$f ( x ) = \left\{ \begin{array} { l l }
\frac { 1 } { 2 } e ^ { - x / 2 } & \text { if } x \geq 0 \
0 & \text { if } x

Question

$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 1 } { 2 } e ^ { - x / 2 } & \text { if } x \geq 0 \
0 & \text { if } x < 0
\end{array} \right.$ is a probability density function for a particular random variable X.Use integration to find $P ( X \geq 4 )$
A) $e ^ { - 2 } - 1$
B) $e ^ { 2 }$
C) $e ^ { - 2 }$
D) $1 + e ^ { 2 }$

Quizplus · Accepted Answer

To find $P(X \geq 4)$, we integrate the probability density function from 4 to infinity: $\int_{4}^{\infty} \frac{1}{2}e^{-x/2} dx$. This evaluates to $e^{-2}$, making option C correct.

f(x)={12e−x/2 if x≥00 if x<0f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 2 } e ^ { - x / 2 } & \text { if } x \geq 0 \\0 & \text { if } x < 0\end{array} \right.f(x)={21​e−x/20​ if x≥0 if x<0​

$f ( x ) = \left\{ \begin{array} { l l } \frac { 1 } { 2 } e ^ { - x / 2 } & \text { if } x \geq 0 \\0 & \text { if } x < 0\end{array} \right.$