A random variable X has probability density function f(x) = $k e ^ { - k x }$ (x ≥ 1) for some constant k. Suppose that $\operatorname { Pr } ( 1 \leq X \leq 2 ) = \frac { 1 } { 4 }$ , what is the value of k?
A) $\frac { 1 } { 2 }$ ln 2
B) $\frac { 3 } { 2 }$ ln 2
C) $\frac { 1 } { 4 }$
D) ln 2
E) none of these

Question

Quizplus · Accepted Answer

To find the value of $k$, we integrate the probability density function $f(x) = k e^{-kx}$ from 1 to 2 and set it equal to $\frac{1}{4}$, as given by the condition $\operatorname{Pr}(1 \leq X \leq 2) = \frac{1}{4}$. The integral of $f(x)$ from 1 to 2 is $\int_{1}^{2} k e^{-kx} dx$, which evaluates to $[-e^{-kx}]$ from 1 to 2, multiplied by $k$. Plugging in the limits, we get $k(e^{-k} - e^{-2k}) = \frac{1}{4}$. Solving this equation for $k$ gives $k = \ln(2)$, as it satisfies the given probability condition.

A Random Variable X Has Probability Density Function F(x) = ke−kxk e ^ { - k x }ke−kx

A Random Variable X Has Probability Density Function F(x) = $k e ^ { - k x }$