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Data from www.centralhudsonlabs.com determined the mean number of insect fragments in 225 -gram chocolate bars was \(14.4,\) but three brands had insect contamination more than twice the average. See the U.S. Food and Drug Administration-Center for Food Safety and Applied Nutrition for Defect Action Levels for food products. Assume that the number of fragments (contaminants) follows a Poisson distribution. (a) If you consume a 225 -gram bar from a brand at the mean contamination level, what is the probability of no insect contaminants? (b) Suppose that you consume a bar that is one-fifth the size tested (45 grams) from a brand at the mean contamination level. What is the probability of no insect contaminants? (c) If you consume seven 28.35 -gram (one-ounce) bars this week from a brand at the mean contamination level, what is the probability that you consume one or more insect fragments in more than one bar? (d) Is the probability of contamination more than twice the mean of 14.4 unusual, or can it be considered typical variation? Explain.

Short Answer

Expert verified
(a) Approximately 5.1 × 10^{-7}. (b) About 5.6%. (c) Approximately 99.85%. (d) More than twice the mean is likely unusual.

Step by step solution

01

Understanding Poisson Distribution

The Poisson distribution is used to model the probability of a number of events happening in a fixed interval of time or space, given the average number of times the event occurs over that interval. The mean number of insect fragments is 14.4 for a 225-gram bar.
02

Probability of No Contaminants in a 225-gram Bar

To find the probability of no insect contaminants in a 225-gram bar, we use the formula for the Poisson probability: \( P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \), where \( \lambda \) is the mean (14.4) and \( x = 0 \).\[P(X = 0) = \frac{e^{-14.4} \cdot 14.4^0}{0!} = e^{-14.4}approximately 5.1 \times 10^{-7}\]This means the probability is very small.
03

Adjusting for a 45-gram Bar

For a 45-gram bar, which is one-fifth the size, the mean number \( \lambda \) becomes \( \frac{14.4}{5} = 2.88 \). Use the Poisson formula again to find \( P(X = 0) \) for \( \lambda = 2.88 \).\[P(X = 0) = \frac{e^{-2.88} \cdot 2.88^0}{0!} = e^{-2.88}approximately 0.056 \]This indicates a 5.6% chance of no contamination in a 45-gram bar.
04

Analyzing Seven 28.35-gram Bars

Each 28.35-gram bar has a mean contamination of \( \frac{14.4}{8} = 1.8 \), as it's one-eighth the size of 225 grams. We analyze seven such bars. First, find the probability of no insects in a single bar: \( P(X = 0) = e^{-1.8} \approx 0.165 \). Now, find the probability of no insects in one or more of these multiple bars using 1 minus the probability of the complement.\[P(\text{1 or more bars have insects}) = 1 - P(\text{zero contaminants in 7 bars}) = 1 - (0.165^7) approximately 0.9985 \]This means there's a very high probability that at least one or more bars will have insect fragments.
05

Evaluating Unusual Contamination

A contamination level more than twice the average is \( 2 \times 14.4 = 28.8 \). To determine if this is unusual, we use the Poisson distribution to calculate \( P(\text{X} < 28.8) \). Since \( 14.4 \) is already quite high, anything more than twice the mean would usually be considered unusual unless specified by regulatory guidelines or calculations indicate otherwise.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with calculating the likelihood of various events. In this exercise, we delve into the realm of probability using the Poisson distribution. This specific distribution is instrumental in modeling rare events, especially in contexts where these events occur independently within a fixed interval of time or space.
The Poisson distribution is defined by its mean, denoted by \( \lambda \), which signifies the average number of occurrences in a given interval. For example, when analyzing insect fragments in chocolate bars, \( \lambda \) was provided as 14.4 for a 225-gram bar. Understanding the math behind this helps in predicting the chance or likelihood of different numbers of contaminants.
Thus, using a Poisson model, you can solve questions like: "What is the probability of no insect contamination in a single chocolate bar?" This requires calculating probabilities for \( x = 0 \) events (i.e., no contaminants), which is achieved using the formula \( P(X = x) = \frac{e^{-\lambda} \cdot \lambda^x}{x!} \).
Contamination Analysis
Contamination analysis involves examining and understanding the distribution and frequency of contaminants within products. In the context of our chocolate bar exercise, this centers around determining the expected number of insect fragments within a given size of chocolate bar. This understanding informs quality control processes and safety guidelines.
The exercise provided different scenarios, such as altering the chocolate bar size, to analyze:
  • The mean level of contaminants when bar sizes alter.
  • The probability of contaminants manifesting at different levels across various sizes of bars.
By scaling down from a 225-gram bar to, say, a 45-gram bar, the mean contamination (\( \lambda \)) also reduces proportionally to 2.88, which is one-fifth of the original size. This scaling helps us understand how contamination levels are distributed and prepares us to estimate contamination in different product sizes.
Statistical Calculations
Statistical calculations are at the heart of probability theory and contamination analysis, and they enable us to quantify uncertainties effectively. In the context of Poisson distribution, these calculations help determine the probability of a specific number of events (contaminants, in this case) over a specified interval.
One of the calculations made is to determine the chances of finding one or more contaminants in multiple smaller bars of chocolate. Say, you have seven 28.35-gram bars; the mean contamination per bar is reduced to 1.8. The calculation uses probabilities from Poisson's formula to gauge the likelihood of these contaminants being present. It involves deriving the probability for no contaminants and adapting it over multiple attempts (bars consumed).
Through these statistical calculations, which involve multiplying probabilities and totaling them in various ways, we unveil insights into product safety and the effectiveness of contamination controls, thereby making informed decisions about product quality and consumer safety.

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