Let X be a continuous random variable with a cumulative distribution function $\mathrm { F } ( \mathrm { x } ) = 1 - \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } ( \mathrm { x } \geq 0 )$ . Find $\operatorname { Pr } ( 1 \leq X \leq 2 )$
A) $\mathrm { e } ^ { - 1 }$ - $e ^ { - 4 }$
B) $\mathrm { e } ^ { - 1 }$ - $e ^ { - 2 }$
C) 1 - $\mathrm { e } ^ { - 1 }$ - $e ^ { - 2 }$
D) $\mathrm { e } ^ { - 1 }$ - $e ^ { - 2 }$ - 2
E) none of these

Question

Quizplus · Accepted Answer

The probability that $X$ is between 1 and 2 is given by $P(1 \leq X \leq 2) = F(2) - F(1)$, where $F(x)$ is the cumulative distribution function of $X$. Substituting the given $F(x) = 1 - e^{-x^2}$ for $x = 2$ and $x = 1$, we get $F(2) = 1 - e^{-4}$ and $F(1) = 1 - e^{-1}$. Therefore, $P(1 \leq X \leq 2) = (1 - e^{-4}) - (1 - e^{-1}) = e^{-1} - e^{-4}$.

Let X Be a Continuous Random Variable with a Cumulative F(x)=1−e−x2(x≥0)\mathrm { F } ( \mathrm { x } ) = 1 - \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } ( \mathrm { x } \geq 0 )F(x)=1−e−x2(x≥0)

Let X Be a Continuous Random Variable with a Cumulative $\mathrm { F } ( \mathrm { x } ) = 1 - \mathrm { e } ^ { - \mathrm { x } ^ { 2 } } ( \mathrm { x } \geq 0 )$