If the random variable x is exponentially distributed with parameter $\lambda$ = 4,then P(x ? 0.25),up to 4 decimal places,is:
A) 0.6321.
B) 0.3679.
C) 0.2500.
D) 0.5000.

Question

If the random variable x is exponentially distributed with parameter $\lambda$ = 4,then P(x 	? 0.25),up to 4 decimal places,is:
A) 0.6321.
B) 0.3679.
C) 0.2500.
D) 0.5000.

Quizplus · Accepted Answer

Since x is exponentially distributed with parameter λ = 4,
the PDF of x is f(x) = λe^(-λx) = 4e^(-4x) for x ≥ 0.
Thus, the CDF of x is F(x) = ∫_0^xf(t)dt = 1 - e^(-4x) for x ≥ 0.
Therefore, P(x ≤ 0.25) = F(0.25) = 1 - e^(-4*0.25) = 1 - e^(-1) ≈ 0.6321.

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

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