设随机变量x服从二项分布B(10,0.1),则E(5X^2+3)=

x服从二项分布B(10,0.1),根据公式EX=nP=10*0.1=1DX=nP(1-P)=10*0.1*0.9=0.9=E(X-EX)^2=E(X^2-2x+1)=0.9E(X^2)=0.9+2EX-1=0.9+2-1=1.9E(5X^2+3)=5E(X^2)+3=5*1.9+3=12.5这个答案绝对正确!E(x^2)咋算的？E(X^2-2x+1)

=E(X^2)-2EX+1=0.9 移向整理
E(X^2)=0.9+2EX-1=1.9