We are given that the random variable $X$ follows an exponential distribution with a mean of 6. For an exponential distribution, the probability density function (PDF) is given by:

$f(x) = \frac{1}{\lambda} e^{-\frac{x}{\lambda}}, \quad x \geq 0$

where $\lambda$ is the mean of the distribution. In this case, $\lambda = 6$.

The probability that the random variable exceeds 5 is represented as $P(X > 5)$. Using the cumulative distribution function (CDF) of the exponential distribution:

$P(X > x) = 1 - F(x) = e^{-\frac{x}{\lambda}}$

Substituting $x = 5$ and $\lambda = 6$:

$P(X > 5) = e^{-\frac{5}{6}}$

Now, let's compute that:

$P(X > 5) = e^{-\frac{5}{6}} \approx e^{-0.8333} \approx 0.435$

So, the probability that the exponentially distributed random variable exceeds 5 is approximately 0.435 or 43.5%.

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Would you like further details on the exponential distribution?

Here are 5 related questions for further exploration:

1. How do you derive the cumulative distribution function (CDF) for the exponential distribution?
2. How would the probability change if the mean of the distribution were different (say 10)?
3. What is the memoryless property of the exponential distribution and how does it affect probabilities?
4. Can you derive the expected value and variance for an exponential distribution?
5. How does the exponential distribution relate to other continuous distributions like the normal distribution?

Tip: The exponential distribution is memoryless, meaning that the probability of an event occurring after some time does not depend on how much time has already passed.