/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 154 Suppose that a healthcare provid... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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) 0.253, (b) 0.302, (c) 0.445, (d) Mean: 2, Variance: 1.731.

Step by step solution

01

Define Variables and Model

Let \(X\) be the random variable representing the number of patients in the sample who fail to adhere. We have a hypergeometric distribution because the sampling is without replacement. The population size is \(N = 500\), the number of successes in the population is \(K = 500 \times 0.1 = 50\), and the sample size is \(n = 20\).
02

Calculate Probability (a)

For part (a), we need to find \(P(X = 2)\) because if \(10\%\) of 20 patients fail, that's 2 patients. Using the hypergeometric probability formula:\[P(X = 2) = \frac{\binom{50}{2} \binom{450}{18}}{\binom{500}{20}}\]Calculate the values:- \(\binom{50}{2} = 1225\)- \(\binom{450}{18} = 558,685,348,788,861\)- \(\binom{500}{20} = 15,511,210,043,586\)Putting it all together:\[P(X=2) = \frac{1225 \times 558,685,348,788,861}{15,511,210,043,586} \approx 0.253\]
03

Calculate Probability (b)

For part (b), we need to find \(P(X < 2)\) which is \(P(X = 0) + P(X = 1)\). Calculate each:- \(P(X = 0) = \frac{\binom{50}{0} \binom{450}{20}}{\binom{500}{20}} = \frac{1 \times 1,013,243,319,989,054}{15,511,210,043,586} \approx 0.065\)- \(P(X = 1) = \frac{\binom{50}{1} \binom{450}{19}}{\binom{500}{20}} = \frac{50 \times 5,947,572,545,998}{15,511,210,043,586} \approx 0.237\)Then add them:\[P(X < 2) \approx 0.065 + 0.237 = 0.302\]
04

Calculate Probability (c)

For part (c), we have \(P(X > 2) = 1 - P(X \leq 2)\). We already have:- \(P(X = 2) \approx 0.253\)- \(P(X < 2) \approx 0.302\)So \(P(X \leq 2) = P(X < 2) + P(X = 2) \approx 0.302 + 0.253 = 0.555\).Then:\[P(X > 2) = 1 - 0.555 = 0.445\]
05

Calculate Mean and Variance

For a hypergeometric distribution, the mean \(\mu = n \left( \frac{K}{N} \right)\) and the variance \(\sigma^2 = n \left( \frac{K}{N} \right) \left( 1 - \frac{K}{N} \right) \frac{N-n}{N-1}\).\[\mu = 20 \times \frac{50}{500} = 2\]\[\sigma^2 = 20 \times \frac{50}{500} \times \frac{450}{500} \times \frac{480}{499}\]Plug the values in:\[\sigma^2 = 20 \times 0.1 \times 0.9 \times 0.962 \approx 1.7306\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Calculation
When dealing with probability calculations, especially in scenarios where you're sampling without replacement like in our case with the patients, hypergeometric distribution is your friend. This distribution helps when determining the likelihood of a certain outcome from a finite population.

Let's focus on understanding what happens when we need to calculate probabilities using this distribution. For instance, to find the probability that exactly 10% of the patients fail to adhere, calculate for exactly 2 patients in the sample, knowing 10% of the population of 500 (i.e., 50 patients) fail to adhere.

We use the hypergeometric probability formula \[ P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} \] where:
  • \(K\) is the number of successes in the population (patients not adhering),
  • \(N\) is the population size,
  • \(n\) is the sample size,
  • \(k\) is the number of successes in the sample.
Plug these into the formula, compute the binomial coefficients, and you'll find the desired probabilities. This approach translates into steps for calculating for fewer than or more than a given threshold.
Mean and Variance
Understanding the mean and variance in a hypergeometric distribution is crucial because it gives insight into the expected outcome and the spread of our sample results.

In the context of our problem, the mean describes the average number of patients expected to fail adherence to their medication schedule. For a hypergeometric distribution, it is given by \[ \mu = n \left( \frac{K}{N} \right) \] where
  • \(n\) is the sample size,
  • \(K\) is the number of successes in the population,
  • \(N\) is the population size.
In this case, the mean turns out to be 2, reflecting that, on average, you'd expect 2 of the 20 sampled patients to not adhere.

Variance in a hypergeometric distribution measures this spread, telling us how much the actual number of non-adhering patients might deviate from this mean. It is calculated as:\[ \sigma^2 = n \left( \frac{K}{N} \right) \left( 1 - \frac{K}{N} \right) \frac{N-n}{N-1} \]Plug in the values for our scenario to get the variance \( \approx 1.7306 \), giving insights into the variability of the potential outcomes.
Random Sampling
Random sampling without replacement is the process used in our problem. It’s a method where each sample drawn is not returned to the population before the next draw, affecting the outcome probabilities.

Imagine you have a jar with 500 marbles, and you want to pick a few without putting them back. Every time you draw a marble, you decrease the total count, which influences the odds of drawing a specific marble on your next turn.

Similarly, when the healthcare provider is selecting patients, once a patient is selected, they aren't returned, influencing subsequent selections.
  • This method keeps changing the probability of subsequent selections, adding to the complexity which the hypergeometric distribution helps manage.
  • The key here is that with each patient selected, the choices remaining change the overall distribution and thus need a special probabilistic treatment.
Random sampling helps ensure that the sample you choose genuinely represents the larger group, thus ensuring accurate adherence analysis results.
Adherence Analysis
Adherence analysis refers to studying how well patients stick to their prescribed medication routines. It's vital because understanding patterns of adherence helps make informed decisions about patient care and resource allocations.

In this exercise, the healthcare provider examines a random sample to gauge the broader adherence behavior in the patient population.

By calculating probabilities for various scenarios (exactly a certain percentage, fewer than expected, or more), healthcare providers gain valuable insights into potential adherence challenges.
  • They can interpret this data as early warning signals and identify groups at risk of poor adherence.
  • The understanding gathered can be used to develop strategies to improve adherence rates across the population, such as personalized reminders or follow-ups.
  • Ultimately, good adherence ensures effective treatment outcomes and optimized health resources.
So, adherence analysis directly feeds into better healthcare management and improved patient outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Messages arrive to a computer server according to a Poisson distribution with a mean rate of 10 per hour. Determine the length of an interval of time such that the probability that no messages arrive during this interval is 0.90 .

A shipment of chemicals arrives in 15 totes. Three of the totes are selected at random without replacement for an inspection of purity. If two of the totes do not conform to purity requirements, what is the probability that at least one of the nonconforming totes is selected in the sample?

Assume that the number of errors along a magnetic recording surface is a Poisson random variable with a mean of one error every \(10^{5}\) bits. A sector of data consists of 4096 eight-bit bytes. (a) What is the probability of more than one error in a sector? (b) What is the mean number of sectors until an error occurs?

Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting, that is, of not obtaining a busy signal. Assume that your calls are independent. (a) What is the probability that your first call that connects is your 10 th call? (b) What is the probability that it requires more than five calls for you to connect? (c) What is the mean number of calls needed to connect?

When a computer disk manufacturer tests a disk, it writes to the disk and then tests it using a certifier. The certifier counts the number of missing pulses or errors. The number of errors on a test area on a disk has a Poisson distribution with \(\lambda=0.2\). (a) What is the expected number of errors per test area? (b) What percentage of test areas have two or fewer errors?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.