高中数学第7题
1个回答
展开全部
a+b+c=1 (1)
a,b,c 成等差数列
a+c =2b (2)
(1)-(2)
3b=1
b=1/3
ie
a+c = 2/3
c= 2/3 - a
E(ξ) = -a +0(b) +c = -a + c = -a + 2/3 -a = 2/3 -2a
E(ξ^2) = a +0(b) +c = a + c = a + 2/3 -a = 2/3
D(ξ) = E(ξ^2) -[E(ξ)]^2 = 2/3 - [ 2/3 -2a]^2
max D(ξ) = 2/3
ans :A
a,b,c 成等差数列
a+c =2b (2)
(1)-(2)
3b=1
b=1/3
ie
a+c = 2/3
c= 2/3 - a
E(ξ) = -a +0(b) +c = -a + c = -a + 2/3 -a = 2/3 -2a
E(ξ^2) = a +0(b) +c = a + c = a + 2/3 -a = 2/3
D(ξ) = E(ξ^2) -[E(ξ)]^2 = 2/3 - [ 2/3 -2a]^2
max D(ξ) = 2/3
ans :A
追问
最后两部不明白
追答
ξ=-1 ,P(ξ=-1)=a
ξ=0 ,P(ξ=0)=b
ξ=1 ,P(ξ=1)=c
E(ξ)
= (-1).P(ξ=-1) +(0).P(ξ=0) +(1).P(ξ=1)
= (-1)a +0(b) +(1)c
=-a+c
E(ξ^2)
= (-1)^2.P(ξ=-1) +(0)^2.P(ξ=0) +(1)^2.P(ξ=1)
=a +0(b) +c
=a+c
D(ξ)
= E(ξ^2) -[E(ξ)]^2
= 2/3 - [ 2/3 -2a]^2
max D(ξ) at a=1/3 = 2/3
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