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A preliminary investigation reported that approximately \(30 \%\) of locally grown poultry were infected with an intestinal parasite that, though not harmful to those consuming the poultry, decrease the usual weight growth rates in the birds. A diet supplement believed to be effective against this parasite was added to the bird's food. Twenty-five birds were examined after having the supplement for at least two weeks, and three birds were still found to be infested with the parasite. a. If the diet supplement is ineffective, what is the probability of observing three or fewer birds infected with the intestinal parasite? b. If in fact the diet supplement was effective and reduced the infection rate to \(10 \%,\) what is the probability observing three or fewer infected birds? a. If the diet supplement is ineffective, what is the probability of observing three or fewer birds infected with the intestinal parasite?

Short Answer

Expert verified
Answer: The probability of observing 3 or fewer infected birds is approximately 0.069% when assuming the diet supplement is ineffective and approximately 64.09% when assuming it is effective.

Step by step solution

01

Determine the parameters for the binomial distribution

Firstly, we need to identify the parameters for the binomial distribution. The number of trials, \(n\), is the number of examined birds, which is 25 in this case. For part a, the probability of success (an infected bird), \(p\), is given as 30%. For part b, the probability of success (an infected bird), \(p\), is given as 10%.
02

Find the probability for Part a

In part a, we assume the diet supplement is ineffective, which means the probability of success (an infected bird) remains 30%. The probability of observing 3 or fewer infected birds can be computed using the cumulative binomial probability formula: $$P(X \leq k) = \sum_{x=0}^{k} \binom{n}{x} p^x (1-p)^{n-x}$$ We are asked to find the probability of observing 3 or fewer infected birds (i.e., \(k\) = 3). Plug in the known values and calculate the probability: $$P(X \leq 3) = \sum_{x=0}^{3} \binom{25}{x} (0.3)^x (0.7)^{25-x}$$ Using a calculator for the binomial probability formula, we obtain: $$P(X \leq 3) \approx 0.00069$$ So, the probability of observing three or fewer birds infected with the intestinal parasite, assuming the diet supplement is ineffective, is approximately 0.00069 or 0.069%.
03

Find the probability for Part b

In part b, we assume the diet supplement is effective, and the infection rate is reduced to 10%. Therefore, \(p\) = 0.1. We use the same cumulative binomial probability formula with the new value of \(p\) and find the probability of observing 3 or fewer infected birds: $$P(X \leq 3) = \sum_{x=0}^{3} \binom{25}{x} (0.1)^x (0.9)^{25-x}$$ Using a calculator for the binomial probability formula, we obtain: $$P(X \leq 3) \approx 0.64088$$ So, if the diet supplement is effective and reduces the infection rate to 10%, the probability of observing three or fewer infected birds is approximately 0.64088 or 64.09%.

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Key Concepts

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

Understanding Probability
Probability is essentially the measure of the likelihood that an event will occur, which can be expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In the context of our exercise, when we talk about the chance of a bird being infected by a parasite, we are referring to the probability of that infection occurring. The higher the probability, the more likely the event is to happen. For example, a probability of 0.3 (or 30%) suggests that there is a 30% chance of any given bird being infected.

When calculating probability, it is important to consider all possible outcomes and how they relate to the event in question. This helps in estimating the risks involved and making more informed decisions, whether it be about poultry health, or other real-world scenarios. The precision of the calculation comes from understanding the nature of the event, and utilizing relevant formulas, such as the cumulative binomial probability formula that was used in the textbook solution, to compute the likelihood of various outcomes.
Cumulative Binomial Probability Explained
Cumulative binomial probability is used to calculate the probability that a binomial random variable falls within a certain range. This concept is useful when you need the probability of an event happening up to a certain number of times, rather than an exact number. In the exercise, we are asked to find the probability of observing three or fewer infected birds, which necessitates summing up the probabilities of observing 0, 1, 2, or 3 infected birds.

The formula for cumulative binomial probability, \[P(X \leq k) = \sum_{x=0}^{k} \binom{n}{x} p^x (1-p)^{n-x}\], calculates the sum of probabilities where \(X\) is the number of successes in \(n\) trials, \(p\) is the probability of a single success, and \(k\) is the number of successes we are up to. Importantly, this cumulative approach is what differentiates it from calculating the probability of a single event. Understanding this concept allows students to tackle a wide variety of problems involving risks or rates of occurrences, such as infection rates in a population.
Infection Rate Implications
Infection rate relates to the frequency at which a disease or parasite is found within a particular group, such as our example of poultry infected by an intestinal parasite. This rate significantly influences probability calculations in healthcare, veterinary, and many other fields, as it is a crucial measurement of how widespread an issue is. When we assume an infection rate, like the added supplementary diet said to reduce the infection rate to 10%, we are making a hypothesis about the effectiveness of a treatment or intervention.

Understanding infection rates and how to apply them in calculations is fundamental for creating strategies to combat the spread of diseases. This concept doesn't just hold importance in academic exercises; it is vital for real-world applications in public health and epidemiology. An accurate measure of the infection rate helps in predicting the likelihood of an outbreak and the potential impact of preventive measures.

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Most popular questions from this chapter

Many employers provide workers with sick/personal days as well as vacation days. Among workers who have taken a sick day when they were not sick, \(49 \%\) say that they needed a break! \(^{13}\) Suppose that a random sample of \(n=12\) workers who took a sick day is selected. Rounding \(49 \%\) to \(p=.5\), find the probabilities of the following events. a. What is the probability that more than six workers say that they took a sick day because they needed a break? b. What is the probability that fewer than five of the workers needed a break? c. What is the probability that exactly 10 of the workers took a sick day because they needed a break?

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If a person is given the choice of an integer from 0 to 9 , is it more likely that he or she will choose an integer near the middle of the sequence than one at either end? a. If the integers are equally likely to be chosen, find the probability distribution for \(x\), the number chosen. b. What is the probability that a person will choose a \(4,5,\) or \(6 ?\) c. What is the probability that a person will not choose a \(4,5,\) or \(6 ?\)

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