For an exponentially distribution random variable X, find the value of ? that satisfies the following relationship: P(X<=3) = 0.9P(X<=4) Solution P(x<=3) = 1- e^(-3) P(x<=4) = 1 - e^(-4) [1- e^-3] = 0.9 (1 - e^-4) 0.1 = e^-3 (1 - e^-) = 0.203 so distribution is P(X=x) = e^-x where =0.203.