Question 1018800
Expected value= x*p(x)
it is (1/6)(1+2+3+4+5+6)=3.5
variance is 
sum of (x-mean)^2/n
(1/6)(2.5^2+1.5^2+0.5^2+0.5^2+1.5^2+2.5^2)=
(1/6)*17.50=2.92
E(x)=3.5
V(x)=2.92