Suppose the length of time, x, that it takes a chimpanzee to solve a simple puzzle is measured by a random variable X that is exponentially distributed with a probability density function $$f ( x ) = \left\{ \begin{array} { l l }
\frac { 2 } { 3 } e ^ { - 2 x / 3 } & \text { if } x \geq 0 \
0 & \text { if } x

Question

Suppose the length of time, x, that it takes a chimpanzee to solve a simple puzzle is measured by a random variable X that is exponentially distributed with a probability density function $$f ( x ) = \left\{ \begin{array} { l l } 
\frac { 2 } { 3 } e ^ { - 2 x / 3 } & \text { if } x \geq 0 \
0 & \text { if } x < 0
\end{array} \right.$$ where x is in minutes. Find the probability that a randomly chosen chimpanzee will take more than 12 minutes to solve the puzzle.
A) $e ^ { - 8 } + 1$
B) $e ^ { - 8 }$
C) $e ^ { 8 }$
D) $e ^ { 8 } - 1$

Quizplus · Accepted Answer

The probability that a randomly chosen chimpanzee will take more than 12 minutes to solve the puzzle is found by integrating the probability density function from 12 to infinity. For an exponential distribution with rate $\lambda = \frac{2}{3}$, the probability that X > x is given by $P(X > x) = e^{-\lambda x}$. Substituting $x = 12$ and $\lambda = \frac{2}{3}$, we get $P(X > 12) = e^{-(2/3) \cdot 12} = e^{-8}$.

Suppose the Length of Time, X, That It Takes a Chimpanzee