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The number of typing errors made by a typist has a Poisson distribution with an average of four errors per page. If more than four errors appear on a given page, the typist must retype the whole page. What is the probability that a randomly selected page does not need to be retyped?

Short Answer

Expert verified
The probability that a page does not need retyping is approximately 0.629.

Step by step solution

01

Define the Poisson Distribution

A Poisson distribution is defined by its average rate (mean) of occurrence, denoted as \( \lambda \). For this problem, the average number of typing errors per page is \( \lambda = 4 \). We're interested in the number of pages that do not need to be retyped, meaning those with 4 or fewer errors.
02

Establish the Condition

The condition for not needing to retype a page is having 4 or fewer errors. Therefore, we need to calculate the probability of \( X \leq 4 \), where \( X \) is the number of errors, modeled as a Poisson random variable.
03

Employ the Poisson Probability Formula

The probability of having \( k \) errors on a page is given by the formula: \[ P(X = k) = \frac{{\lambda^k e^{-\lambda}}}{{k!}} \]. Here, \( e \approx 2.71828 \), \( \lambda = 4 \) and \( k = 0, 1, 2, 3, \text{ or } 4 \).
04

Calculate Individual Probabilities for k = 0 to 4

Calculate each necessary probability: - \( P(X = 0) = \frac{{4^0 e^{-4}}}{{0!}} = e^{-4} \)- \( P(X = 1) = \frac{{4^1 e^{-4}}}{{1!}} = 4e^{-4} \)- \( P(X = 2) = \frac{{4^2 e^{-4}}}{{2!}} = 8e^{-4} \)- \( P(X = 3) = \frac{{4^3 e^{-4}}}{{3!}} = \frac{{64}}{6}e^{-4} \)- \( P(X = 4) = \frac{{4^4 e^{-4}}}{{4!}} = \frac{{256}}{24}e^{-4} \)
05

Sum the Probabilities

Now, sum these probabilities to find \( P(X \leq 4) \): \[ P(X \leq 4) = e^{-4} + 4e^{-4} + 8e^{-4} + \frac{{64}}{6}e^{-4} + \frac{{256}}{24}e^{-4} \] \[ = e^{-4}(1 + 4 + 8 + \frac{64}{6} + \frac{256}{24}) \]
06

Simplify the Expression

Calculate the final expression: \[ 1 + 4 + 8 + \frac{64}{6} + \frac{256}{24} \approx 1 + 4 + 8 + 10.67 + 10.67 = 34.34 \] Thus, \[ P(X \leq 4) = e^{-4} \times 34.34 \] Using \( e^{-4} \approx 0.01832 \), \[ P(X \leq 4) \approx 0.629 \]
07

Conclusion

Therefore, the probability that a randomly selected page does not need to be retyped, i.e., has 4 or fewer errors, is approximately 0.629.

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Key Concepts

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

Typing Errors
When a typist is working on documents, each page can have a varying number of typing errors. While some pages might have no mistakes at all, others could feature several errors. These mistakes can range from simple spelling errors to punctuation problems. Understanding the nature and frequency of such typing errors is essential, particularly when assessing the work output of a typist over a period of time.

In statistical terms, the occurrence of typing errors per page in this scenario is modeled using the Poisson distribution. This type of distribution helps us predict the probability of how many errors we can expect on average per page. In our exercise, the average number of typing errors per page is set at 4. If a page contains no more than 4 errors, it is considered satisfactory and does not need retyping. However, should a page exceed this number, the typist is required to redo the entire page, ensuring the final document is as error-free as possible.
Probability Calculation
The Poisson distribution is a powerful tool for probability calculation. Concretely, it is used to predict the probability of a given number of events happening in a fixed interval of time or space. The parameters we work with include the average rate of occurrence, denoted by \( \lambda \). In this problem, \( \lambda = 4 \) because we have an average of 4 typing errors per page.

To find out how likely it is that a page does not require retyping, we calculate the probability that a page has 4 or fewer errors using the Poisson probability formula. This is given by:
  • The probability mass function: \[ P(X = k) = \frac{{\lambda^k e^{-\lambda}}}{{k!}} \]
  • Where \( e \approx 2.71828 \), \( \lambda = 4 \), and \( k \) is the number of errors, varying from 0, 1, 2, 3, to 4.
Each probability is calculated individually and then summed to provide the total probability of a page having 4 or fewer errors.
Retyping the Page
Imagine typing a document and being just one or two mistakes away from having to start over—in the context of our problem, that’s what retyping the page entails. This necessity arises only when a page has more than 4 typing errors. Thus, the probability calculation we did helps the typist understand how often they might expect to retype, based solely on error frequency.

The calculated probability, approximately 0.629, is the likelihood that a randomly selected page will not need to be retyped. This insight is crucial—especially in high-volume typing tasks—because it can significantly impact workflow and efficiency.

Retyping can be time-consuming, so knowing the chances of needing to redo pages can help typists manage their time better, allow them to plan ahead, and also encourage them to adopt better typing practices to reduce the occurrence of errors initially.

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Most popular questions from this chapter

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