题目
【题目】设随机变量X的分布律为X-202P0.40.30.3求:E(X), E(X^2) , E(3X^2+5)
【题目】设随机变量X的分布律为X-202P0.40.30.3求:E(X), E(X^2) , E(3X^2+5)
题目解答
答案
【解析】0.2,2.8,13.4
解析
步骤 1：计算E(X)
E(X) = Σ[xi * P(xi)] = (-2) * 0.4 + 0 * 0.3 + 2 * 0.3 = -0.8 + 0 + 0.6 = -0.2
步骤 2：计算E(X^2)
E(X^2) = Σ[xi^2 * P(xi)] = (-2)^2 * 0.4 + 0^2 * 0.3 + 2^2 * 0.3 = 4 * 0.4 + 0 + 4 * 0.3 = 1.6 + 1.2 = 2.8
步骤 3：计算E(3X^2+5)
E(3X^2+5) = 3E(X^2) + 5 = 3 * 2.8 + 5 = 8.4 + 5 = 13.4
E(X) = Σ[xi * P(xi)] = (-2) * 0.4 + 0 * 0.3 + 2 * 0.3 = -0.8 + 0 + 0.6 = -0.2
步骤 2：计算E(X^2)
E(X^2) = Σ[xi^2 * P(xi)] = (-2)^2 * 0.4 + 0^2 * 0.3 + 2^2 * 0.3 = 4 * 0.4 + 0 + 4 * 0.3 = 1.6 + 1.2 = 2.8
步骤 3：计算E(3X^2+5)
E(3X^2+5) = 3E(X^2) + 5 = 3 * 2.8 + 5 = 8.4 + 5 = 13.4