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The time between the arrival of electronic messages at your computer is exponentially distributed with a mean of two hours. (a) What is the probability that you do not receive a message during a two- hour period? (b) If you have not had a message in the last four hours, what is the probability that you do not receive a message in the next two hours? (c) What is the expected time between your fifth and sixth messages?

Short Answer

Expert verified
(a) \(0.3679\); (b) \(0.3679\); (c) 2 hours.

Step by step solution

01

Understand the Distribution

The time between arrivals is exponentially distributed with a mean of 2 hours. The exponential distribution is characterized by the parameter \( \lambda \), where the mean is \( \frac{1}{\lambda} \). Thus, \( \lambda = \frac{1}{2} \text{ per hour} \).
02

Step 2(a): Calculate Probability for No Message in Two Hours

For an exponential distribution, the probability that no event occurs in time \( t \) is given by the formula \( P(T > t) = e^{-\lambda t} \). For \( t = 2 \) hours, we calculate: \[ P(T > 2) = e^{-\frac{1}{2} \times 2} = e^{-1} \approx 0.3679 \]
03

Step 3(b): Apply Memoryless Property for Four Hours Without Message

The exponential distribution is memoryless, meaning the probability of no message in the next 2 hours is the same regardless of how much time has already passed. Therefore, the probability is \[ P(T > 2) = e^{-1} \approx 0.3679 \]
04

Step 4(c): Expected Time Between Fifth and Sixth Messages

Each inter-arrival time in an exponential distribution is identically distributed with a mean of \( 2 \) hours, regardless of count. Therefore, the expected time between the fifth and sixth message is \( \text{Expected Time} = \frac{1}{\frac{1}{2}} = 2 \text{ hours}. \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Memoryless Property
In the context of the exponential distribution, the memoryless property is a fascinating feature. It suggests that the probability of an event happening in the future is independent of past events. This means that even if certain time has already passed without the event occurring, the probability for its occurrence remains unchanged in predictive terms.
For example, if you haven't received a message in the last four hours, this property implies that the chances of not receiving a message in the next two hours are still the same as originally calculated.
The exponential distribution is the only continuous distribution with this memoryless trait. This characteristic is incredibly convenient for certain calculations and makes this distribution ideal for modeling random events like message arrivals.
Probability Calculation
Calculating probabilities using the exponential distribution involves a specific formula. This distribution is defined using a parameter known as \( \lambda \), which is the rate at which events occur. Given that the mean time between events is the inverse of \( \lambda \), you can determine probability using this relationship.
The formula for the probability that no event occurs in a given time \( t \) is
  • \( P(T > t) = e^{-\lambda t} \)
For instance, when \( t = 2 \) hours, with a \( \lambda \) of 0.5 per hour, substitute these values into the formula to get \( e^{-(0.5 \times 2)} = e^{-1} \), which equals about 0.3679. This calculation tells us the likelihood of not receiving any messages over that time period.
Understanding and applying this formula is critical for solving a variety of real-world problems involving exponentially distributed inter-times.
Expected Time
The expected time in an exponential distribution refers to the mean time between events. Because the exponential distribution represents the "waiting time" until the next event, this expected time is an intuitive metric.
Given that each event is identically and independently distributed, the expected time for any inter-arrival period remains constant. It doesn’t matter which specific interval you're evaluating; it always equals the inverse of \( \lambda \), or the given mean.
For instance, if you have five messages and you are calculating the expected time between the fifth and sixth message, this is simply the mean of the distribution. With a mean of 2 hours, as stated, the expected inter-arrival time between any two consecutive messages remains 2 hours. This consistency is a hallmark of each interval in the exponential sequence and confirms the stationary nature of exponential processes.

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