(a) The marginal distribution of X is
Pr(X = x) = ∑ Pr(X = x, Y = y)
… = 0.0625 + 0.0625 + 0.0625 + 0.0625
… = 0.25
That is, the first equality follows from the law of total probability, with the sum taken over y from {0, 5, 10, 15}. Each probability Pr(X = x, Y = y) is given in the table to be 0.0625.
Similarly, the marginal distribution of Y is
Pr(Y = y) = 0.25
(b) Yes, they're independent because
Pr(X = x, Y = y) = 0.0625,
and
Pr(X = x) Pr(Y = y) = 0.25 • 0.25 = 0.0625.
(c) The mean of X is
E[X] = ∑ x Pr(X = x)
… = 0.25 ∑ x
… = 0.25 (0 + 5 + 10 + 15)
… = 7.5
and you would find the same mean for Y,
E[Y] = 7.5
The variance of X is
V[X] = E[X^2] - E[X]^2
… = (∑ x^2 Pr(X = x)) - 7.5^2
… = 0.25 (∑ x^2) - 56.25
… = 0.25 (0^2 + 5^2 + 10^2 + 15^2) - 56.25
… = 31.25
and similarly,
V[Y] = 31.25
(each sum is taken with x and y from {0, 5, 10, 15})