Problem 40 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

JOURNALISM Suppose that the number of typographical errors on a page of a local newspaper follows a Poisson distribution with an average of \(2.5\) errors per page. a. Find the probability that a randomly selected page is free from typographical errors. b. Find the probability that a randomly selected page has at least one typographical error. c. Find the probability that a randomly selected page has at least three typographical errors. d. Find the probability that a randomly selected page has fewer than three typographical errors.

Step-by-Step Solution

Answer

a. \(0.0821\), b. \(0.9179\), c. \(0.4560\), d. \(0.5439\)

Step 1: Understand the Poisson distribution

The Poisson distribution is used to model the number of events (typographical errors) happening within a fixed interval (one page). The formula for the Poisson probability is: \[ P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!} \] where \(\lambda\) is the average number of events per interval (2.5 errors per page) and \(k\) is the actual number of events.

Step 2: Probability of zero errors (part a)

For part (a), find the probability that a page has zero errors: \(k=0\). Use the formula: \[ P(X=0) = \frac{2.5^0 e^{-2.5}}{0!} = e^{-2.5} \] Use a calculator to find: \( e^{-2.5} \approx 0.0821 \).

Step 3: Probability of at least one error (part b)

For part (b), find the probability that a page has at least one error. This is the complement of having zero errors. Use: \[ P(X \geq 1) = 1 - P(X=0) = 1 - 0.0821 \approx 0.9179 \]

Step 4: Probability of at least three errors (part c)

For part (c), find the probability that a page has at least three errors: Use the complement principle: \[ P(X \geq 3) = 1 - (P(X=0) + P(X=1) + P(X=2)) \] Calculate each term: \[ P(X=1) = \frac{2.5^1 e^{-2.5}}{1!} = 2.5 e^{-2.5} \approx 0.2053 \] \[ P(X=2) = \frac{2.5^2 e^{-2.5}}{2!} = 3.125 e^{-2.5} \approx 0.2566 \] Sum the probabilities: \[ P(X \geq 3) = 1 - (0.0821 + 0.2053 + 0.2566) \approx 0.4560 \]

Step 5: Probability of fewer than three errors (part d)

For part (d), find the probability that a page has fewer than three errors: Use: \[ P(X < 3) = P(X=0) + P(X=1) + P(X=2) \] Sum the probabilities already calculated: \[ P(X < 3) = 0.0821 + 0.2053 + 0.2566 \approx 0.5439 \]

Key Concepts

headline of the respective core concept

Understanding probability theory is essential when dealing with statistical problems like our Poisson distribution example.
Probability theory deals with the likelihood that events will occur. In simple terms, it measures how probable it is to get a certain result.
This is fundamental for understanding typographical errors in a newspaper. We model these probabilities to estimate real-world events.
One important concept is the

Typographical Errors

Typographical errors are mistakes made during the typing process. In the context of our newspaper example, these errors are the events we are measuring.
The number of typographical errors can vary from page to page, but we assume here it follows a specific statistical pattern.
By understanding how often these errors occur on average, we can use models like the Poisson distribution to make predictions about their occurrence on any given page.
In our example, \(\lambda = 2.5\), suggesting that on average, there are 2.5 typographical errors per page.

Complement Principle

The Complement Principle is a useful tool in probability theory, particularly when calculating the probability of an event not happening.
It states that the probability of an event happening is equal to 1 minus the probability that it does not happen.
For example, the probability of having at least one error on a page is the complement of having zero errors.
This can be mathematically represented as:
\[ P(X \geq 1) = 1 - P(X = 0)\].
This simplification is beneficial as calculating the complement event is often more straightforward.

Mathematical Calculation

Mathematical calculations help us quantify probabilities using specific formulas. In the Poisson distribution, we employ the following formula:
\[ P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]
Here, \(\lambda\) is the average number of events (errors), and \(k\) is the actual number of events.
In part (a), for zero errors, this looks like: \[ P(X = 0) = \frac{2.5^0 e^{-2.5}}{0!} = e^{-2.5}\].
Using a calculator, we find \(P(X = 0) \approx 0.0821\), helping us understand the exact likelihood.

Applied Calculus

Applied calculus assists in solving these probability problems by integrating various mathematical tools.
We use exponential functions like \(e^{-2.5}\) which arise naturally from the Poisson process.
These calculations allow us to understand and predict typographical errors in real-life scenarios.
For instance, integrating the probability of zero, one, and two errors, \(P(X < 3) = P(X=0) + P(X=1) + P(X=2)\), explains comprehensively the likelihood of fewer than three errors occurring.
This shows how different areas of mathematics work together to solve practical problems.