每天供电80,则对应的X=80/100=0.8,于是问题转为求P(X>0.8)
P(X≤0.8)=∫p(x)dx 积分限（-∞,0.8)
=∫12x(1-x)(1-x) dx （0,0.8)
=12∫x-2x^2+x^3 dx (0,0.8)
=12 (x^2/2-2x^3/3+x^4/4) x=0.8
=12(0.32-0.3413+0.1024)
=12*0.0811
=0.9732
于是P(X>0.8)=1-P(X≤0.8)=1-0.9732=0.0278
即该城市缺电的概率为0.0278
同理解2
X=90/100=0.9
求P(X