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Let \(X\), the number of flaws on the surface of a randomly selected carpet of a particular type, have a Poisson distribution with parameter \(\lambda=5 .\) Use Appendix Table A.2 to compute the following probabilities: a. \(P(X \leq 8)\) b. \(P(X=8)\) c. \(P(9 \leq X)\) d. \(P(5 \leq X \leq 8)\) e. \(P(5

Short Answer

Expert verified
a. 0.9329, b. 0.0663, c. 0.0671, d. 0.4924, e. 0.2507

Step by step solution

01

Understanding Poisson Distribution

The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space. In this exercise, the number of flaws on a carpet is defined by a Poisson distribution with parameter \( \lambda = 5 \). This parameter represents the average number of flaws per carpet.
02

Using Table A.2 for Cumulative Probability

Appendix Table A.2 provides cumulative probabilities for Poisson distributions. For \( P(X \leq k) \), directly use the value from the cumulative table for Poisson distribution at \( k \).
03

Calculating P(X ≤ 8)

From Appendix Table A.2, find the cumulative probability \( P(X \leq 8) \) for \( \lambda = 5 \). According to the table, \( P(X \leq 8) = 0.9329 \).
04

Calculating P(X = 8)

To find \( P(X = 8) \), use the difference between cumulative probabilities: \( P(X \leq 8) - P(X \leq 7) \). From the table, \( P(X \leq 7) = 0.8666 \). So, \( P(X = 8) = 0.9329 - 0.8666 = 0.0663 \).
05

Calculating P(9 ≤ X)

To determine \( P(9 \leq X) \), calculate \( 1 - P(X \leq 8) \). We have \( P(X \leq 8) = 0.9329 \), therefore \( P(9 \leq X) = 1 - 0.9329 = 0.0671 \).
06

Calculating P(5 ≤ X ≤ 8)

Subtract the cumulative probability \( P(X \leq 4) \) from \( P(X \leq 8) \). From the table, \( P(X \leq 4) = 0.4405 \). Thus, \( P(5 \leq X \leq 8) = 0.9329 - 0.4405 = 0.4924 \).
07

Calculating P(5 < X < 8)

Subtract \( P(X \leq 5) \) from \( P(X \leq 7) \). From the table, \( P(X \leq 5) = 0.6159 \) and \( P(X \leq 7) = 0.8666 \). So, \( P(5 < X < 8) = 0.8666 - 0.6159 = 0.2507 \).

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Key Concepts

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

Cumulative Probability
The concept of cumulative probability in a Poisson distribution is crucial for understanding how probabilities add up. When working with cumulative probabilities, you're looking at the probability of obtaining a number of events less than or equal to a certain value. In the case of the Poisson distribution, cumulative probability is the total probability of observing up to a specific number of events.

For instance, if you need to calculate the probability of observing up to 8 flaws on a carpet, denoted as \( P(X \leq 8) \), you could simply refer to the cumulative probability table for this distribution. This process involves finding this value directly in a predesigned table, which saves a significant amount of calculation time.

This table lists cumulative probabilities for different numbers of events and parameters, helping to simplify complex calculations.
Parameter Lambda
The parameter \(\lambda\), also known as lambda, is a defining feature of the Poisson distribution. It represents the average rate at which an event occurs within a fixed interval. In our carpet example, \(\lambda = 5\) specifies that, on average, 5 flaws are expected per carpet.

Understanding lambda is crucial because it helps determine the shape and probabilities of the Poisson distribution.
  • A higher \(\lambda\) means more events are expected, leading to a wider spread of probabilities across values.
  • Conversely, a lower \(\lambda\) indicates fewer expected events, concentrating probabilities towards smaller numbers.

Lambdas are typically derived from historical data or observational studies and play a key role in predictive analyses.
Probability Calculation
Calculating probabilities for specific or ranges of values in a Poisson distribution involves a few key methods. Let's break it down:

1. **Exact Probability**: To find the probability of a specific number of events, such as exactly 8 flaws \(P(X = 8)\), you subtract the cumulative probability of the previous number \(P(X \leq 7)\) from \(P(X \leq 8)\). This results in the probability for just 8 events.

2. **Range Probability**: For a range, like \(P(5 \leq X \leq 8)\), you subtract the cumulative probability up to just before the range begins \(P(X \leq 4)\) from the cumulative probability up to the end of the range \(P(X \leq 8)\). This method gives the cumulative probability within those bounds.

Breaking down these calculations helps understand how differences between cumulative probabilities reveal specific event outcomes.
Appendix Table Usage
Appendix tables for Poisson distributions are valuable resources for quickly finding probabilities without having to calculate them manually each time. These tables display cumulative probabilities for different values of \(k\) when given a specific \(\lambda\).

The process of using an appendix table is straightforward:
  • Locate the column or row that corresponds with the given \(\lambda\).
  • Find the value \(k\) (like 8 in \(P(X \leq 8)\)) to directly obtain the cumulative probability.

Using these tables minimizes computational errors and saves time, especially when integrating into larger calculations or when needing probabilities frequently. This method is extremely helpful in academic settings or quick decision-making scenarios.

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Most popular questions from this chapter

An article in the Los Angeles Times (Dec. 3 , 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives computer boards in lots of five. Two boards are selected from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair \((1,2)\) represents the selection of boards 1 and 2 for inspection. a. List the ten different possible outcomes. b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define \(X\) to be the number of defective boards observed among those inspected. Find the probability distribution of \(X\). c. Let \(F(x)\) denote the cdf of \(X\). First determine \(F(0)=P(X \leq 0), F(1)\), and \(F(2)\), and then obtain \(F(x)\) for all other \(x\).

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

Let \(X=\) the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or \(h(X)=1 / X\) dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that \(1 / E(X)=E(1 / X)\).]

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a chemical, which it sells to customers in 5-lb containers. Let \(X=\) the number of containers ordered by a randomly chosen customer, and suppose that \(X\) has pmf \begin{tabular}{l|llll} \(x\) & 1 & 2 & 3 & 4 \\ \hline\(p(x)\) & \(.2\) & \(.4\) & \(.3\) & \(.1\) \end{tabular} Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of \(X\).]

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