Question

概率论，第4题，求解

Answer 1

X~N(3, 2^2)，正态分布 μ=3， σ=2

f(x) = 1/( 2sqrt(2π) ) e^( (x-3)^2/8)

P{x1<X<x2} = Φ ((x2-μ)/σ) - Φ ((x1-μ)/σ)
Φ (-x) = 1 - Φ (x)

(1) P{2<X<5} = Φ(1) - Φ(-1/2) = Φ(1) + Φ(1/2) - 1
(2) P{-2<X<7} = Φ(2) - Φ(-5/2) = Φ(2) + Φ(5/2) - 1
(3) P1 = P{-2<X<2} = Φ (-1/2) - Φ (-5/2) = 1 - Φ(1/2) + Φ(5/2) - 1 =Φ(5/2) - Φ(1/2)
P{|X|>2} = 1 - P1 = 1 + Φ(1/2) - Φ(5/2)
(4) P{X>3} = Φ(0) = 1/2
(5) 显然
P{X>c} = 1-Φ((c-3)/2)
P{X<=c} = Φ((c-3)/2)
P{X>c} = P{X<=c}
Φ((c-3)/2) =1/2
(c-3)/2 = 0
c = 3