To solve this problem, we need to fit a Poisson distribution. The Poisson distribution is a good model for the number of errors on a page because these events can be considered rare and independently occurring within a fixed interval (in this case, per 5 pages).
First, let’s determine the average number of errors per page. We have the information that there are on average 2 errors per 5 pages.
**Step 1: Calculate the average number of errors per page (λ):**
\[
\lambda = \frac{2 \text{ errors}}{5 \text{ pages}} = 0.4 \text{ errors per page}
\]
**Step 2: Poisson Probability Mass Function:**
The Poisson probability mass function is given by:
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
where:
- \( \lambda \) is the average number of errors per page (0.4),
- \( k \) is the number of errors,
- \( e \) is the base of the natural logarithm (approximately equal to 2.71828).
**Step 3: Calculate the probabilities for 0, 1, 2, 3, and 4 errors using λ = 0.4:**
- For \( k = 0 \):
\[
P(X = 0) = \frac{e^{-0.4} \cdot 0.4^0}{0!} = e^{-0.4} \approx 0.6703
\]
- For \( k = 1 \):
\[
P(X = 1) = \frac{e^{-0.4} \cdot 0.4^1}{1!} = 0.4 \cdot e^{-0.4} \approx 0.2681
\]
- For \( k = 2 \):
\[
P(X = 2) = \frac{e^{-0.4} \cdot 0.4^2}{2!} = 0.08 \cdot e^{-0.4} \approx 0.0536
\]
- For \( k = 3 \):
\[
P(X = 3) = \frac{e^{-0.4} \cdot 0.4^3}{3!} = 0.01067 \cdot e^{-0.4} \approx 0.0072
\]
- For \( k = 4 \):
\[
P(X = 4) = \frac{e^{-0.4} \cdot 0.4^4}{4!} = 0.001707 \cdot e^{-0.4} \approx 0.0007
\]
**Step 4: Calculate the expected number of pages with 0, 1, 2, 3, and 4 errors in 1000 pages:**
- Expected number of pages with 0 errors:
\[
E(X=0) = 1000 \times 0.6703 \approx 670.3
\]
- Expected number of pages with 1 error:
\[
E(X=1) = 1000 \times 0.2681 \approx 268.1
\]
- Expected number of pages with 2 errors:
\[
E(X=2) = 1000 \times 0.0536 \approx 53.6
\]
- Expected number of pages with 3 errors:
\[
E(X=3) = 1000 \times 0.0072 \approx 7.2
\]
- Expected number of pages with 4 errors:
\[
E(X=4) = 1000 \times 0.0007 \approx 0.7
\]
**Summary:**
In 1000 pages, based on the Poisson distribution with an average of 0.4 errors per page, we would expect approximately:
- 670 pages with 0 errors,
- 268 pages with 1 error,
- 54 pages with 2 errors,
- 7 pages with 3 errors,
- 1 page with 4 errors.
These values summarize the expected distribution of errors in the book's first print.
After correcting 50 pages of a book, the proofreader finds that there are, on the average, 2 errors per 5 pages. How many pages would one expect to find with 0, 1, 2, 3, and 4 errors in 1000 pages of the first print of the book? Fit a Poisson distribution.
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