Problem 143 - Applied Statistics and Probability for Engineers

Question

Errors caused by contamination on optical disks occur at the rate of one error every \(10^{5}\) bits. Assume that the errors follow a Poisson distribution. (a) What is the mean number of bits until five errors occur? (b) What is the standard deviation of the number of bits until five errors occur? (c) The error-correcting code might be ineffective if there are three or more errors within \(10^{5}\) bits. What is the probability of this event?

Step-by-Step Solution

Answer

(a) \(5 \times 10^5\), (b) \(223606.8\) bits, (c) 0.08.

Step 1: Define the underlining properties of the Poisson distribution

The problem states that errors follow a Poisson distribution, which is defined by a rate \( \lambda \). Given there is one error every \(10^5\) bits, \( \lambda = 1 \) per \(10^5\) bits. The mean number of errors in a certain number of trials or bits can be computed using this rate.

Step 2: Find the mean number of bits until five errors occur

The situation where we look at the number of trials until a fixed number of successes (in this case, errors) occurs follows a Gamma distribution. The expected number of bits until five errors is \( 5 \times 10^5 \), because each error is expected to occur every \(10^5\) bits. Thus, the mean number of bits until five errors is \(5 \times 10^5\).

Step 3: Calculate the standard deviation of the number of bits until five errors occur

For a Gamma distribution with shape parameter \(k = 5\) and rate \( \lambda = 1/10^5 \), the standard deviation of bits until \(k\) errors occur is given by \(\sqrt{k} \cdot (1/\lambda) = \sqrt{5} \times 10^5\). Calculating this gives approximately \(223606.8\) bits.

Step 4: Find the probability of three or more errors within \(10^5\) bits

Use the Poisson probability formula: \( P(X=k) = \frac{{e^{-\lambda} \cdot \lambda^k}}{{k!}} \) where \(\lambda = 1\) for \(10^5\) bits. We need to find \(P(X \geq 3) = 1 - (P(X=0) + P(X=1) + P(X=2))\). Compute these probabilities: \(P(X=0) = e^{-1}\approx 0.368, P(X=1) = e^{-1}(1^1/1!) \approx 0.368, P(X=2) = e^{-1}(1^2/2!) \approx 0.184\). Thus, \(P(X \geq 3) = 1 - (0.368 + 0.368 + 0.184) = 0.08\).

Key Concepts

Gamma distribution

A Gamma distribution is a two-parameter family of continuous probability distributions. It often appears in contexts where we are interested in modeling the time taken until a certain number of events occur. In this problem, we are interested in finding the number of bits until five errors occur on an optical disk.
A Gamma distribution is characterized by a shape parameter, often denoted as \( k \), and a rate parameter \( \lambda \). In the context of this problem, \( k = 5 \) as we are looking for the occurrence of five errors. The rate \( \lambda \) is given as \( 1/10^5 \) because one error appears every \( 10^5 \) bits on average.
The mean or expected value for a Gamma distribution is given by \( \frac{k}{\lambda} \), which results in \( 5 \times 10^5 \) for our problem. This indicates the expected number of bits until five errors occur.

Error rate

The error rate deals with the frequency at which errors occur. In many practical situations, including our exercise, this rate provides critical information to understand the reliability of a certain process.
In the given problem, the error rate is described as one error for every \( 10^5 \) bits. This establishes our Poisson process with a rate parameter \( \lambda = 1 \) per \( 10^5 \) bits, which helps determine both the mean and probability regarding error occurrence within a defined space.
Understanding the error rate helps in planning how much error correction might be required to maintain data integrity, and it was especially useful here to calculate further expectations.

Mean and standard deviation

Mean and standard deviation are statistical measures that provide insights on the distribution of data. The mean gives us an average value, whereas the standard deviation tells us about the variation from this average.
In this exercise, the mean number of bits until five errors occur was calculated using the expression for a Gamma distribution as \( 5 \times 10^5 \) bits. For standard deviation, which gives a sense of the spread of values around the mean, the formula \( \sqrt{k} \cdot \frac{1}{\lambda} \) was used, yielding approximately \( 223606.8 \) bits.
These statistics are crucial for understanding the variability and expectation of failures, allowing for better system design and error correction strategies.

Probability of events

Probability helps to estimate the likelihood of an event occurring within a specified context. In this exercise, calculating the probability of observing three or more errors within \( 10^5 \) bits is required.
Using the Poisson probability formula, probabilities for zero, one, and two errors were determined before calculating the sum of these probabilities and subtracting from one. The calculated probability of at least three errors within this span was found to be 0.08. This means there’s an 8% chance that the error-correcting code might be ineffective.
Such probabilistic assessments help engineers and data scientists understand the risks and design systems accordingly to mitigate potential failures.