**4. Correct. The answer is true.**

The probability function of X is
P(x) = 1/6, x = 1,2,3,4,5,6. We know that E[H(X)] = ∑ H(x)p(x) where H(X) is a function of X. In
this particular case, H(X) = X^{2}. First of all, we have to calculate
E[X^{2}] in order to calculate E[2X^{2 }-
5].

E[X^{2}] = 1^{2}*(1/6)
+ 2^{2}*(1/6) + 3^{2}*(1/6) + 4^{2}*(1/6) + 5^{2}*(1/6)
+ 6^{2}*(1/6) = 91/6

Then, by applying some theorems on expectations, we have that:

-E[2X^{2 }– 5] = 2E(X^{2}) - 5
= 2*(91/6) - 5 = 76/3 ≈ 25 of US$ 2,500.