Question

概率统计设连续型随机变量X的概率密度为f(x)=kx^α,0

Answer 1

积分[0,1]kx^αdx=[0,1][k/(α+1)]x^(α+1)=k/(α+1)=1 (1)

积分[0,1]xf(x)dx=积分[0,1]kx^(α+1)dx=[0,1][k/(α+2)]x^(α+2)=0.75 (2)

(1)/(2)得：(α+2)/(α+1)=1/0.75=4/3.

解得：α=2.

k/(α+1)=k/3=1、k=3.