Question

求一个概率论与数理统计的问题

谢谢大佬！！！！

Answer 1

f(x)
=(2/π)(cosx)^2 ; |x|<π/2
=0 ; elsewhere
E(X)
=∫(-π/2 -> π/2 ) xf(x) dx
=∫(-π/2 -> π/2 ) (2/π)x(cosx)^2 dx
=0
E(X^2)

=∫(-π/2 -> π/2 ) x^2. f(x) dx
=∫(-π/2 -> π/2 ) (2/π)x^2.(cosx)^2 dx
=(4/π) ∫(0 ->π/2) x^2.(cosx)^2 dx
=(2/π) ∫(0 ->π/2) x^2.(1+ cos2x) dx
=(2/π)[ (1/3)x^3]|(0 ->π/2) + (2/π) ∫(0 ->π/2) x^2.cos2x dx
=(1/12)π^2 +(1/π) ∫(0 ->π/2) x^2.dsin2x
=(1/12)π^2 +(1/π) [ x^2.sin2x ]|(0 ->π/2) -(2/π) ∫(0 ->π/2) xsin2x dx
=(1/12)π^2 +0 + (1/π) ∫(0 ->π/2) xdcos2x
=(1/12)π^2 + (1/π) [ x.cos2x]|(0 ->π/2) -(1/π) ∫(0 ->π/2) cos2x dx
=(1/12)π^2 - 1/2 - [1/(2π)][sin2x]|(0 ->π/2)
=(1/12)π^2 - 1/2
D(X)
=E(X^2)-[E(X)]^2
=(1/12)π^2 - 1/2

Answer 2

考察F分布的定义和定义式：设X₁服从自由度为m的χ²分布，X₂服从自由度为n的χ²分布，且X₁、X₂相互独立，则称变量F=(X₁/m)/(X₂/n)服从F分布，其中第一自由度为m，第二自由度为n。
因为Xi服从标准正态分布，所以Xi²~χ²(1)，根据χ²的可加性，1~4的∑Xi²服从χ²(4)，5~9的∑Xi²服从χ²(5)，前面的系数5/4正式F分布中的n/m，所以服从F(4,5)

Answer 3

如下图所示