回答:
#Var = sigma ^ 2 = int(x-mu)^ 2f(x)dx# 可以写成:
#sigma ^ 2 = intx ^ 2f(x)dx-2mu ^ 2 + mu ^ 2 = intx ^ 2f(x)dx-mu ^ 2#
#sigma_0 ^ 2 = 3int_-1 ^ 1 x ^ 4dx = 3/5 x ^ 5 _- 1 ^ 1 = 6/5#
说明:
我假设这个问题意味着说
#f(x)= 3x ^ 2“for”-1 <x <1; 0“否则”#
找出方差?
#Var = sigma ^ 2 = int(x-mu)^ 2f(x)dx#
扩大:
#sigma ^ 2 = intx ^ 2f(x)dx-2mucancel(intxf(x)dx)^ mu + mu ^ 2cancel(intf(x)dx)^ 1#
#sigma ^ 2 = intx ^ 2f(x)dx-2mu ^ 2 + mu ^ 2 = intx ^ 2f(x)dx-mu ^ 2#
替代
#sigma ^ 2 = 3int_-1 ^ 1 x ^ 2 * x ^ 2dx -mu ^ 2 = sigma_0 ^ 2 + mu ^ 2#
哪里, #sigma_0 ^ 2 = 3int_-1 ^ 1 x ^ 4dx# 和 #mu = 3int_-1 ^ 1 x ^ 3dx#
所以让我们计算一下 #sigma_0 ^ 2“和”mu#
通过对称 #亩= 0# 让我们看看:
#mu = 3int_-1 ^ 1 x ^ 3dx = 3 / 4x ^ 4 _- 1 ^ 1 = 3/4 1-1#
#sigma_0 ^ 2 = 3int_-1 ^ 1 x ^ 4dx = 3/5 x ^ 5 _- 1 ^ 1 = 6/5#