Exponential cdf: P(X>2), mean, variance

The distribution function F(x) = 1 − e⁻²ˣ (x ≥ 0) is that of an exponential distribution with rate λ = 2. Its pdf is

f(x) = dF/dx = 2e⁻²ˣ, x ≥ 0.

a) P(X > 2)

P(X > 2) = 1 − F(2) = 1 − (1 − e⁻⁴) = e⁻⁴ = 0.0183

b) Mean and variance

E(X)  = ∫₀^∞ x · 2e⁻²ˣ dx = 1/2 = 0.5
E(X²) = ∫₀^∞ x² · 2e⁻²ˣ dx = 2/λ² = 2/4 = 0.5
Var   = E(X²) − [E(X)]² = 0.5 − 0.25 = 0.25

(Integration by parts, or the standard results mean = 1/λ and variance = 1/λ².)

Answer: P(X > 2) = e⁻⁴ = 0.0183, mean = 0.5, variance = 0.25