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Suppose that a healthcare provider selects 20 patients randomly (without replacement) from among 500 to evaluate adherence to a medication schedule. Suppose that \(10 \%\) of the 500 patients fail to adhere with the schedule. Determine the following: a. Probability that exactly \(10 \%\) of the patients in the sample fail to adhere. b. Probability that fewer than \(10 \%\) of the patients in the sample fail to adhere. c. Probability that more than \(10 \%\) of the patients in the sample fail to adhere. d. Mean and variance of the number of patients in the sample who fail to adhere.

Short Answer

Expert verified
a. Use hypergeometric formula for exactly 2 non-adhering; b. Calculate for X < 2; c. Compute 1 - P(X ≤ 2); d. Mean = 2, variance from the formula.

Step by step solution

01

Identify Parameters of the Hypergeometric Distribution

In this problem, we have a total of 500 patients (population size, \( N = 500 \)), of which 10% fail to adhere (successes in population, \( K = 0.10 \times 500 = 50 \)). We select a sample of 20 patients (sample size, \( n = 20 \)). We aim to find the probabilities related to the number of non-adhering patients in the sample, which follows a hypergeometric distribution.
02

Probability Exactly 10% Non-Adhering Patients

We need to calculate the probability of exactly 2 non-adhering patients in our sample of 20 (since 10% of 20 is 2). The probability is given by the hypergeometric distribution formula: \[ P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}} {\binom{N}{n}} \]Substitute \( K = 50, N = 500, n = 20 \), and \( k = 2 \):\[ P(X = 2) = \frac{{\binom{50}{2} \binom{450}{18}}}{\binom{500}{20}} \]
03

Probability Fewer Than 10% Non-Adhering Patients

To find the probability of fewer than 2 non-adhering patients, calculate \( P(X < 2) \), which is:\[ P(X < 2) = P(X = 0) + P(X = 1) \]Compute each probability using the hypergeometric distribution formula and sum them:\[ P(X = 0) = \frac{{\binom{50}{0} \binom{450}{20}}}{\binom{500}{20}} \] \[ P(X = 1) = \frac{{\binom{50}{1} \binom{450}{19}}}{\binom{500}{20}} \] Add these results to get \( P(X < 2) \).
04

Probability More Than 10% Non-Adhering Patients

To find the probability of more than 2 non-adhering patients, use:\[ P(X > 2) = 1 - P(X \leq 2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)] \]We have already computed \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) in the previous steps. Subtract their sum from 1 to get \( P(X > 2) \).
05

Calculate Mean and Variance

The mean (\( \mu \)) and variance (\( \sigma^2 \)) of a hypergeometric distribution can be found using:\[ \mu = n \cdot \frac{K}{N} = 20 \cdot \frac{50}{500} = 2 \]\[ \sigma^2 = n \left(\frac{K}{N}\right) \left(1 - \frac{K}{N}\right) \left( \frac{N - n}{N-1} \right) \]Substitute \( n = 20, K = 50, N = 500 \) to find:\[ \sigma^2 = 20 \cdot \frac{50}{500} \cdot \left(1 - \frac{50}{500}\right) \cdot \frac{480}{499} \]

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Key Concepts

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

Probability Calculation
In the realm of probability theory, one intriguing aspect is calculating the likelihood of a specific number of successes in a sample.
When dealing with a population and trying to understand how certain traits are distributed within a smaller sample, the hypergeometric distribution can be quite handy. This is particularly the case when you're selecting items without replacement, such as patients in a medical study.
  • The probability of a given number of successes in a sample is determined based on the total population, the number of successes in the population, the sample size, and the desired number of successes in the sample.
In the provided exercise, the healthcare provider wants to determine how many patients, out of the sample of 20, stick to their medication schedule.
We use the hypergeometric distribution to calculate the probability of finding exactly 0, 1, 2 (and so on) non-adhering patients in the study. The key here is to use the formula \(P(X = k) = \frac{{\binom{K}{k} \binom{N-K}{n-k}}} {\binom{N}{n}}\), where each component signifies a part of our population or sample. You plug in the values for \(K\), \(N\), \(n\), and \(k\), and you get your probability. Understanding this probability helps researchers and statisticians make informed decisions about the likelihood of various outcomes.
Random Sampling
Random sampling is a technique used to ensure that every individual or item in a population has an equal chance of being selected in the sample. This is critical for the integrity of the study. In our exercise, the health care provider randomly chooses 20 patients from a total of 500.
This random selection minimizes bias and helps ensure that the sample is representative of the broader population.
  • Random sampling is quite common in sectors like healthcare, market research, and scientific studies.
  • It allows for more accurate estimates and conclusions about the entire population.
  • The randomness in selection ensures that every possible combination of patients has an equal probability of being chosen.
For our scenario, since sampling is done without replacement, the hypergeometric distribution is applicable. This makes the calculations practical and relevant as it accounts for the diminishing number of elements once a member is chosen.
Mean and Variance
The mean and variance are fundamental concepts in statistics that provide insights into the average outcome and the distribution of data in a sample or population. For a hypergeometric distribution, these values give us a deeper understanding of the likelihood of certain events.
  • The **mean** (\( \mu \)) indicates the expected number of successes in the sample, calculated as \( \mu = n \cdot \frac{K}{N} \).
  • In our exercise, this is 2, meaning on average, 2 patients out of 20 are expected to not adhere to their medication.
Variance measures the spread of the data, showing how much the number of successes might differ from the mean. For our hypergeometric setting, the variance (\( \sigma^2 \)) is given by:\[ \sigma^2 = n \left(\frac{K}{N}\right) \left(1 - \frac{K}{N}\right) \left(\frac{N - n}{N-1}\right)\]Plugging our values from the problem, we gain a measure that helps us understand the diversity of patient adherence across samples. This information is vital when making predictions or decisions based on the data, as it accounts for the inherent variability in the sample.

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