = 1 4 P \{ X+Y \ge 0 \} = \frac {1}{4} P{ X+Y≥0}=41
B. P { X − Y ≥ 0 } = 1 4 P \{ X-Y \ge 0 \} = \frac {1}{4} P{ X−Y≥0}=41
C. P { min ( X , Y ) ≥ 0 } = 1 4 P \{ \min (X,Y) \ge 0 \} = \frac {1}{4} P{ min(X,Y)≥0}=41
D. P { max ( X , Y ) ≥ 0 } = 1 4 P \{ \max (X,Y) \ge 0 \} = \frac {1}{4} P{ max(X,Y)≥0}=41
【正确答案:C】 -
设随机变量X,Y独立同分布,均服从标准正态分布,则有()。
A. P { X + Y ≥ 0 } = 1 4 P \{ X+Y \ge 0 \} = \frac {1}{4} P{ X+Y≥0}=41
B. P { X − Y ≥ 0 } = 1 4 P \{ X-Y \ge 0 \} = \frac {1}{4} P{ X−Y≥0}=41
C. P { min ( X , Y ) ≥ 0 } = 1 4 P \{ \min (X,Y) \ge 0 \} = \frac {1}{4} P{ min(X,Y)≥0}=41
D. P { max ( X , Y ) ≥ 0 } = 1 4 P \{ \max (X,Y) \ge 0 \} = \frac {1}{4} P{ max(X,Y)≥0}=41
【正确答案:C】 -
设随机变量X与Y相互独立同分布,均服从(0,4)上的均匀分布,则P{1< min(X, Y) < 3}=()
A. 1 8 \frac {1}{8} 81
B. 1 4 \frac {1}{4} 41
C. 1 6 \frac {1}{6} 61
D. 1 2 \frac {1}{2} 21
【正确答案:D】