/*! 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 8 An article in The Canadian Entom... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 8

An article in The Canadian Entomologist (Harcourt et al., 1977 , Vol. 109 , pp. \(1521-1534\) ) reported on the life of the alfalfa weevil from eggs to adulthood. The following table shows the number of larvae that survived at each stage of development from eggs to adults. $$ \begin{array}{cccccc} & \text { Early } & \text { Late } & \text { Pre- } & \text { Late } & \\ \text { Eggs } & \text { Larvae } & \text { Larvae } & \text { pupae } & \text { Pupae } & \text { Adults } \\ 421 & 412 & 306 & 45 & 35 & 31 \end{array} $$ a. What is the probability an egg survives to adulthood? b. What is the probability of survival to adulthood given survival to the late larvae stage? c. What stage has the lowest probability of survival to the next stage?

Short Answer

Expert verified
a. \( \frac{31}{421} \); b. \( \frac{31}{306} \); c. Late larvae to pre-pupae (\( \frac{45}{306} \)).

Step by step solution

01

Determine Total Eggs and Surviving Adults

The total number of eggs is given as 421, and the number of adults that survived is 31.
02

Calculate Probability of Egg Surviving to Adulthood

The probability an egg survives to adulthood is calculated by dividing the number of adults by the number of eggs: \( P(\text{egg to adult}) = \frac{31}{421} \).
03

Calculate Probability of Adult Given Late Larvae Survival

For this, we focus on individuals that survive to the late larvae stage. There are 306 late larvae, leading to 31 adults, so the probability is \( \frac{31}{306} \).
04

Determine Survival Probabilities Between Stages

Calculate the probability for each stage transition: - Eggs to early larvae: \( \frac{412}{421} \)- Early larvae to late larvae: \( \frac{306}{412} \) - Late larvae to pre-pupae: \( \frac{45}{306} \) - Pre-pupae to late pupae: \( \frac{35}{45} \) - Late pupae to adults: \( \frac{31}{35} \).
05

Identify Stage with Lowest Survival Probability

From the calculations:- Late larvae to pre-pupae has the lowest survival probability of \( \frac{45}{306} \).

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 91影视!

Key Concepts

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

Survival Analysis
Survival analysis is a statistical method used to study the time until an event of interest occurs. In biological studies, this event often corresponds to a specific life stage or an outcome like death. In the case of the alfalfa weevil, survival analysis helps us understand at which stage most larvae fail to progress to the next stage. By analyzing survival probabilities across different life cycle stages, researchers can pinpoint where interventions might be most needed to improve survival rates. These evaluations are essential in ecological and evolutionary biology, offering insights into species' survival mechanisms and challenges.
Life Cycle Stages
The life cycle of the alfalfa weevil consists of several stages: egg, early larvae, late larvae, pre-pupae, late pupae, and adult. Each stage presents unique challenges and vulnerabilities. For instance, environmental conditions, predation, and resource availability can impact survival. Evaluating the number of individuals transitioning from one stage to the next provides valuable data for calculating probabilities of survival. For the alfalfa weevil, understanding how many eggs progress to adults can inform studies on pest population dynamics and control strategies. Recognizing the weakest link in the life cycle, such as the shift from late larvae to pre-pupae, underscores the potential for targeted interventions.
Probability Calculation
Probability calculation in the context of life cycle studies involves determining the likelihood of survival from one stage to another. Consider the probability that an egg will survive to become an adult alfalfa weevil. This is calculated by dividing the number of surviving adults by the total number of eggs not fully developed into adults. Here, the formula is: \( P(\text{egg to adult}) = \frac{31}{421} \), providing a clear view of survival chances. Similarly, the conditional probability that a late larva will reach adulthood, given it survives up to the late larvae stage, is found using: \( \frac{31}{306} \). These calculations are foundational to biostatistics, helping quantify vital rates in biological populations.
Biostatistics
Biostatistics is the application of statistical principles to biological studies, offering a toolkit to address variability and uncertainty. It encompasses a range of methods including regression analysis, hypothesis testing, and probability calculation. In the context of the alfalfa weevil's life stages, biostatistics enables the evaluation of survival rates and identification of the most critical transition points. By applying biostatistical methods, scientists can effectively model life cycle processes and predict outcomes under different scenarios. This approach enhances our understanding of natural phenomena and informs conservation and management practices by highlighting trends in population dynamics and identifying areas for intervention.

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

Samples of skin experiencing desquamation are analyzed for both moisture and melanin content. The results from 100 skin samples are as follows: $$ \begin{array}{lccc} & & \text { Melanin Content } \\ & & \text { High } & \text { Low } \\ \text { Moisture } & \text { High } & 13 & 7 \\ \text { content } & \text { Low } & 48 & 32 \end{array} $$ Let \(A\) denote the event that a sample has low melanin content, and let \(B\) denote the event that a sample has high moisture content. Determine the following probabilities: a. \(P(A)\) b. \(P(B)\) c. \(P(A \mid B)\) d. \(P(B \mid A)\)

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 samples are summarized as follows: $$\begin{array}{lccc} & & {\text { Conforms }} \\ & & \text { Yes } & \text { No } \\ & 1 & 22 & 8 \\\& 2 & 25 & 5 \\\\\text { Supplier } & 2 & 30 & 10\end{array}$$ Let \(A\) denote the event that a sample is from supplier \(1,\) and let \(B\) denote the event that a sample conforms to specifications. Determine the number of samples in \(A^{\prime} \cap B, B^{\prime},\) and \(A \cup B\).

2.4.11 The article "Clinical and Radiographic Outcomes of Four Different Treatment Strategies in Patients with Early Rheumatoid Arthritis" [Arthritis \& Rheumatism (2005, Vol. 52, pp. \(3381-3390\) ) ] considered four treatment groups. The groups consisted of patients with different drug therapies (such as prednisone and infliximab): sequential monotherapy (group 1), step-up combination therapy (group 2), initial combination therapy (group 3), or initial combination therapy with infliximab (group 4). Radiographs of hands and feet were used to evaluate disease progression. The number of patients without progression of joint damage was 76 of 114 patients \((67 \%), 82\) of 112 patients \((73 \%), 104\) of 120 patients \((87 \%),\) and 113 of 121 patients \((93 \%)\) in groups \(1-4\) respectively. Suppose that a patient is selected randomly. Let \(A\) denote the event that the patient is in group \(1,\) and let \(B\) denote the event that there is no progression. Determine the following probabilities: a. \(P(A \cup B)\) b. \(P\left(A^{\prime} \cup B^{\prime}\right)\) c. \(P\left(A \cup B^{\prime}\right)\)

A computer system uses passwords that contain exactly eight characters, and each character is one of the 26 lowercase letters \((a-z)\) or 26 uppercase letters \((A-Z)\) or 10 integers \((0-9)\) Let \(\Omega\) denote the set of all possible passwords. Suppose that all passwords in \(\Omega\) are equally likely. Determine the probability for each of the following: a. Password contains all lowercase letters given that it contains only letters b. Password contains at least 1 uppercase letter given that it contains only letters c. Password contains only even numbers given that it contains all numbers

A credit card contains 16 digits. It also contains the month and year of expiration. Suppose there are 1 million credit card holders with unique card numbers. A hacker randomly selects a 16-digit credit card number. a. What is the probability that it belongs to a user? b. Suppose a hacker has a \(25 \%\) chance of correctly guessing the year your card expires and randomly selects 1 of the 12 months. What is the probability that the hacker correctly selects the month and vear of expiration?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation
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.