We compute μ_{X} and \sigma^{2}_{X} from Equations (4.20) and (4.21), substituting) λ = 2. We obtain μ_{X} = 0.25, \sigma^{2}_{X} = 0.0625. Using Equation (4.19), we find that

μ_{X}  =  \frac{1}{λ}        (4.20)
\sigma^{2}_{X}  =  \frac{1}{λ²}        (4.21)

F(x)= P(X\leq x) \begin{cases} 1 – e^{-\lambda x} & x\gt 0 \\ 0 & x\leq 0 \\ \end{cases}        (4.19)

P(X ≤ 0.2) = 1 – e^{-4(0.2)} = 0.5507