第1个回答 2018-10-29
由样本的性质知Xi~b（1，p）(i=1,...,n)，且X1，X2，。。。Xn相互独立,所以Xi的分布律为 P{Xi=xi}=p^xi (1-p)^(1-xi ) (xi=0,1; i=1,...,n) (1)P{(X1,...,Xn)=(x1,...,xn)}=P{X1=x1}...P{Xn=xn}=p^x1(1-p)^(1-x1)...p^xn(1-p)^(1-xn) =p^∑xi (1-p)^(n-∑xi) (2)∑Xi即n次试验中成功(即Xi=1)的次数,故∑Xi~b（n，p）(二项分布),分布律就不用我帮你写了吧.
第2个回答 2018-11-10
第3个回答 2018-10-29
(1)
f(x)
=kx ; 0≤x<3
=2 - x/2 ; 3≤x<4
=0 ; elsewhere
∫(0->3) kx dx +∫(3->4) [ 2- x/2] dx =1
(1/2)k [x^2]|(0->3) + [ 2x -(1/4)x^2]|(3->4) =1
(9/2)k + [( 8 - 4) -(6- 9/4 ) ] = 1
(9/2)k + 1/4 =1
(9/2)k = 3/4
k=3/2
(2)
f(x)
=(3/2)x ; 0≤x<3
=2 - x/2 ; 3≤x<4
=0 ; elsewhere
F(x)
=0 : x<0
= (3/4)x^2 ; 0≤x<3
=2x - (1/4)x^2 -3 ; 3≤x<4
=1 ; x≥4
(1)
P(1<X≤7/2)
=F(7/2) - F(1)
= 2(7/2) - (1/4)(7/2)^2 -3 -(3/4)(1)^2
= 7 - 49/16 -3 - 3/4
=3/16本回答被网友采纳
f(x)
=kx ; 0≤x<3
=2 - x/2 ; 3≤x<4
=0 ; elsewhere
∫(0->3) kx dx +∫(3->4) [ 2- x/2] dx =1
(1/2)k [x^2]|(0->3) + [ 2x -(1/4)x^2]|(3->4) =1
(9/2)k + [( 8 - 4) -(6- 9/4 ) ] = 1
(9/2)k + 1/4 =1
(9/2)k = 3/4
k=3/2
(2)
f(x)
=(3/2)x ; 0≤x<3
=2 - x/2 ; 3≤x<4
=0 ; elsewhere
F(x)
=0 : x<0
= (3/4)x^2 ; 0≤x<3
=2x - (1/4)x^2 -3 ; 3≤x<4
=1 ; x≥4
(1)
P(1<X≤7/2)
=F(7/2) - F(1)
= 2(7/2) - (1/4)(7/2)^2 -3 -(3/4)(1)^2
= 7 - 49/16 -3 - 3/4
=3/16本回答被网友采纳
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