Continuous Random Variables

$k\displaystyle\int_0^4 x\,dx = k\left[\tfrac{x^2}{2}\right]_0^4 = 8k = 1$, so $k = \tfrac18$.

$E(X) = \tfrac{a+b}{2} = 6$ and $\operatorname{Var}(X) = \tfrac{(b-a)^2}{12} = \tfrac{64}{12} = 5.33$.

$F(2) - F(1) = \tfrac{8}{27} - \tfrac{1}{27} = \tfrac{7}{27} = 0.259$.

$f(x) = F'(x) = \tfrac{3x^2}{27} = \tfrac{x^2}{9}$ on that interval, and 0 elsewhere.

$E(X) = \tfrac19\displaystyle\int_0^3 x^3\,dx = \tfrac19 \times \tfrac{81}{4} = \tfrac94 = 2.25$.

$\lambda = \tfrac15 = 0.2$, so $P(X > 8) = e^{-0.2 \times 8} = e^{-1.6} = 0.2019$.