Random Variables

One of the values must occur, so the total probability is 1. This is what determines any unknown constant.

$\dfrac{1+2+3+4+5+6}{6} = 3.5$. Note the die can never actually show 3.5 — an expectation is an average.

$0(0.5) + 1(0.3) + 2(0.2) = 0.7$.

$E(X^2) = 0 + 0.3 + 0.8 = 1.1$, so $\operatorname{Var}(X) = 1.1 - 0.49 = 0.61$.

$\operatorname{Var}(aX+b) = a^2\operatorname{Var}(X)$ with $a = -2$, so $(-2)^2 \times 4 = 16$. The constant 5 has no effect and the sign of $a$ disappears.

$P(X = 3) = F(3) - F(2) = 0.7 - 0.3 = 0.4$. Individual probabilities are differences of the cumulative function.