If the random variable X is exponentially distributed with parameter $\lambda$ = 3,then the probability P(X $\ge$ 2)equals:
A) 0.3333
B) 0.5000
C) 0.6667
D) 0.0025

Question

Quizplus · Accepted Answer

The cumulative distribution function (CDF) of an exponential distribution with parameter lambda is given by:
F(x) = 1 - e^(-lambda*x)
Thus, for lambda = 3, we have:
P(X >= 2) = 1 - P(X < 2) = 1 - F(2) = 1 - (1 - e^(-3*2)) = e^(-6) = 0.0025
Therefore, the answer is D.

If the Random Variable X Is Exponentially Distributed with Parameter λ\lambdaλ

If the Random Variable X Is Exponentially Distributed with Parameter $\lambda$