Question

Suppose that f (x )=e^-(x-4) for x>4. Determine x such that P(X<x)=0.9

Answer 1

$P(X<x)=\int_0^xf(x)dx=\int_4^xe^{-(x-4)}dx=1-e^{-(4-x)}$

$P(X<x)=0.9\to 1-e^{-(x-4)}=0.9\to e^{-(x-4)}=0.1\to$

$\to -(x-4)=\ln(0.1)\to x=4-\ln(0.1)\approx 6.3026.$