...第二小问和第三小问的具体的积分步骤。不知道他是怎么积分出来的...
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发布时间：2024-10-24 02:32
共1个回答
热心网友
时间：2024-11-18 20:47
正态分布
f(x) = [1/√(2πσ)] e^.[ -(x-μ)^2/(2σ^2)]
∫(-无穷->+无穷) f(x) dx =1
(2)
E(X)
=k^2.∫(-无穷->+无穷) x^2. e^(-k^2.x^2) dx
=-(1/2) ∫(-无穷->+无穷) x de^(-k^2.x^2)
=-(1/2) [x.e^(-k^2.x^2)]|(-无穷->+无穷) +(1/2)∫(-无穷->+无穷) e^(-k^2.x^2) dx
=0 +(1/2) ∫(-无穷->+无穷) e^(-k^2.x^2) dx
μ=0 , σ^2=1/(2k^2)
=(1/2)√(2π) .( k/√2 ) { [1/√(2π).( 1/((√2.k ))] ∫(-无穷->+无穷) e^(-k^2.x^2) dx }
=(1/2)√(2π) .[1/(√2.k )] (1)
=(1/2) [√π/k]
