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Chapter 4: Problem 4

Raw materials are studied for contamination. Suppose that the number of particles of contamination per pound of material is a Poisson random variable with a mean of 0.01 particle per pound. a. What is the expected number of pounds of material required to obtain 15 particles of contamination? b. What is the standard deviation of the pounds of materials required to obtain 15 particles of contamination?

Short Answer

Expert verified
a. 1500 pounds; b. 387.298 pounds.

Step by step solution

01

Understanding Poisson Distribution

The problem states that the number of particles of contamination per pound is given by a Poisson distribution with a mean (位) of 0.01 particles per pound. This means the average rate of contamination is 0.01 particles per pound of material.
02

Setting Up the Situation for Part (a)

To find the expected number of pounds of material required to obtain 15 particles, we apply the concept of waiting time for a Poisson process, which is modeled by the Exponential distribution. The parameter for the Exponential distribution is the reciprocal of the Poisson mean rate, i.e., 1/位.
03

Calculating the Expected Pounds for 15 Particles

Given the Poisson parameter 位 = 0.01, we need 15 contaminating particles. For such a process, the waiting time is given by a Gamma distribution with shape parameter 15 and rate parameter 位 = 0.01. The expected value for a Gamma distribution is given by shape/位. Thus, the expected number of pounds is 15/0.01 = 1500 pounds.
04

Solving Part (b): Standard Deviation Interpretation

In a similar context, the standard deviation for the Gamma distribution, which applies due to the waiting times in a Poisson process, is given by the square root of the variance of the amount of material (shape/位^2). For 15 particles, the standard deviation is sqrt(15/0.01^2) = sqrt(15/0.0001) = sqrt(150000) = 387.298.

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Key Concepts

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

Exponential Distribution
The Exponential Distribution is a continuous probability distribution often associated with waiting times between events in a Poisson process.

In this context of contamination, it describes the time or amount until a specific number of particles are observed.
  • Exponential distributions are characterized by a mean rate 位, which is the average rate of occurrence of the events.
  • The probability density function is given by \( f(x;位) = 位e^{-位x} \) for \(x \geq 0\).
  • The mean or expected value of an Exponential distribution is \( \frac{1}{位} \).

Understanding this distribution is essential when dealing with processes where events happen continuously and independently at a constant average rate.
It helps predict how much material you need to process before encountering a certain number of contamination particles.
Gamma Distribution
The Gamma Distribution generalizes the Exponential Distribution by describing the waiting time for multiple events. It arises naturally in the process of observing a specified number of occurrences of a Poisson process.

In the problem, the need for 15 particles is modeled using a Gamma distribution.
  • The Gamma distribution has two parameters: the shape parameter (often denoted as 'k'), which represents the number of events, and a rate parameter (位) corresponding to the Poisson process's mean rate.
  • The probability density function for the Gamma Distribution is \( f(x; k, 位) = \frac{位^k x^{k-1} e^{-位x}}{(k-1)!} \) for \(x \geq 0\).
  • The expected value, or mean, of a Gamma distribution is \( \frac{k}{位} \).

This distribution is incredibly useful for predicting how much material is needed when aiming to reach a targeted count of contamination particles.
For this problem, the calculations tell us one should expect 1500 pounds of material to encounter 15 particles.
Expected Value
The Expected Value is a crucial concept that gives the average outcome or long-term mean of a probability distribution. It helps to predict the average result if an experiment were repeated many times.

In probability distributions, the expected value (or mean) provides a central tendency.
  • For the Poisson distribution, the expected value is equal to the mean 位.
  • For an Exponential distribution, it is \( \frac{1}{位} \).
  • For a Gamma distribution, it is given as \( \frac{k}{位} \) where 'k' is the shape parameter, representing the event count.

In the context of the exercise, determining the expected value helps understand how many pounds of material are likely needed to find 15 contamination particles.
This represents a crucial part of risk and resource management in controlling material quality.
Standard Deviation
Standard Deviation is a measure of how spread out the values within a distribution are. It provides insight into the variability of outcomes around the expected value.

  • Standard deviation is the square root of the variance, providing a scale for data dispersion.
  • For a Gamma distribution, the standard deviation is expressed as \( \sqrt{\frac{k}{位^2}} \).
  • It is often used alongside the mean to give context to the expected outcome's variability.

In this exercise, calculating the standard deviation of the material required helps assess the reliability and precision of the estimated pounds.
A standard deviation of 387.298 pounds indicates a significant spread, meaning that while 1500 pounds is an expected average, actual values could vary noticeably around this mean. It鈥檚 important for precision in predicting resource usage and planning.

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