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Chapter 4: Problem 43

An experiment is designed to test the potency of a drug on 20 rats. Previous animal studies have shown that a \(10-\mathrm{mg}\) dose of the drug is lethal \(5 \%\) of the time within the first 4 hours; of the animals alive at 4 hours, \(10 \%\) will die in the next 4 hours. Suppose 2 rats die in the first 4 hours. What is the probability that 2 or fewer rats will die in the next 4 hours?

Short Answer

Expert verified
The probability is approximately 0.749.

Step by step solution

01

Understand the Problem

We have an initial group of 20 rats and know the probabilities of death during two 4-hour periods. In the first 4-hour period, 5% of the rats are expected to die. In the next phase, for those alive, 10% may die in the following 4 hours. We need to calculate the probability that 2 or fewer of the remaining rats die in the next 4 hours.
02

Calculate Remaining Rats after 4 Hours

Since 5% of the 20 rats are expected to die in the first 4 hours, we calculate: \(0.05 \times 20 = 1\) rat. However, it's given that 2 rats die in this period, so there are: \(20 - 2 = 18\) rats remaining after 4 hours.
03

Set Up Binomial Probability Distribution

For the remaining 18 rats, the probability of a single rat dying in the next 4 hours is 0.10. We are interested in finding the probability that 2 or fewer rats die. Thus, use the binomial distribution formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( n = 18 \), \( k \) ranges from 0 to 2, and \( p = 0.10 \).
04

Find Probability for Exact Outcomes

Calculate \( P(X=0) \), \( P(X=1) \), and \( P(X=2) \) using the binomial formula. - \( P(X=0) = \binom{18}{0} (0.10)^0 (0.90)^{18} \)- \( P(X=1) = \binom{18}{1} (0.10)^1 (0.90)^{17} \)- \( P(X=2) = \binom{18}{2} (0.10)^2 (0.90)^{16} \)
05

Calculation and Total Probability

Compute the individual probabilities:- \( P(X=0) \approx 0.150 \)- \( P(X=1) \approx 0.300 \)- \( P(X=2) \approx 0.299 \)Add these probabilities to find the total probability that 2 or fewer rats die: \[ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) \approx 0.150 + 0.300 + 0.299 = 0.749 \]
06

Conclusion

The probability that 2 or fewer rats will die in the next 4 hours is approximately \(0.749\).

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Key Concepts

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

Probability Distribution
Probability distributions are essential in biostatistics to predict the likelihood of different outcomes in an experiment. A probability distribution assigns a probability to each possible outcome of a random variable. The most common types of probability distributions are the normal distribution, binomial distribution, and Poisson distribution. Each type is used based on the characteristics of the data being analyzed. In the context of the drug potency experiment, we focus on the binomial distribution. It is used because we are dealing with a fixed number of trials (the 18 remaining rats), two possible outcomes for each trial (a rat either dies or it doesn't), and a constant probability of success (death) in each trial. Such conditions are ideal for binomial distribution analysis. Understanding how these distributions work helps us make informed predictions about the experimental outcomes, aiding in statistical analysis.
Binomial Distribution
The binomial distribution is a type of probability distribution used in statistics when a fixed number of trials are conducted, and each trial has only two possible outcomes. In this drug experiment, each rat either succumbs to the drug's side effects or survives, making the binomial distribution appropriate.The formula for the binomial probability is:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where:
  • \( n \) is the number of trials \( (18 \text{ rats })\)
  • \( k \) is the number of successful outcomes (deaths) we are interested in
  • \( p \) is the probability of success on an individual trial (0.10)
In this problem, we were tasked to find the probability that 0, 1, or 2 rats die. To solve this, we calculated the probability for each scenario and summed them up. It's crucial because it helps predict how many rats might die within the specified time frame. Binomial distribution gives a clear, mathematical structure to predict outcomes in experiments similar to this one.
Experimental Design
Experimental design is a fundamental aspect of conducting scientific research. It lays the groundwork for collecting valid, reliable data and drawing meaningful conclusions. In biostatistics, this involves choosing the right sample size, structuring the study, and ensuring it can accurately test hypotheses. For the drug potency experiment, the design involves:
  • Defining the number of subjects (20 rats) to ensure statistically significant results.
  • Establishing time intervals for observation (two periods of 4 hours each).
  • Calculating probabilities for outcomes within these intervals to measure the drug's effect.
Designing an experiment is a meticulous process that ensures results are valid and applicable. This experiment exemplifies a basic experimental design but emphasizes key principles: control over experimental variables, specified measurement criteria, and clear objectives for measurement. By structuring experiments well, researchers can confidently apply statistical tools, like those used for analyzing probability distributions, and derive credible conclusions.

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Most popular questions from this chapter

Suppose the number of admissions to the emergency room at a small hospital follows a Poisson distribution but the incidence rate changes on different days of the week. On a weekday there are on average two admissions per day, while on a weekend day there is on average one admission per day. What is the probability of at least one admission on a Wednesday?

One interesting phenomenon of bacteriuria is that there is a "turnover"; that is, if bacteriuria is measured on the same woman at two different points in time, the results are not necessarily the same. Assume that \(20 \%\) of all women who are bacteriuric at time 0 are again bacteriuric at time 1 (1 year later), whereas only 4.2\% of women who were not bacteriuric at time 0 are bacteriuric at time 1 . Let \(X\) be the random variable representing the number of bacteriuric events over the two time periods for 1 woman and still assume that the probability that a woman will be positive for bacteriuria at any one exam is \(5 \%\) What is the variance of \(X ?\)

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One interesting phenomenon of bacteriuria is that there is a "turnover"; that is, if bacteriuria is measured on the same woman at two different points in time, the results are not necessarily the same. Assume that \(20 \%\) of all women who are bacteriuric at time 0 are again bacteriuric at time 1 (1 year later), whereas only 4.2\% of women who were not bacteriuric at time 0 are bacteriuric at time 1 . Let \(X\) be the random variable representing the number of bacteriuric events over the two time periods for 1 woman and still assume that the probability that a woman will be positive for bacteriuria at any one exam is \(5 \%\) What is the mean of \(X ?\)
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