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In a Poisson process, events occur continuously and independently at a constant average rate. An important characteristic of such processes is the mean time between events. This is particularly useful in determining how often you can expect events to take place.
To calculate the mean time between events, you need to know the rate, denoted by the Greek letter \( \lambda \). The rate \( \lambda \) represents the average number of events per unit time. In our example, \( \lambda \) is 3 counts per minute, indicating that, on average, you observe 3 log-ons to the network each minute.
The mean time between events is simply the reciprocal of \( \lambda \). Therefore, you find it using the formula:

\[ \text{Mean time between events} = \frac{1}{\lambda} \]

For our case, it's \( \frac{1}{3} \) minutes or approximately 20 seconds. This tells us that, on average, there will be around 20 seconds between each log-on. Understanding this concept is vital for planning capacity in network management or similar contexts.

The time between successive events in a Poisson process follows what is known as an exponential distribution. This distribution is key to understanding how time gaps between events are statistically spread out.
The exponential distribution is characterized by its mean and standard deviation, both equal to \( \frac{1}{\lambda} \). This directly relates the rate of events to the expected dispersion of time intervals. In the example of the computer network, both the mean and standard deviation of the time between log-ons are \( \frac{1}{3} \) minutes.
Some noteworthy properties of exponential distribution include:

• The distribution is memoryless, meaning the probability of an event occurring in future time periods is independent of past history.
• It is suitable for modeling scenarios where events occur continuously and independently.

Recognizing when to apply the exponential distribution helps greatly in analyzing and predicting time-dependent processes, from network log-ons to radioactive decay.

When dealing with Poisson processes and exponential distributions, calculating probabilities is often necessary to make predictions or decisions.
In our problem, we need to determine the time \( x \) such that there's a 95% probability of at least one log-on occurring before \( x \) minutes pass. The formula to calculate this probability \( P(X \leq x) \) is:

\[ 1 - e^{-\lambda x} \]

Setting this equal to the desired probability, 0.95, you solve for \( x \):

\[ 1 - e^{-3x} = 0.95 \]

\( e^{-3x} = 0.05 \).
By taking the natural logarithm and rearranging, you compute:

\[ x = -\frac{\ln(0.05)}{3} \]

This gives us \( x \approx 0.766 \) minutes, meaning that within approximately 46 seconds, there's a 95% chance of observing at least one log-on. Mastering these probability calculations enables you to assess the likelihood of events and optimize processes based on their time distribution.