The mean number of errors per page made by a member of the word processing pool for a large company is thought to be 2.72.7 with the number of errors distributed according to a Poisson distribution. If a page is​ examined, what is the probability that more than two errors will be​ observed?
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Part 1
The probability that more than two errors will be observed is enter your response here.
​(Round to four decimal places as​ needed.)

To solve this problem, we will use the Poisson distribution formula, which is:

\[
P(X = k) = \frac{{\lambda^k e^{-\lambda}}}{k!}
\]

Where:
- \( \lambda \) is the average number of errors per page (which is 2.7 in this case),
- \( k \) is the number of errors,
- \( e \) is Euler's number (approximately 2.71828),
- \( k! \) is the factorial of \( k \).

We are asked to find the probability that more than 2 errors are observed, i.e., \( P(X > 2) \). This is equal to:

\[
P(X > 2) = 1 - P(X \leq 2)
\]

So first, we need to calculate \( P(X \leq 2) \), which is the sum of the probabilities for 0, 1, and 2 errors:

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

Now, let's calculate each of these:

1. \( P(X = 0) = \frac{{2.7^0 e^{-2.7}}}{0!} = e^{-2.7} \)
2. \( P(X = 1) = \frac{{2.7^1 e^{-2.7}}}{1!} = 2.7 e^{-2.7} \)
3. \( P(X = 2) = \frac{{2.7^2 e^{-2.7}}}{2!} = \frac{{2.7^2}}{2} e^{-2.7} \)

Now, we can sum these probabilities and subtract them from 1 to find \( P(X > 2) \).

Let me calculate that for you.The probability that more than two errors will be observed on a page is approximately **0.5064** (rounded to four decimal places).

Would you like further details or have any questions about this? Here are 5 related questions to explore further:

1. How would the probability change if the mean number of errors was different?
2. What is the probability of observing exactly three errors on a page?
3. How do we compute cumulative probabilities for other values of \( k \) in a Poisson distribution?
4. Can the Poisson distribution be used to model other types of events?
5. How does the Poisson distribution compare to the binomial distribution?

**Tip:** When solving Poisson distribution problems, it helps to remember that it models rare events over a fixed interval or space, like errors per page or arrivals per hour.

The mean number of errors per page made by a member of the word processing pool for a large company is thought to be 2.7 with the number of errors distributed according to a Poisson distribution. If a page is examined, what is the probability that more than two errors will be observed?

Learn how to calculate the probability of observing more than two errors on a page using the Poisson distribution formula with a mean of 2.7 errors per page. This detailed step-by-step solution covers key probability concepts, including cumulative probabilities and Poisson distribution, suitable for students in Grades 10-12.

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