If the random variable X is exponentially distributed with parameter $\lambda$= 1)5,then P(2 $\leq$ X $\leq$4),up to 4 decimal places,is
A) 0.6667
B) 0.0473
C) 0.5000
D) 0.2500

Question

Quizplus · Accepted Answer

The probability for an exponentially distributed random variable X with parameter $\lambda$ is given by $P(a \leq X \leq b) = e^{-\lambda a} - e^{-\lambda b}$. Here, $\lambda = 5$, so $P(2 \leq X \leq 4) = e^{-5 \times 2} - e^{-5 \times 4} = e^{-10} - e^{-20} \approx 0.0473$.

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

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