Chapter 16: Problem 14
Suppose a 200 -page book has, on the average, one misprint every 10 pages. On about how many pages would you expect to find 2 misprints?
Short Answer
Expert verified
About 1 page will have 2 misprints.
Step by step solution
01
- Identify the given information and the goal
The book has 200 pages and, on average, one misprint every 10 pages. We need to find the number of pages expected to have 2 misprints.
02
- Calculate the average number of misprints per page
Since there is one misprint every 10 pages, the average number of misprints per page is given by \(\frac{1}{10}\).
03
- Use the Poisson distribution for calculation
For calculating the number of pages with 2 misprints, use the Poisson distribution. The probability mass function for a Poisson-distributed random variable with mean \(\text{λ}\) is given by \(P(X=k) = \frac{e^{-\text{λ}} \text{λ}^k}{k!}\). Here, \(\text{λ} = \frac{200}{10} = 20\) (average number of misprints in the whole book).
04
- Calculate the number of pages with exactly 2 misprints
To find how many pages have exactly 2 misprints, set \(k = 2\). We need the probability of finding 2 misprints on any given page. First, find the average misprints per page: \(\text{λ}_{\text{page}} = \frac{1}{10}\). Using the Poisson formula for a single page: \(P(X = 2) = \frac{e^{-\frac{1}{10}} (\frac{1}{10})^2}{2!}\).
05
- Calculate the Poisson probability
Substitute the values into the Poisson formula: \(P(X = 2) = \frac{e^{-\frac{1}{10}} (\frac{1}{10})^2}{2!} = \frac{e^{-0.1} (0.01)}{2}\).\(e^{-0.1} ≈ 0.905\) for simplicity. So, \(P(X = 2) = \frac{0.905 \times 0.01}{2} = 0.004525\).
06
- Calculate the total expected pages with 2 misprints
Multiply the probability by the total number of pages: \(0.004525 \times 200 = 0.905\). Therefore, about 0.905 pages are expected to have exactly 2 misprints.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Average Number of Misprints
Understanding the average number of misprints is key to solving problems related to the Poisson distribution.
Here, the book has an average of one misprint for every 10 pages. This means that over larger intervals, this average can help estimate the likelihood of multiple misprints occurring.
In this problem, you need the average number of misprints per page. Since there's one misprint every 10 pages, divide 1 by 10 to get \(\frac{1}{10}\) misprints per page.
This will be handy when working with the Poisson distribution's probability mass function.
Here, the book has an average of one misprint for every 10 pages. This means that over larger intervals, this average can help estimate the likelihood of multiple misprints occurring.
In this problem, you need the average number of misprints per page. Since there's one misprint every 10 pages, divide 1 by 10 to get \(\frac{1}{10}\) misprints per page.
This will be handy when working with the Poisson distribution's probability mass function.
Probability Mass Function
The Poisson distribution is ideal for estimating the number of events (like misprints) over a continuous interval.
The formula for the Poisson distribution's probability mass function (PMF) is:
\[P(X=k) = \frac{e^{-\text{λ}} \text{λ}^k} {k!} \]
Here, \(λ\) is the average rate, \(k\) is the number of events, and \(e\) is approximately 2.71828.
In this scenario, the average rate of misprints per page is \(λ_{\text{page}} = \frac{1}{10} = 0.1\). We substitute this into the PMF formula to find the probability of 2 misprints on any given page.
The formula for the Poisson distribution's probability mass function (PMF) is:
\[P(X=k) = \frac{e^{-\text{λ}} \text{λ}^k} {k!} \]
Here, \(λ\) is the average rate, \(k\) is the number of events, and \(e\) is approximately 2.71828.
In this scenario, the average rate of misprints per page is \(λ_{\text{page}} = \frac{1}{10} = 0.1\). We substitute this into the PMF formula to find the probability of 2 misprints on any given page.
Expected Value Calculation
The expected value helps estimate the number of occurrences over several trials.
Here, it calculates the number of pages with exactly 2 misprints in a 200-page book.
First, calculate the probability of 2 misprints on one page using: \[P(X=2) = \frac{e^{-0.1} (0.1)^2}{2!} \]
After calculating for simplicity, \(e^{-0.1} \) is around 0.905.
So, the probability is: \[P(X=2) = \frac{0.905 \times 0.01}{2} \]
This gives: \[P(X=2) \approx 0.004525. \]
Multiply this by 200 for the total expected value: 0.004525 x 200 = 0.905 pages expected to have exactly 2 misprints.
Here, it calculates the number of pages with exactly 2 misprints in a 200-page book.
First, calculate the probability of 2 misprints on one page using: \[P(X=2) = \frac{e^{-0.1} (0.1)^2}{2!} \]
After calculating for simplicity, \(e^{-0.1} \) is around 0.905.
So, the probability is: \[P(X=2) = \frac{0.905 \times 0.01}{2} \]
This gives: \[P(X=2) \approx 0.004525. \]
Multiply this by 200 for the total expected value: 0.004525 x 200 = 0.905 pages expected to have exactly 2 misprints.
Number of Misprints Per Page
The focus of this problem lies in determining the distribution of misprints across pages.
Knowing the average number of misprints per page \( \text{λ}_{\text{page}} = 0.1 \) allows calculation of specific probabilities.
This part laid the groundwork for applying the Poisson distribution effectively.
Every page essentially has a tiny average of misprints, making it a good fit for this statistical model.
Use this information to predict occurrences (like 2 misprints) might appear on any given page in the book.
Knowing the average number of misprints per page \( \text{λ}_{\text{page}} = 0.1 \) allows calculation of specific probabilities.
This part laid the groundwork for applying the Poisson distribution effectively.
Every page essentially has a tiny average of misprints, making it a good fit for this statistical model.
Use this information to predict occurrences (like 2 misprints) might appear on any given page in the book.