/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Let \(X\) be the number of mater... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodology for Probabilistic Life Prediction of MultipleAnomaly Materials" (Amer. Inst. of Aeronautics and Astronautics \(J ., 2006: 787-793\) ) proposes a Poisson distribution for \(X\). Suppose that \(\mu=4\). a. Compute both \(P(X \leq 4)\) and \(P(X<4)\). b. Compute \(P(4 \leq X \leq 8)\). c. Compute \(P(8 \leq X)\). d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

Short Answer

Expert verified
a. 0.6299 and 0.4335; b. 0.4749; c. 0.1192; d. 0.7977.

Step by step solution

01

Understanding Poisson Distribution

The Poisson distribution is used to model the probability of a given number of events happening in a fixed interval of time or space. It is defined by the parameter \( \mu \), the mean number of occurrences. Here, \( \mu = 4 \). The probability mass function is given by: \[P(X=k) = \frac{\mu^k e^{-\mu}}{k!}\] where \( k \) is the number of events.
02

Calculate P(X ≤ 4) for Poisson Distribution

We need to calculate \( P(X \leq 4) \), which is the cumulative probability of \( X \) being 0, 1, 2, 3, or 4. \[P(X \leq 4) = \sum_{k=0}^{4} \frac{4^k e^{-4}}{k!}\] Calculating each term individually and summing them yields approximately:- \( P(X=0) = 0.0183 \)- \( P(X=1) = 0.0733 \)- \( P(X=2) = 0.1465 \)- \( P(X=3) = 0.1954 \)- \( P(X=4) = 0.1954 \)Thus, \( P(X \leq 4) \approx 0.6299 \).
03

Determine P(X < 4)

The probability \( P(X < 4) \) is the probability of having 0, 1, 2, or 3 anomalies:\[P(X < 4) = \sum_{k=0}^{3} \frac{4^k e^{-4}}{k!}\] Using earlier calculations, \( P(X<4) = 0.0183 + 0.0733 + 0.1465 + 0.1954 \approx 0.4335 \).
04

Compute P(4 ≤ X ≤ 8)

The probability \( P(4 \leq X \leq 8) \) is calculated as:\[P(4 \leq X \leq 8) = P(X \leq 8) - P(X < 4)\]First, find \( P(X \leq 8) \):\[P(X \leq 8) = \sum_{k=0}^{8} \frac{4^k e^{-4}}{k!}\]Plug values to get \( \approxeq 0.9084 \). Thus:\[P(4 \leq X \leq 8) = 0.9084 - 0.4335 = 0.4749\].
05

Calculate P(8 ≤ X)

The probability \( P(8 \leq X) \) is given by the complement rule:\[P(8 \leq X) = 1 - P(X \leq 7)\].Calculate \( P(X \leq 7) \) using the cumulative probabilities up to 7:\[P(X \leq 7) \approx 0.8808 \]Thus, \[P(8 \leq X) = 1 - 0.8808 = 0.1192\].
06

Probability of X within One Standard Deviation

The standard deviation for a Poisson distribution is \( \sigma = \sqrt{\mu} \). Here \( \sigma = \sqrt{4} = 2 \).We need the probability that \( 2 \leq X \leq 6 \) (mean = 4) within one standard deviation unit:\[P(2 \leq X \leq 6) = P(X \leq 6) - P(X < 2) \].First, find \( P(X \leq 6) \approx 0.8893 \)Second, find \( P(X < 2) \approx 0.0916 \).Thus, \( P(2 \leq X \leq 6) = 0.8893 - 0.0916 \approx 0.7977 \).

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Key Concepts

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

Cumulative Probability
Cumulative probability in the context of the Poisson distribution refers to the probability that a random variable, say \(X\), takes a value less than or equal to a specified value \(k\). This is represented as \(P(X \leq k)\). For a given \(\mu\), which is the mean of the distribution, this is calculated by summing the individual probabilities of \(X\) from 0 up to \(k\). In formula terms, you have \[P(X \leq k) = \sum_{i=0}^{k} P(X=i)\].

This sum is what we call the cumulative probability function. It's especially useful for understanding the behavior of anomalies, as in our example involving a gas-turbine disk. Each term in the sum is calculated using the Poisson probability mass function, and the overall probability provides a comprehensive picture of the likelihood that \(X\) will be at or below a certain value.
Probability Mass Function
The probability mass function (PMF) is a fundamental concept in probability theory. It provides the probability that a discrete random variable is exactly equal to some value. In the Poisson distribution context, where we model the number of events in a given time interval, the PMF is given by:

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

Here, \(k\) is the number of occurrences, \(\mu\) is the average rate, and the exponential term \(e^{-\mu}\) represents the Poisson process decay.

Understanding the PMF helps in calculating specific probabilities, such as the ones we found for \(P(X=0), P(X=1),\) and so on. Each calculation tells us how likely a specific number of anomalies is and contributes to the larger picture provided by cumulative probabilities.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of a Poisson distribution, the standard deviation is particularly straightforward to compute because it equals the square root of the mean, \(\mu\). Thus, \(\sigma = \sqrt{\mu}\). This simplicity arises because the Poisson distribution models independent events occurring at a constant average rate.

In our case, \(\mu = 4\), so \(\sigma = 2\). This means that most of the values of \(X\) are likely to fall within \(4 \pm 2\), which enables us to calculate probabilities within one standard deviation of the mean. This calculation provides insights into the "spread" of anomalies around the average number.
Parameter Estimation
Parameter estimation in the context of a Poisson distribution involves determining the mean number of occurrences, \(\mu\). This parameter is crucial as it underlies the calculations of both the probability mass function and cumulative probabilities. Accurately estimating \(\mu\) is key, since it directly influences the predictive capability of the Poisson model.

For estimating \(\mu\), you typically rely on observed data, where the average number of events observed over several trials or periods gives you an estimate of \(\mu\). This estimate helps in predicting future occurrences and in setting expectations based on the Poisson model. Understanding how to estimate \(\mu\) carefully ensures the application of the Poisson distribution is aligned with real-world event frequencies.

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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.

Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let \(X\) be the number among the first 6 examined that have a defective compressor. Compute the following: a. \(P(X=5)\) b. \(P(X \leq 4)\) c. The probability that \(X\) exceeds its mean value by more than 1 standard deviation. d. Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If \(X\) is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) \(P(X \leq 5)\) than to use the hypergeometric pmf.

Suppose that the number of plants of a particular type found in a rectangular sampling region (called a quadrat by ecologists) in a certain geographic area is an rv \(X\) with pmf $$ p(x)= \begin{cases}\mathrm{c} / x^{3} & x=1,2,3, \ldots \\ 0 & \text { otherwise }\end{cases} $$ Is \(E(X)\) finite? Justify your answer (this is another distribution that statisticians would call heavy-tailed). 35\. A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf \begin{tabular}{l|llllll} \(x\) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(\frac{1}{15}\) & \(\frac{2}{15}\) & \(\frac{3}{15}\) & \(\frac{4}{15}\) & \(\frac{3}{15}\) & \(\frac{2}{15}\) \end{tabular} Suppose the store owner actually pays \(\$ 2.00\) for each copy of the magazine and the price to customers is \(\$ 4.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand \(X\), and then compute the expected revenue.]

If the sample space \(S\) is an infinite set, does this necessarily imply that any rv \(X\) defined from \(\&\) will have an infinite set of possible values? If yes, say why. If no, give an example.

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number \(X\) has a Poisson distribution with parameter \(\mu=2\). (Suggested in "Average Sample Number for Semi-Curtailed Sampling Using the Poisson Distribution," J. Quality Technology, 1983: 126-129.) a. What is the probability that a disk has exactly one missing pulse? b. What is the probability that a disk has at least two missing pulses? c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

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