Hướng dẫn giải đề thi Xác suất thống kê docx

P(AB) = 0 P(AB) > 0 P(A) > 0 P(B) > 0 x, y ∈ [0; 1] |x − y|  1 0  P(A|B), P(A)  1 |P(A) − P(A|B)|  1. P(A|B) = P(AB) P(B)    P(A) − P(AB) P(B)     1. |P(A)P(B) − P(AB)|  P(B). |P(A)P(B) − P(AB)|  P(A). |P(A)P(B) − P(AB)|  min(P(A), P(B)). a = 2 π F (π/12) = π/12  −∞ f(x)dx = −π/2  −∞ f(x)dx + π/12  −π/2 f(x)dx = π/12  −π/2 f(x)dx = a π/12  −π/2 cos 2 xdx = 2 π  7π 24 + 1 8  = 7 12 + 1 4π . E(X) = +∞  −∞ xf(x)dx = a π/2  −π/2 x cos 2 xdx = 2 π π/2  −π/2 x cos 2 xdx. π/2  −π/2 x cos 2 xdx = 0 E(X) = 0 f(x, y) = ke − 1 2 (2x 2 −4x+2+y 2 ) = ke −(x−1) 2 e − y 2 2 . f(t) = 1 σ √ 2π e − (t−µ) 2 2σ 2 N(µ, σ 2 ) +∞  −∞ f(t)dt = 1 +∞  −∞ e − (t−µ) 2 2σ 2 dt = σ √ 2π +∞  −∞ e −(x−1) 2 dx = √ π +∞  −∞ e − y 2 2 dy = √ 2π +∞  −∞ +∞  −∞ f(x, y)dxdy = 1 1 = +∞  −∞ +∞  −∞ f(x, y)dxdy = k +∞  −∞ +∞  −∞ e −(x−1) 2 e − y 2 2 dxdy = k +∞  −∞ e −(x−1) 2 dx +∞  −∞ e − y 2 2 dy = k √ π × √ 2π = k √ 2π. k = 1 √ 2π X f X (x) = +∞  −∞ f(x, y)dy = k +∞  −∞ e −(x−1) 2 e − y 2 2 dy = ke −(x−1) 2 +∞  −∞ e − y 2 2 dy = k √ 2πe −(x−1) 2 = 1 √ π e −(x−1) 2 . X ∼ N(1, (1/ √ 2) 2 ) P(X  1+ √ 2) = 1−P(X < 1+ √ 2) = 1 − Φ  1 + √ 2 − 1 1/ √ 2  = 1 −Φ(2) = 1 −0, 9773 = 0, 0227 X ∼ N(1, (1/ √ 2) 2 ) E(X) = 1 Var(X) = 1/2 P(1 − √ 5 < X < 1 + √ 5) = P(|X − 1| < √ 5)  1 − 1 2( √ 5) 2 = 9 10 . • 14 2 +∞  −∞ e −(x−1) 2 dx = √ π +∞  −∞ e − y 2 2 dy = √ 2π +∞  −∞ e − (t−µ) 2 2σ 2 dt = σ √ 2π 3 • 1b 10 Var(X n ) = 9− 13 2n −(3− 3 2n ) 2  9 n {X n } {X n }