Yes, E(-x) is -100. No, it is not necessary for x to have negative values for this to make sense. E(f(x))= where the sum (or integral for continuous distributions) is taken over all possible values of x For example, suppose x has value 1 with prob 1/2 and value 2 with prob 1/2. Then E(x)= p(1)(-1)+ p(2)(-2)= (1/2)(-1)+ (1/2)(-2)= -3/2.for c, E(-x) would equal -100, right? I think this is assuming that the pmf still exists for x having negative values, and that the probabilities are the same...is this correct?
using the same equation as part a, i found V(-x) to equal 15 as well, since squaring a negative number still results in a positive one, and so the std. dev of (-x) would be sqrt(15)?