If the random variable X is exponentially distributed with parameter $\lambda$ = 1.5,then the probability P(2 $\le$ X $\le$ 4),up to 4 decimal places,is
A) 0.6667
B) 0.0473
C) 0.5000
D) 0.2500

Question

Quizplus · Accepted Answer

We know that for an exponentially distributed random variable X with parameter λ, the probability density function is given by:
f(x) = λe^(-λx)     for x ≥ 0

Then the cumulative distribution function (CDF) of X is:
F(x) = P(X ≤ x) = ∫[0,x] {λe^(-λt)} dt = 1 - e^(-λx) for x ≥ 0

Using this, we can find the required probability as:
P(2 ≤ X ≤ 4) = F(4) - F(2)
= [1 - e^(-1.5*4)] - [1 - e^(-1.5*2)]
= e^(-3) - e^(-1)
≈ 0.0473 (rounded to 4 decimal places)

Thus, the best choice is B.

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

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