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An internal study at Lahey Electronics, a large software development company, revealed the mean time for an internal e-mail message to arrive at its destination was 2 seconds. Further, the distribution of the arrival times followed the Poisson distribution. a. What is the probability a message takes exactly 1 second to arrive at its destination? b. What is the probability it takes more than 4 seconds to arrive at its destination? c. What is the probability it takes virtually no time, i.e., "zero" seconds?

Short Answer

Expert verified
a) 0.2706 b) 0.0527 c) 0.1353

Step by step solution

01

Understanding the Poisson Distribution

The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space. Given the average rate (mean), \( \lambda \), we can find the probability of a given number of events occurring (in this case, e-mails arriving). In our problem, \( \lambda = 2 \) seconds.
02

Formula for Poisson Probability

The probability of observing \( x \) events in a Poisson distribution is given by the formula: \[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]where \( e \) is approximately 2.71828, \( \lambda \) is the average number of events (mean), and \( x \) is the number of events.
03

Calculate Probability for Exactly 1 Second

Using the Poisson formula, find \( P(X = 1) \) with \( \lambda = 2 \).\[ P(X = 1) = \frac{e^{-2} \times 2^1}{1!} = \frac{2e^{-2}}{1} = 2 \times 0.1353 \approx 0.2706 \]
04

Calculate Probability for More than 4 Seconds

To find \( P(X > 4) \), we calculate the cumulative probability of \( X \leq 4 \) and subtract from 1.\[ P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \]Each calculated as:\[ P(X = 0) = \frac{e^{-2} \times 2^0}{0!} = 0.1353 \]\[ P(X = 2) = \frac{e^{-2} \times 2^2}{2!} = 0.2706 \]\[ P(X = 3) = \frac{e^{-2} \times 2^3}{3!} = 0.1804 \]\[ P(X = 4) = \frac{e^{-2} \times 2^4}{4!} = 0.0902 \]Summing these probabilities gives \( P(X \leq 4) = 0.9473 \).Then, \( P(X > 4) = 1 - 0.9473 = 0.0527 \).
05

Calculate Probability for Zero Seconds

Using the Poisson formula for \( x = 0 \):\[ P(X = 0) = \frac{e^{-2} \times 2^0}{0!} = 0.1353 \].

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Key Concepts

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

Probability Calculation
When dealing with events occurring over a fixed interval, such as the time for an email to arrive, the Poisson distribution is a useful tool for calculating probabilities. The Poisson distribution helps us determine the likelihood of a given number of events (e.g., emails arriving) happening in a specific time frame or space.
In the context of email transmission at Lahey Electronics, the problem requires calculating probabilities based on this distribution. Probability calculations with the Poisson distribution use the formula: \[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \] where:
  • \( e \) is the base of natural logarithms, approximately 2.71828.
  • \( \lambda \) represents the mean or average rate of occurrence, which is 2 seconds here.
  • \( x \) is the number of events or occurrences we want the likelihood of.
With this formula, you can find the probability of an email taking exactly a particular number of seconds to arrive by substituting \( x \) with the desired time.
For example, to find the probability of an email taking exactly 1 second to arrive, use \( x = 1 \): \[ P(X = 1) = \frac{ e^{-2} \times 2^1}{1!} = 0.2706 \] This means there's a 27.06% chance that an email will take exactly 1 second to reach its destination.
Mean Arrival Time
The mean arrival time, denoted by \( \lambda \) in the Poisson distribution, is an essential parameter. In Poisson processes, \( \lambda \) gives us the expected number of occurrences over the specified interval. For this case, \( \lambda = 2\), indicating that, on average, it takes 2 seconds for an email to arrive.
One of the primary reasons \( \lambda \) is so crucial lies in its direct influence on probability calculations. The larger the mean arrival time, the more spread out the possible outcomes for an event will be. Conversely, a smaller mean arrival time indicates that events are more tightly grouped around the mean.
Knowing the mean arrival time allows us to better predict and understand the behavior of a system over time. In Lahey Electronics' case, it aids in managing and optimizing their email transmission system by providing insights into expected performance. Companies can then adjust their infrastructures as necessary to deal with potential delays in communication effectively.
Email Transmission Time
Email transmission time is a critical aspect in communication processes, especially in a corporate environment like Lahey Electronics. The Poisson distribution is helpful in modeling such times because it reflects how these transmissions occur over time, factoring in the randomness and frequency of messages.
Understanding the transmission time helps in determining the efficiency of the email system. Littles' known delays, like the probability of an email taking more than four seconds to arrive, can be analyzed to enhance system performance. This kind of analysis provides a clearer view of potential bottlenecks in the email system.
The Poisson probability calculation for more than four seconds involves finding the complementary probability of the cumulative distribution function for times up to and including four seconds:
  • Compute \( P(X \leq 4) \) by summing probabilities from zero to four seconds.
  • Subtract this result from 1 to find \( P(X > 4) \), which is 0.0527.
Thus, there's a 5.27% chance that an email will take longer than four seconds to reach its destination. Tackling delays in email transmission time can significantly enhance overall communication efficiency within organizations.

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