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Antibiotic Resist ance. According to CDC estimates, at least \(2.8\) million people in the United States are sickened each year with antibiotic-resistant infections, and at least 35,000 die as a result. Antibiotic resistance occurs when disease-causing microbes become resistant to antibiotic drug therapy. Because this resistance is typically genetic and transferred to the next generations of microbes, it is a very serious public health problem. Of the infections considered most serious by the CDC, gonorrhea has an estimated \(1.14\) million new cases occurring annually, and approximately \(50 \%\) of those cases are resistant to any antibiotic. \({ }^{7}\) A public health clinic in California sees eight patients with gonorrhea in a given week. a. What is the distribution of \(X\), the number of these eight cases that are resistant to any antibiotic? b. What are the mean and standard deviation of \(X\) ? c. Find the probability that exactly one of the cases is resistant to any antibiotic. What is the probability that at least one case is resistant to any antibiotic? (Hint: It is easier to first find the probability that exactly zero of the eight cases were resistant.)

Short Answer

Expert verified
a) \(X \sim \text{Binomial}(n=8, p=0.5)\) b) Mean = 4, SD \(\approx 1.41\) c) \(P(X=1) = 0.03125\), \(P(X \geq 1) \approx 0.996\).

Step by step solution

01

Understanding Distribution Type

The exercise involves a scenario where there are two outcomes for each patient: either their infection is resistant to antibiotics or it is not. This can be modeled with a Binomial distribution, where the number of trials is 8 (the number of patients) and the probability of success (resistance) per trial is 0.5.
02

Parameters of the Binomial Distribution

For a Binomial distribution, we need to know the number of trials and the probability of success. Here, we have 8 trials (patients) and the probability of a case being resistant, \(p\), is 0.5 (50%). Thus, \(X \sim \text{Binomial}(n=8, p=0.5)\).
03

Calculating the Mean and Standard Deviation

For a binomial distribution \(X \sim \text{Binomial}(n, p)\), the mean \(\mu\) is calculated as \(\mu = np\) and the standard deviation \(\sigma\) is \(\sigma = \sqrt{np(1-p)}\). Here, \(\mu = 8 \times 0.5 = 4\) and \(\sigma = \sqrt{8 \times 0.5 \times 0.5} = \sqrt{2} \approx 1.41\).
04

Probability of Exactly One Resistant Case

The probability that exactly one case is resistant to antibiotics can be calculated using the binomial probability formula: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\). Substituting \(n = 8\), \(k = 1\), and \(p = 0.5\), we find \(P(X=1) = \binom{8}{1} (0.5)^1 (0.5)^{7} = 8 \times 0.5^8 = 0.03125\).
05

Probability of At Least One Resistant Case

To find the probability of at least one resistant case, it is often simpler to first find the probability of zero resistant cases and subtract this from 1. Using the binomial formula: \(P(X=0) = \binom{8}{0} (0.5)^0 (0.5)^8 = 1 \times 0.5^8 = 0.00390625\). Thus, \(P(X \geq 1) = 1 - P(X=0) = 1 - 0.00390625 = 0.99609375\).

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Key Concepts

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

Antibiotic Resistance
Antibiotic resistance is a growing concern in the realm of public health, making it difficult to treat certain infections effectively. When bacteria or other microbes become resistant, they no longer respond to the drugs designed to kill them, allowing infections to persist and potentially spread. This resistance typically arises through genetic mutations and can be passed on to subsequent generations, complicating treatment options.
For instance, gonorrhea is an infection where resistance is particularly concerning. With 1.14 million new cases annually and around 50% showing resistance to antibiotics, it represents a significant challenge for healthcare providers. Understanding and addressing antibiotic resistance is crucial to controlling the spread of resistant infections.
Probability Calculation
Calculating probabilities is key in understanding how likely a specific event is to occur. In this scenario, we're dealing with a Binomial distribution, which is appropriate because we're considering two possible outcomes for each patient: either their infection is resistant or it isn't.
  • The number of trials (oldsymbol{n}) is 8, one for each patient.
  • The probability of success (oldsymbol{p}), or in this case, resistance, is 0.5 or 50%.
The Binomial distribution formula helps calculate the probability of a fixed number of successes:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Using this formula, we computed the probability of exactly one resistant case as 0.03125 and the probability of at least one resistant case as 0.99609375.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of a Binomial distribution, it offers insights into how much the number of resistant cases might differ from the expected value (mean).
To calculate the standard deviation for a Binomial distribution, we use the formula:
\[ \sigma = \sqrt{np(1-p)} \]For our problem, the standard deviation is approximately 1.41, showing us that while the mean is 4 cases, there is variability indicating how cases could be distributed around this mean.
Public Health Statistics
Public health statistics are vital in understanding and addressing the impact of diseases on populations. They provide a quantitative basis for decision-making and policy development. For antibiotic resistance, such statistics help illuminate the scale and urgency of the issue.
The Centers for Disease Control and Prevention (CDC) highlighted that 2.8 million Americans are affected by antibiotic-resistant infections each year, with 35,000 resulting in death. These numbers stress the critical need for effective public health strategies to reduce the spread of resistant infections and mitigate their impact on communities across the nation.
By using data-driven insights, public health officials can design targeted interventions to curb the rise of antibiotic resistance, ensuring better health outcomes now and into the future.

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