Small question about exponential distribution

Question

If $X\sim exp(0.001)$ then:
$P(800<X<1200)=F_X(1200)-F_X(800)=1-e^{-1.2}-(1-e^{-0.8})=-e^{-1.2}+e^{-0.8}$ ??

Thank you!

Answer 1

Close, but for the sign: $F(x) = 1 - exp(-x/\beta)$. In your parameterization, $\beta = 1/0.001$ so that $F(x) = 1 - exp(-0.001x)$