/*! 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 79 Let \(X\), the number of flaws o... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\), the number of flaws on the surface of a randomly selected boiler of a certain 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
Use Poisson cumulative table: a. Check P(X ≤ 8), b. Table P(X=8), c. Use complement 1-P(X≤8), d. Cumulative difference P(X≤8)-P(X≤4), e. Difference P(X≤7)-P(X≤5).

Step by step solution

01

Understanding the Poisson distribution

The Poisson distribution is used to model the number of events occurring in a fixed interval of time or space. For this problem, the number of flaws, \( X \), on the surface of a boiler, follows a Poisson distribution with mean \( \lambda = 5 \). The probability of observing \( k \) events in a Poisson distribution is given by \(P(X=k) = \frac{{e^{-\lambda} \lambda^k}}{{k!}}\).Here, the appendix table provides cumulative probabilities and probabilities for specific values of \( X \).
02

Calculating P(X ≤ 8)

To find \( P(X \leq 8) \), we sum the probabilities from \( X = 0 \) to \( X = 8 \) using appendix tables or cumulative distribution functions:\[P(X \leq 8) = \sum_{k=0}^{8} P(X = k)\]Using Appendix Table A.2, look up the cumulative probability \( P(X \leq 8) \) for a Poisson distribution with \( \lambda = 5 \), which provides the cumulative value directly.
03

Calculating P(X = 8)

The probability \( P(X = 8) \) is directly provided in the appendix tables. Using the formula for the Poisson probability,\[P(X = 8) = \frac{{e^{-5} \cdot 5^8}}{{8!}}\]Use the appendix table to find \( P(X = 8) \) or use the Poisson formula directly with \( \lambda = 5 \).
04

Calculating P(9 ≤ X)

To find \( P(9 \leq X) \), we use the complementary probability:\[P(9 \leq X) = 1 - P(X \leq 8)\]Since we already found \( P(X \leq 8) \) in Step 2, subtract this value from 1 to get \( P(9 \leq X) \).
05

Calculating P(5 ≤ X ≤ 8)

For \( P(5 \leq X \leq 8) \), we calculate:\[P(5 \leq X \leq 8) = P(X \leq 8) - P(X < 5)\]Where \( P(X < 5) = P(X \leq 4) \). Use the appendix table to find these cumulative probabilities and subtract them.
06

Calculating P(5 < X < 8)

Finally, we find \( P(5 < X < 8) \) using:\[P(5 < X < 8) = P(X < 8) - P(X \leq 5)\]With \( P(X < 8) = P(X \leq 7) \), get these cumulative probabilities from the appendix and subtract accordingly.

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Key Concepts

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

Probability Calculations
Probability calculations involve finding the likelihood of different outcomes from a specific distribution, such as the Poisson distribution. In our scenario, we are dealing with the distribution of flaws on boilers.
To calculate probabilities for a particular event, like finding exactly 8 flaws (i.e., \(P(X=8)\)), we use the Poisson probability formula. It evaluates the chance of a precise number of events happening in a fixed interval.
Essentially, probability calculations give us a way to predict and make decisions based on the likelihood of certain outcomes. When probabilities are listed in tables, they often show values for individual events, which help us in Poisson calculations.
Cumulative Distribution Functions
Cumulative Distribution Functions (CDFs) are crucial tools in probability that sum probabilities for all outcomes up to a certain value. For a Poisson distribution, like in the boiler flaw exercise, a CDF helps calculate pooled probabilities for events like \(P(X \leq 8)\).
The CDF provides the probability that the variable takes a value less than or equal to a target number. This is beneficial because it can simplify calculations—you don't have to find each individual probability separately and add them up. Instead, the CDF provides a direct solution from tables or equations, streamlining the process.
  • CDF utility: Quickly find probabilities for various scenarios.
  • Express probabilities as readily grouped values, increasing ease for problems like \(P(9 \leq X)\).
Poisson Formula
The Poisson formula is designed to calculate the probabilities of a given number of events occurring in a fixed interval. For the Poisson distribution, this probability is calculated as:
\[P(X=k) = \frac{{e^{-\lambda} \lambda^k}}{{k!}}\]This formula requires:
  • \(\lambda\): The average rate (mean) of occurrence, which is 5 in the boiler defect example.
  • \(k\): The exact number of occurrences you want to find the probability for.
This formula assumes events occur independently, which fits perfectly with many real-world scenarios like technical failures or natural phenomena. The use of the Poisson formula simplifies probability calculations for exact occurrences.
Complementary Probability
Complementary probability involves finding the probability of the opposite of your event of interest. For example, instead of calculating \(P(9 \leq X)\) directly, it's often easier to find \(P(X \leq 8)\) first and subtract it from one, i.e., \(1 - P(X \leq 8)\).
This method is particularly useful when it's cumbersome to calculate large or complex probabilities directly. By using complementary probability, we take advantage of what we already know about probability (that it sums to 1), making calculations more straightforward and less error-prone for broader ranges or tail-end probabilities.
  • Makes it easy to derive difficult probabilities.
  • Saves time by using previously calculated probabilities.

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Most popular questions from this chapter

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is \(5 \%\). Let \(X=\) the number of defective boards in a random sample of size \(n=25\), so \(X \sim \operatorname{Bin}(25, .05)\). a. Determine \(P(X \leq 2)\). b. Determine \(P(X \geq 5)\). c. Determine \(P(1 \leq X \leq 4)\). d. What is the probability that none of the 25 boards is defective? e. Calculate the expected value and standard deviation of \(X\).

Let \(X\) be the damage incurred (in \$) in a certain type of accident during a given year. Possible \(X\) values are 0,1000 , 5000 , and 10000 , with probabilities .8, .1, .08, and \(.02\), respectively. A particular company offers a \(\$ 500\) deductible policy. If the company wishes its expected profit to be \(\$ 100\), what premium amount should it charge?

Consider a deck consisting of seven cards, marked \(1,2, \ldots\), 7. Three of these cards are selected at random. Define an rv \(W\) by \(W=\) the sum of the resulting numbers, and compute the pmf of \(W\). Then compute \(\mu\) and \(\sigma^{2}\). [Hint: Consider outcomes as unordered, so that \((1,3,7)\) and \((3,1,7)\) are not different outcomes. Then there are 35 outcomes, and they can be listed. (This type of rv actually arises in connection with a hypothesis test called Wilcoxon's rank-sum test, in which there is an \(x\) sample and a \(y\) sample and \(W\) is the sum of the ranks of the \(x\) 's in the combined sample.)]

If \(a \leq X \leq b\), show that \(a \leq E(X) \leq b\).

The article "Reliability-Based Service-Life Assessment of Aging Concrete Structures" (J. Structural Engr., 1993: \(1600-1621\) ) suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is .5 year. a. How many loads can be expected to occur during a 2year period? b. What is the probability that more than five loads occur during a 2-year period? c. How long must a time period be so that the probability of no loads occurring during that period is at most .1?

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