/*! 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 152 An article in Mathematical Biosc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An article in Mathematical Biosciences ["Influence of Delayed Viral Production on Viral Dynamics in HIV-1 Infected Patients" \((1998,\) Vol. \(152(2),\) pp. \(143-163)]\) considered the time delay between the initial infection by immunodeficiency virus type 1 (HIV-1) and the formation of productively infected cells. In the simulation model, the time delay is approximated by a gamma distribution with parameters \(r=4\) and \(1 / \lambda=0.25\) days. Determine the following: (a) Mean and variance of time delay (b) Probability that a time delay is more than half a day (c) Probability that a time delay is between one-half and one day

Short Answer

Expert verified
(a) Mean: 1 day, Variance: 0.25 days²; (b) Calculate using gamma CDF; (c) Calculate using gamma CDF.

Step by step solution

01

Understand the Gamma Distribution

The gamma distribution is defined by two parameters: the shape parameter \(r\) and the rate parameter \(\lambda\). The article provides these as \(r=4\) and \(\frac{1}{\lambda} = 0.25\) days. Thus, the rate \(\lambda\) is \(\frac{1}{0.25} = 4\).
02

Calculate Mean and Variance

For the gamma distribution, the mean \(\mu\) is given by \(\frac{r}{\lambda}\) and the variance \(\sigma^2\) is given by \(\frac{r}{\lambda^2}\). Using the given parameters: - Mean \(\mu = \frac{4}{4} = 1\) day - Variance \(\sigma^2 = \frac{4}{4^2} = \frac{4}{16} = 0.25\) days².
03

Calculate Probability for More than Half a Day

For the gamma distribution, the cumulative distribution function (CDF) can be used. We need \(P(X > 0.5) = 1 - P(X \leq 0.5)\). Using a standard gamma distribution table or calculator for \(P(X \leq 0.5)\), compute this probability, given a gamma distribution with \(r=4\) and \(\lambda=4\).
04

Calculate Probability Between Half and One Day

We want \(P(0.5 < X < 1) = P(X < 1) - P(X \leq 0.5)\). Using a gamma distribution table or software, find \(P(X < 1)\) and use the previously computed \(P(X \leq 0.5)\). Subtract to obtain the desired probability.

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 working with the Gamma Distribution, calculating probabilities involves understanding its cumulative nature. Specifically, we utilize the cumulative distribution function (CDF), which helps us determine the probability that a random variable is less than or equal to a certain value. In the given problem, we are interested in the probabilities concerning specific time delays modeled by the Gamma Distribution.

### Calculating Specific ProbabilitiesTo find the probability of a time delay being more than half a day, we use:
  • 1 minus the CDF for 0.5 days. This probability is expressed as \(P(X > 0.5) = 1 - P(X \leq 0.5)\).
For something more complex, like the probability of a time delay between half a day and one day, it's calculated as:
  • The CDF for 1 day minus the CDF for 0.5 days: \(P(0.5 < X < 1) = P(X < 1) - P(X \leq 0.5)\).
Such calculations might require the use of statistical software or a standard gamma distribution table as these can't be easily solved by hand for non-integer values.

Understanding the CDF in the Gamma Distribution context allows us to evaluate these probabilities effectively, ensuring we can describe the likelihood of specific events in a real-world scenario.
Mean and Variance
Mean and variance are core concepts when working with any probability distribution, including the gamma distribution. They provide crucial insights into the distribution's behavior, such as its central tendency and the spread of possible values. For our exercise, parameters play an essential role in determining these characteristics.

### Calculating the MeanFor the gamma distribution, the mean \( \mu \) is derived from its parameters by the formula:
  • \( \mu = \frac{r}{\lambda} \)
Given the problem's parameters \(r=4\) and \(\lambda=4\), the mean calculates as:
  • \( \mu = \frac{4}{4} = 1 \) day.
This tells us that on average, the time delay is expected to be 1 day.

### Calculating the VarianceVariance \( \sigma^2 \) is another metric providing insights into the spread of the distribution. It's calculated by:
  • \( \sigma^2 = \frac{r}{\lambda^2} \)
Plugging in the values, we get:
  • \( \sigma^2 = \frac{4}{4^2} = \frac{4}{16} = 0.25 \) days².
This shows that while the average time is 1 day, there's a moderate spread of values around this mean.
Cumulative Distribution Function
The cumulative distribution function (CDF) is an essential tool when working with the gamma distribution to understand the cumulative probability up to a certain threshold. It helps us compute the probability of a variable being less than or equal to a specific value.

### What is the CDF?The CDF for a random variable \(X\) gives us the probability that \(X\) will take a value less than or equal to \(x\). Mathematically, it is defined as:
  • \( F(x) = P(X \leq x) \)
For our problem involving gamma distribution, the CDF helps in computing probabilities over different intervals and is especially useful for handling ranges not covered directly in discrete distributions.

### Utilizing the CDF in the Gamma DistributionFor this particular type of distribution, calculating probabilities involves using the CDF to find the difference between two probabilities:\ \( P(a < X < b) = F(b) - F(a) \). For example, in our exercise, to find the probability between 0.5 and 1 day, we compute:\ \( P(0.5 < X < 1) = F(1) - F(0.5) \).

This process often requires statistical tables or computational methods, as exact mathematical expressions for the CDF in case of the gamma distribution for non-standard values need complex integration.

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

The life of a semiconductor laser at a constant power is normally distributed with a mean of 7000 hours and a standard deviation of 600 hours. (a) What is the probability that a laser fails before 5800 hours? (b) What is the life in hours that \(90 \%\) of the lasers exceed? (c) What should the mean life equal for \(99 \%\) of the lasers to exceed 10,000 hours before failure? (d) A product contains three lasers, and the product fails if any of the lasers fails. Assume that the lasers fail independently. What should the mean life equal for \(99 \%\) of the products to exceed 10,000 hours before failure?

The life (in hours) of a computer processing unit (CPU) is modeled by a Weibull distribution with parameters \(\beta=3\) and \(\delta=900\) hours. Determine (a) and (b): (a) Mean life of the CPU. (b) Variance of the life of the CPU. (c) What is the probability that the CPU fails before 500 hours?

An article in Chemosphere ["Statistical Evaluations Reflecting the Skewness in the Distribution of TCDD Levels in Human Adipose Tissue" \((1987,\) Vol. \(16(8),\) pp. \(2135-\) 2140) ] concluded that the levels of 2,3,7,8 -TCDD (colorless persistent environmental contaminants with no distinguishable odor at room temperature) in human adipose tissue has a lognormal distribution (based on empirical evidence from North America). The mean and variance of this lognormal distribution in the USA are 8 and \(21,\) respectively. Let \(X\) denote this lognormal random variable. Determine the following: (a) \(P(2000

Assume that the life of a roller bearing follows a Weibull distribution with parameters \(\beta=2\) and \(\delta=10,000\) hours. (a) Determine the probability that a bearing lasts at least 8000 hours. (b) Determine the mean time until failure of a bearing. (c) If 10 bearings are in use and failures occur independently, what is the probability that all 10 bearings last at least 8000 hours?

The length of an injection-molded plastic case that holds magnetic tape is normally distributed with a length of 90.2 millimeters and a standard deviation of 0.1 millimeter. (a) What is the probability that a part is longer than 90.3 millimeters or shorter than 89.7 millimeters? (b) What should the process mean be set at to obtain the highest number of parts between 89.7 and 90.3 millimeters? (c) If parts that are not between 89.7 and 90.3 millimeters are scrapped, what is the yield for the process mean that you selected in part (b)? Assume that the process is centered so that the mean is 90 millimeters and the standard deviation is 0.1 millimeter. Suppose that 10 cases are measured, and they are assumed to be independent. (d) What is the probability that all 10 cases are between 89.7 and 90.3 millimeters? (e) What is the expected number of the 10 cases that are between 89.7 and 90.3 millimeters?

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.