/*! 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 69 The report "Twitter in Higher Ed... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The report "Twitter in Higher Education: Usage Habits and Trends of Today's College Faculty" (Magna Publications, September 2009) describes results of a survey of nearly 2000 college faculty. The report indicates the following: \- \(30.7 \%\) reported that they use Twitter and \(69.3 \%\) said that they did not use Twitter. \- Of those who use Twitter, \(39.9 \%\) said they sometimes use Twitter to communicate with students. \- Of those who use Twitter, \(27.5 \%\) said that they sometimes use Twitter as a learning tool in the classroom. Consider the chance experiment that selects one of the study participants at random and define the following events: \(T=\) event that selected faculty member uses Twitter \(C=\) event that selected faculty member sometimes uses Twitter to communicate with students \(L=\) event that selected faculty member sometimes uses Twitter as a learning tool in the classroom a. Use the given information to determine the following probabilities: i. \(\quad P(T)\) ii. \(P\left(T^{C}\right)\) iii. \(P(C \mid T)\) iv. \(P(L \mid T)\) v. \(P(C \cap T)\) b. Interpret each of the probabilities computed in Part (a). c. What proportion of the faculty surveyed sometimes use Twitter to communicate with students? [Hint: Use the law of total probability to find \(P(C) .]\) d. What proportion of faculty surveyed sometimes use Twitter as a learning tool in the classroom?

Short Answer

Expert verified
The requested probabilities are as follows: i. \(P(T) = 0.307\), ii. \(P(T^{C}) = 0.693\), iii. \(P(C|T) = 0.399\), iv. \(P(L|T) = 0.275\), v. \(P(C \cap T) = 0.122493\). The proportion of faculty surveyed sometimes uses twitter to communicate with students is \(0.122493\) (12.2493%)

Step by step solution

01

Calculate probabilities of individual events

Using the information from the survey, you can determine that: \n- \(P(T) = 0.307\) (This is given by the percentage of faculty that use Twitter), - \(P(T^{C}) = 1 - P(T) = 0.693\) (This is calculated based on the fact that either a person uses Twitter or does not)
02

Calculate conditional probabilities

Also given are: - \(P(C|T) = 0.399\) (This is the percentage of Twitter users who use it to communicate with students), - \(P(L|T) = 0.275\) (This is the percentage of Twitter users who use it as a learning tool)
03

Calculate the probability of the intersection of events

The intersection of two events (in this case, Twitter usage and communication with students) is given by: \(P(C \cap T) = P(T) \cdot P(C|T) = 0.307 \cdot 0.399 = 0.122493\) (rounded to six decimal places)
04

Compute probabilities using law of total probability

The law of total probability can be used to find the probability of an event by considering all the ways that it could happen:- \(P(C) = P(T) \cdot P(C|T) + P(T^C) \cdot P(C|T^C)\). Because we don't know \(P(C|T^C)\), we'll assume that if the faculty member does not use twitter, he/she cannot communicate via twitter. Therefore, \(P(C|T^C) = 0\). Then the equation becomes: \(P(C) = 0.307 \cdot 0.399 = 0.122493\).

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 Theory
Probability theory is a branch of mathematics focused on analyzing random events. The fundamental idea is to quantify the likelihood of different outcomes—in this case, the behaviors of college faculty regarding Twitter use. Understanding probability helps us to make informed predictions or decisions when faced with uncertainty.

Key terms include the probability of an event, which is a measure from 0 to 1 indicating how likely the event is to occur, with 0 being impossible and 1 being certain. In our exercise, the event 'T', which signifies using Twitter, has a probability of 0.307, meaning it is a likely but not guaranteed event for a randomly selected faculty member.

Another vital term is the sample space, representing all possible outcomes. Here, the sample space includes all faculty members surveyed, dividing them into users and non-users of Twitter. By analyzing these probabilities through practical examples, students can develop a keen understanding of how theoretical probability applies to real-world scenarios.
Conditional Probability
Conditional probability is the probability of an event occurring, given that another event has already occurred. Represented as (P(A|B)), it reads as 'the probability of event A given event B.'

For instance, the exercise asks for the probability that a faculty member uses Twitter to communicate with students, given that they already use Twitter, denoted as (P(C|T)). This is given by 0.399, which means almost 40% of Twitter-using faculty communicate with students through the platform. Grasping conditional probability is crucial for interpreting data where certain conditions or criteria are focused on.

Calculating Conditional Probability

In calculations, the formula often used is: (P(A|B) = \[\frac{P(A \bigcap B)}{P(B)}\]). In terms of the survey, we think of the probability of two events—using Twitter and communicating with students—both happening, and then we divide by the probability of the given condition (using Twitter).
Law of Total Probability
The law of total probability allows us to calculate the probability of an event based on the probabilities of a set of related events. It's particularly effective when dealing with conditional probabilities. In the presented exercise, it's used to determine the proportion of faculty who use Twitter for communication, considering both users and non-users of Twitter.

To find the overall probability (P(C)) of a faculty member using Twitter to communicate, we sum the probability of communicating while being a Twitter user (P(C \bigcap T)) and while not being a Twitter user (P(C \bigcap T^C)). Since faculty cannot communicate via Twitter if they don't use it, (P(C|T^C)) is 0, and the equation simplifies. Applying the law of total probability simplifies complex problems into manageable parts and is a powerful tool for comprehensive statistical analysis.

Application of the Law

The formula looks like this: (P(C) = P(T) \bigcdot \[\frac{P(C \bigcap T)}{P(T)}\] + P(T^C) \bigcdot \[\frac{P(C \bigcap T^C)}{P(T^C)}\]). Since the second part of the equation is zero, we only consider the first part to find the overall probability of faculty using Twitter for student communication.

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

A family consisting of three people- \(\mathrm{P}_{1}, \mathrm{P}_{2}\), and \(\mathrm{P}_{3}\) -belongs to a medical clinic that always has a physician at each of stations 1,2, and \(3 .\) During a certain week, each member of the family visits the clinic exactly once and is randomly assigned to a station. One experimental outcome is \((1,2,1)\), which means that \(\mathrm{P}_{1}\) is assigned to station \(1, \mathrm{P}_{2}\) to station 2, and \(\mathrm{P}_{3}\) to station \(1 .\) a. List the 27 possible outcomes. (Hint: First list the nine outcomes in which \(\mathrm{P}_{1}\) goes to station 1, then the nine in which \(\mathrm{P}_{1}\) goes to station 2, and finally the nine in which \(\mathrm{P}_{1}\) goes to station 3 ; a tree diagram might help.) b. List all outcomes in the event \(A\), that all three people go to the same station. c. List all outcomes in the event \(B\), that all three people go to different stations. d. List all outcomes in the event \(C\), that no one goes to station 2 . e. Identify outcomes in each of the following events: \(B^{C}, C^{C}, A \cup B, A \cap B, A \cap C\).

Consider a Venn diagram picturing two events \(A\) and \(B\) that are not disjoint. a. Shade the event \((A \cup B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cap B^{C} .\) How are these two events related? b. Shade the event \((A \cap B)^{C} .\) On a separate Venn diagram shade the event \(A^{C} \cup B^{C} .\) How are these two events related? (Note: These two relationships together are called DeMorgan's laws.)

The paper "Good for Women, Good for Men, Bad for People: Simpson's Paradox and the Importance of Sex-Spedfic Analysis in Observational Studies" (Journal of Women's Health and Gender-Based Medicine [2001]: \(867-872\) ) described the results of a medical study in which one treatment was shown to be better for men and better for women than a competing treatment. However, if the data for men and women are combined, it appears as though the competing treatment is better. To see how this can happen, consider the accompanying data tables constructed from information in the paper. Subjects in the study were given either Treatment \(\mathrm{A}\) or Treatment \(\mathrm{B}\), and survival was noted. Let \(S\) be the event that a patient selected at random survives, \(A\) be the event that a patient selected at random received Treatment \(\mathrm{A}\), and \(B\) be the event that a patient selected at random received Treatment \(\mathrm{B}\). a. The following table summarizes data for men and women combined: $$ \begin{array}{l|ccc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 215 & 85 & \mathbf{3 0 0} \\ \text { Treatment B } & 241 & 59 & \mathbf{3 0 0} \\ \text { Total } & \mathbf{4 5 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? b. Now consider the summary data for the men who participated in the study: $$ \begin{array}{l|rrr} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 120 & 80 & \mathbf{2 0 0} \\ \text { Treatment B } & 20 & 20 & 40 \\ \text { Total } & \mathbf{1 4 0} & \mathbf{1 0 0} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? c. Now consider the summary data for the women who participated in the study: $$ \begin{array}{l|rrc} & \text { Survived } & \text { Died } & \text { Total } \\ \hline \text { Treatment A } & 95 & 5 & \mathbf{1 0 0} \\ \text { Treatment B } & 221 & 39 & \mathbf{2 6 0} \\ \text { Total } & \mathbf{3 1 6} & \mathbf{1 4 4} & \\ \hline \end{array} $$ i. Find \(P(S)\). ii. Find \(P(S \mid A)\). iii. Find \(P(S \mid B)\). iv. Which treatment appears to be better? d. You should have noticed from Parts (b) and (c) that for both men and women, Treatment \(A\) appears to be better. But in Part (a), when the data for men and women are combined, it looks like Treatment \(\mathrm{B}\) is better. This is an example of what is called Simpson's paradox. Write a brief explanation of why this apparent inconsistency occurs for this data set. (Hint: Do men and women respond similarly to the two treatments?)

A new model of laptop computer can be ordered with one of three screen sizes ( 10 inches, 12 inches, 15 inches) and one of four hard drive sizes \((50 \mathrm{~GB}, 100 \mathrm{~GB}\), \(150 \mathrm{~GB}\), and \(200 \mathrm{~GB}\) ). Consider the chance experiment in which a laptop order is selected and the screen size and hard drive size are recorded. a. Display possible outcomes using a tree diagram. b. Let \(A\) be the event that the order is for a laptop with a screen size of 12 inches or smaller. Let \(B\) be the event that the order is for a laptop with a hard drive size of at most \(100 \mathrm{~GB}\). What outcomes are in \(A^{C}\) ? In \(A \cup B ?\) In \(A \cap B\) ? c. Let \(C\) denote the event that the order is for a laptop with a \(200 \mathrm{~GB}\) hard drive. Are \(A\) and \(C\) disjoint events? Are \(B\) and \(C\) disjoint?

Two individuals, \(A\) and \(B\), are finalists for a chess championship. They will play a sequence of games, each of which can result in a win for \(A\), a win for \(\mathrm{B}\), or a draw. Suppose that the outcomes of successive games are independent, with \(P(\) A wins game \()=.3\), \(P(\) B wins game \()=.2\), and \(P(\) draw \()=.5\). Each time a player wins a game, he earns 1 point and his opponent earns no points. The first player to win 5 points wins the championship. For the sake of simplicity, assume that the championship will end in a draw if both players obtain 5 points at the same time. a. What is the probability that \(A\) wins the championship in just five games? b. What is the probability that it takes just five games to obtain a champion? c. If a draw earns a half-point for each player, describe how you would perform a simulation to estimate \(\mathrm{P}(\mathrm{A}\) wins the championship). d. If neither player earns any points from a draw, would the simulation in Part (c) take longer to perform? Explain your reasoning.

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.