Question

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, decreased 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?

Short answer

Answer: The probability of observing three or fewer infected birds is approximately 0.0005 when the diet supplement is ineffective and approximately 0.5663 when the diet supplement is effective.

Step-by-step solution

Identify the parameters for the binomial distribution

In this case, the infection rate remains unchanged at 30% (\(p = 0.3\)). We have 25 trials (birds examined), so \(n = 25\). We are asked to find the probability of observing three or fewer infected birds, which means \(x = 0, 1, 2, 3\).

Calculate the probability of observing three or fewer infected birds

To find the probability, we can use the following formula for the binomial distribution: \({P(x)} = \binom{n}{x} p^x(1-p)^{(n-x)}\). In our case, we need to calculate \(P(x \leq 3)\), which means summing the probabilities for \(x = 0, 1, 2, 3\). Thus, \(P(x \leq 3) = \sum_{x=0}^{3}{\binom{25}{x} 0.3^x 0.7^{(25-x)}}\). To calculate this sum, use: \({P(x = 0)} = \binom{25}{0} 0.3^0 0.7^{(25-0)}\) \({P(x = 1)} = \binom{25}{1} 0.3^1 0.7^{(25-1)}\) \({P(x = 2)} = \binom{25}{2} 0.3^2 0.7^{(25-2)}\) \({P(x = 3)} = \binom{25}{3} 0.3^3 0.7^{(25-3)}\) Finally, add the probabilities: \(P(x \leq 3) = P(x=0) + P(x=1) + P(x=2) + P(x=3)\). After calculating all probabilities, you will get the result: \(P(x\leq3) \approx 0.0005\) #b. Probability of observing three or fewer infected birds if the supplement is effective#

Identify the parameters for the binomial distribution

In this case, the infection rate is reduced to 10% (\(p = 0.1\)), and we again have 25 trials (birds examined): \(n=25\). We are asked to find the probability of observing three or fewer infected birds, which means \(x=0,1,2,3\).

Calculate the probability of observing three or fewer infected birds

Using the same binomial distribution formula, we now calculate \(P(x \leq 3)\) with the new infection rate: \(P(x \leq 3) = \sum_{x=0}^{3}{\binom{25}{x} 0.1^x 0.9^{(25-x)}}\). To calculate this sum, use: \({P(x = 0)} = \binom{25}{0} 0.1^0 0.9^{(25-0)}\) \({P(x = 1)} = \binom{25}{1} 0.1^1 0.9^{(25-1)}\) \({P(x = 2)} = \binom{25}{2} 0.1^2 0.9^{(25-2)}\) \({P(x = 3)} = \binom{25}{3} 0.1^3 0.9^{(25-3)}\) Finally, add the probabilities: \(P(x \leq 3) = P(x=0) + P(x=1) + P(x=2) + P(x=3)\). After calculating all probabilities, you will get the result: \(P(x\leq3) \approx 0.5663\) In conclusion, if the diet supplement is ineffective, the probability of observing three or fewer infected birds is approximately 0.0005. If the supplement is effective, with a reduced infection rate of 10%, the probability of observing three or fewer infected birds is approximately 0.5663.

Key concepts

Probability Calculation

Understanding the probability calculation using the binomial distribution is essential in determining the chances of a given outcome. The binomial distribution is particularly useful when you deal with scenarios that have two possible results, often referred to as "success" and "failure." In our context, a "success" is a bird being infected with the parasite.

The probability calculation in a binomial distribution consists of several key steps:

• Identify the probability of a single success. In our exercise, if the supplement is ineffective, this is the initial infection rate of 30% or 0.3.
• Determine the number of trials. Here, that is the 25 birds being examined.
• Define the number of successful outcomes. We are considering three or fewer birds still being infected, so our solution includes calculating probabilities for 0, 1, 2, and 3 infected birds.
• Calculate each probability using the binomial formula: \[P(x) = \binom{n}{x} p^x (1-p)^{(n-x)}\]In this formula, \(\binom{n}{x}\) is the number of combinations of \(n\) items taken \(x\) at a time.
• Finally, sum these probabilities to find the overall probability of observing up to three infected birds.

By following these steps, we can calculate the probability of observing a certain number of infected birds based on whether the supplement is effective or not.

Infection Rate

The infection rate is a crucial parameter in assessing the success of interventions, such as a diet supplement in this case. The rate essentially reflects the proportion of birds that are infected out of the total number studied.

In the exercise, we deal with two different infection rates:

• Initial Infection Rate: Initially, it's reported that about 30% of the birds are infected. This is denoted as \(p = 0.3\) for the probability of a bird being infected before the diet supplement is administered.
• Reduced Infection Rate: If the diet supplement is effective, we expect the infection rate to drop. For part (b) of the problem, it assumes that the supplement reduces the infection rate to 10%, or \(p = 0.1\). This signifies a significant potential impact of the supplement on reducing infection levels.

Understanding these rates helps in evaluating the product's efficacy and in making informed decisions in healthcare and agricultural practices.

Diet Supplement Efficacy

Determining the efficacy of the diet supplement hinges on its ability to reduce the infection rate among the birds. This is important both for improving the birds' health and for ensuring their growth rates are not adversely affected by the parasite.

Evaluation involves comparing two scenarios:

• Supplement Ineffective: If no reduction in infection rate is observed, we continue with the initial 30% rate. The low probability (approximately 0.0005) of observing three or fewer infected birds serves as an indicator of the supplement's ineffectiveness.
• Supplement Effective: The expected reduction in infection rate to 10% suggests a high probability (around 0.5663) of three or fewer birds still being infected after using the supplement for two weeks. This significant increase in probability highlights the supplement's potential to substantially lower infection levels.

Analyzing these probabilities allows researchers to assess the supplement's impact and make informed decisions about its continued use and development.