Stats problems

1a:
E(x|theta=2) = 3.4
E(x^2|theta=2) = 13.4
Var(x|theta=2) = E(x^2|theta=2) - {[E(x|theta=2)]^2} = 1.84

1b:
L(theta=1|x=1, 2, 3, 4, 5) = p(x=1|theta=1)*p(x=21|theta=1)*p(x=3|theta=1)*p(x=4|theta=1)*p(x=5|theta=1) = 0.000158
L(theta=2|x=1, 2, 3, 4, 5) = p(x=1|theta=2)*p(x=2|theta=2)*p(x=3|theta=2)*p(x=4|theta=2)*p(x=5|theta=2) = 0.000158
L(theta=3|x=1, 2, 3, 4, 5) = p(x=1|theta=3)*p(x=2|theta=3)*p(x=3|theta=3)*p(x=4|theta=3)*p(x=5|theta=3) = 0.000190
L(theta=4|x=1, 2, 3, 4, 5) = p(x=1|theta=4)*p(x=2|theta=4)*p(x=3|theta=4)*p(x=4|theta=4)*p(x=5|theta=4) = 0.000047

Hence, the max. likelihood estimator of theta is 3

2a:
Pr(X>2|X>1) = Pr(X>2) / Pr(X>1) = Pr(X>1) or 2*Pr(X>2)
Hence,
Pr(X>1)=0.5, and thus mean = 1
Pr(X>2) = 0.25, and thus variance = 2.1981

Pr(X>3) = Pr(Z>1.3490) = 0.0887