/*! 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 82 Consider writing onto a computer... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(\lambda=.2\). (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?

Short Answer

Expert verified
a. 0.1637 b. 0.0176 c. 0.6703

Step by step solution

01

Understand the Poisson Distribution

The Poisson distribution is used to model the number of events (or pulses, in this case) happening in a fixed interval of time or space. It is defined by the parameter \( \lambda \), which is the average number of occurrences in the interval. Here, \( \lambda = 0.2 \).
02

Calculate Probability for Exactly One Missing Pulse

We use the Poisson probability mass function for this part: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \].Substitute \( k = 1 \) and \( \lambda = 0.2 \):\[ P(X=1) = \frac{e^{-0.2} \times 0.2^1}{1!} = e^{-0.2} \times 0.2 \].Compute to get:\[ P(X=1) \approx 0.1637 \].
03

Calculate Probability for At Least Two Missing Pulses

To find \( P(X \geq 2) \), it is easier to use the complement rule: \[ P(X \geq 2) = 1 - P(X < 2) \],which equals \( P(X=0) + P(X=1) \). First calculate \( P(X=0) \):\[ P(X=0) = \frac{e^{-0.2} \times 0.2^0}{0!} = e^{-0.2} \approx 0.8187 \].Then use the previously calculated \( P(X=1) \):\[ P(X \geq 2) = 1 - (P(X=0) + P(X=1)) = 1 - (0.8187 + 0.1637) \].Finally, compute:\[ P(X \geq 2) \approx 0.0176 \].
04

Calculate Probability for Neither Disk Having Missing Pulses

For two independent disks, the probability that neither has a missing pulse is \( P(X=0) \times P(Y=0) \), where Y is the pulse count for the second disk: \[ P(X=0 \text{ and } Y=0) = (e^{-0.2})(e^{-0.2}) = (0.8187)(0.8187) \].Compute to find:\[ P(X=0 \text{ and } Y=0) \approx 0.6703 \].

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) is a fundamental tool in the world of statistics, especially when dealing with discrete random variables like those found in a Poisson distribution. The PMF helps us determine the likelihood of a particular event occurring in a defined scenario. In the context of the Poisson distribution, it is defined mathematically as follows: \[ P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!} \] This equation gives the probability of observing exactly \( k \) events (e.g., missing pulses on a disk) within a fixed time frame or spatial area.
  • \( e \) is the base of the natural logarithm, approximately equal to 2.71828.
  • \( \lambda \) is the average number of events in the interval, known as the event rate.
  • \( k! \) (k factorial) is the product of all positive integers up to \( k \).
When we apply the PMF to solve problems involving the Poisson distribution, we can easily calculate the probability of getting exactly one missing pulse, or any specific number of pulses, as in the exercise example.
Complement Rule
The Complement Rule is a powerful concept that simplifies probability calculations. It leverages the fact that the total probability of all possible outcomes of a random process is 1. If you want to find the probability of event A not occurring, you can take the complement of event A and find its probability using the equation: \[ P(A^c) = 1 - P(A) \] In the context of finding the probability of having at least two missing pulses, the approach involves using this rule extensively.
First, we calculate the probabilities of fewer missing pulses, such as zero or one, and use their sum to find the probability of at least two missing pulses.
  • Calculate for zero pulses: \( P(X=0) \).
  • Calculate for one pulse: \( P(X=1) \).
  • Subtract these from 1 to find at least two: \( P(X \geq 2) = 1 - (P(X=0) + P(X=1)) \).
This method can save a lot of calculating time compared to finding each possible number of missing pulses directly.
Independent Events
Understanding the concept of Independent Events is crucial in probability theory. Independent events imply that the occurrence of one event does not affect the probability of another event occurring. In simpler terms, the outcome of one event does not influence the other.
In exercises like part (c) of the original problem, evaluating two or more independent events often leads to a multiplication of probabilities. Here, we're dealing with two disks being tested for missing pulses—both independently:
  • The probability of neither disk having a missing pulse is found by multiplying their individual probabilities: \( P(X=0) \times P(Y=0) \).
  • In this case, each disk has a probability of zero missing pulses, which was previously calculated as \( e^{-0.2} \).
  • The solution is then: \( (0.8187)(0.8187) \), demonstrating how independent events are multiplied to find combined probabilities.
This property of independence allows us to compute compound probabilities with ease, especially in scenarios involving multiple trials or observations.

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