To solve the problem, we need to understand a few key concepts about the exponential distribution.

Exponential Distribution Properties:

1. Standard Deviation (σ): For an exponential distribution, the standard deviation (σ) is equal to the mean (μ).
2. Mean (μ): This is the expected time for one event (or arrival) to occur.

Given that the standard deviation $\sigma$ is 1111 minutes, we can determine that the mean $\mu$ is also 1111 minutes.

Expected Time for Two Visitors:

Since the arrival times of the visitors are independent, the expected time for the second visitor to arrive after the first visitor will be the sum of the expected times for each visitor. For an exponential distribution, the expected time for one visitor is $\mu$.

Therefore, the expected time for two visitors to arrive is: $\text{Expected time for two visitors} = \mu + \mu = 2 \times \mu = 2 \times 1111 = 2222 \text{ minutes}$

Final Answer:

The expected time for two visitors to arrive is 2222 minutes.

Would you like more details on this topic or have any other questions?

Here are 5 related questions:

1. How would the expected time change if three visitors were expected to arrive?
2. What is the variance of the time between visitor arrivals?
3. Can you explain how the exponential distribution relates to the Poisson process?
4. What are the characteristics of an exponential distribution?
5. How would the expected time change if the standard deviation was halved?

Tip: For exponential distributions, the memoryless property means that the probability of an event occurring in the future is independent of how much time has already elapsed.