The problem is about determining the probability of waiting more than 25 minutes between successive speeders, given that the waiting time follows an exponential distribution with a mean of 15 minutes.

Step 1: Define the exponential distribution

The probability density function (PDF) for an exponential distribution is given by:

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

where:

• $\lambda$ is the rate parameter (the reciprocal of the mean waiting time),
• $t$ is the waiting time.

The mean waiting time is 15 minutes, so $\lambda$ can be calculated as:

$\lambda = \frac{1}{\text{mean}} = \frac{1}{15} \text{ (per minute)}$

Step 2: Determine the probability of waiting more than 25 minutes

We are asked to find the probability that the waiting time $T$ exceeds 25 minutes, i.e., $P(T > 25)$.

For an exponential distribution, the cumulative distribution function (CDF) for $P(T \leq t)$ is:

$P(T \leq t) = 1 - e^{-\lambda t}$

Thus, the probability of waiting more than 25 minutes is:

$P(T > 25) = 1 - P(T \leq 25) = e^{-\lambda \cdot 25}$

Substitute $\lambda = \frac{1}{15}$:

$P(T > 25) = e^{-\frac{25}{15}} = e^{-\frac{5}{3}}$

Step 3: Calculate the value

Now calculate $e^{-\frac{5}{3}}$:

$P(T > 25) \approx e^{-1.6667} \approx 0.1889$

Final Answer:

The probability of waiting more than 25 minutes between successive speeders is approximately 0.1889 (or about 18.89%).

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the probability of waiting more than 30 minutes between speeders?
2. How does the probability change if the mean time is reduced to 10 minutes?
3. What is the expected waiting time between speeders?
4. How do you calculate the median waiting time for an exponential distribution?
5. What is the probability of waiting between 10 and 20 minutes between speeders?

Tip: Exponential distributions are memoryless, meaning the probability of waiting a given time is independent of how much time has already passed.