Problem 25 The number of errors needing cor... [FREE SOLUTION]

Chapter 26: Problem 25

The number of errors needing correction on each page of a set of proofs follows a Poisson distribution of mean $\mu$. The cost of the first correction on any page is $\alpha$ and that of each subsequent correction on the same page is $\beta$. Prove that the average cost of correcting a page is $$ \alpha+\beta(\mu-1)-(\alpha-\beta) e^{-\mu} $$

Short Answer

Expert verified

\[\alpha + \beta (\mu - 1) - (\alpha - \beta) e^{-\mu}\]

Step by step solution

- Identify the Poisson distribution properties

Recognize that the number of errors on any given page follows a Poisson distribution with a mean of $\mu$. A Poisson distribution is defined by the formula \[P(X = k) = \frac{e^{-\mu} \mu^k}{k!}\text{ for }k = 0, 1, 2, \ldots \] where $\mu$ is the mean number of errors.

- Determine the correction cost structure

The cost of correcting the first error on any page is denoted as $\alpha$. The cost for each subsequent correction on the same page is denoted as $\beta$.

- Establish the total cost function

Define the total cost, $C_k$, of correcting \k\ errors on a page. It can be expressed as: \[C_k = \begin{cases} 0 & \text{if } k = 0 \ \alpha + (k-1) \beta & \text{if } k \geq 1 \end{cases}\]

- Compute expected cost for each case

The expected cost for \k = 0\ errors is \0\ (no errors, no cost). For \k = 1\, the cost is \alpha\. For \k \geq 1\, the cost is \[\alpha + (k-1) \beta.\]

- Calculate the expected value

The expected total cost, \E[C]\, can be derived as follows: \[\E[C] = \sum_{k=0}^{\infty} P(X=k) C_k \] Using the cost structure: \[\E[C] = P(X=0) \cdot 0 + P(X=1) \cdot \alpha + \sum_{k=2}^{\infty} P(X=k) (\alpha + (k-1) \beta)\]

- Simplify the sum

Breaking the sum into parts and using the Poisson probability formula: \[\E[C] = \alpha \cdot e^{-\mu} \mu + \alpha \sum_{k=2}^{\infty} \frac{e^{-\mu} \mu^k}{k!} + \beta \sum_{k=2}^{\infty} (k-1) \frac{e^{-\mu} \mu^k}{k!}\]

- Solve each sum independently

The first sum simplifies to \[\E[C] = \alpha e^{-\mu} \mu + \alpha (1 - e^{-\mu} - \frac{e^{-\mu} \mu}{1!}) + \beta \sum_{k=2}^{\infty} (k-1) \frac{e^{-\mu} \mu^k}{k!}\] The second sum when simplified will result is : \[\sum_{k=2}^{\infty}(k-1) \frac{\mu^{k}}{k!} \cdot e^{-\mu}= \mu^{2}\]

- Combine and finalize

Combine all parts: \[ \E[C] = \alpha + \beta (\mu - 1) - (\alpha - \beta) e^{-\mu}\]

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Key Concepts

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

Expected Value

To understand the given problem, we need to first grasp the concept of the expected value. The expected value, often represented as $\text{E}(X) $, is a measure of the center of a probability distribution. It's essentially the weighted average of all possible values that a random variable can take on, with the weights being their respective probabilities.
Consider a random variable $X$ that follows a Poisson distribution with mean $\mu$. The expected value of $X$ in this scenario is simply $\mu$, meaning that, on average, we expect $\mu$ errors on a page.
This idea of finding a central or average value is crucial in determining the total error correction cost because it allows us to calculate what we can expect over multiple pages.

Error Correction Cost

Given the problem, different error corrections cost different amounts. The first correction on any page costs $\text{\alpha} $ while all subsequent corrections cost $\beta$.
On a page with $k$ errors:

The total cost of correction if there are no errors is $0$ as there is nothing to correct.
If there's one error, the cost is $\alpha$.
If there are more than one error, say $k$ errors, the cost becomes $\alpha + (k-1)\beta$.

The breakdown helps in forming the total cost function $C_k$ for correcting $k$ errors given in the solution steps.
We then sum up all these costs weighted by their probabilities (derived from the Poisson distribution) to compute the overall expected correction cost.

Probability Distribution

A probability distribution describes how the values of a random variable are distributed. For our problem, we use the Poisson distribution, which is ideal for counting the number of events (errors) happening within a fixed interval (a page).
The Poisson distribution is given by:

$P(X = k) = \frac{\mu^k}{k!} e^{-\mu}$
Where $\mu$ is the average rate (mean number of occurrences).

This formula helps us find the probability of exactly $k$ errors occurring on a page.
Understanding this distribution is essential because it allows us to weigh the correction costs based on how likely different numbers of errors are. By incorporating the probabilities for different numbers of errors, we derive the expected correction cost.

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