If the random variable X is exponentially distributed with parameter $\lambda$ = 1.75, then P(1.5 $\le$ X $\le$ 3.8), up to 4 decimal places, is:
A)0.0711.
B)0.0473.
C)0.1184.
D)0.4739.

Question

If the random variable X is exponentially distributed with parameter $\lambda$ = 1.75, then P(1.5 $\le$ X $\le$  3.8), up to 4 decimal places, is:
A)0.0711.
B)0.0473.
C)0.1184.
D)0.4739.

Quizplus · Accepted Answer

The probability density function of an exponential distribution is given by f(x) = λe^(-λx). The cumulative distribution function is given by F(x) = 1 - e^(-λx). So, P(1.5 ≤ X ≤ 3.8) = F(3.8) - F(1.5) = (1 - e^(-1.75*3.8)) - (1 - e^(-1.75*1.5)) = 0.0711.

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

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