/*! 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 44 The Associated Press (San Luis O... [FREE SOLUTION] | 91Ó°ÊÓ

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The Associated Press (San Luis Obispo TelegramTribune, August 23,1995 ) reported on the results of mass screening of schoolchildren for tuberculosis (TB). For Santa Clara County, California, the proportion of all tested kindergartners who were found to have TB was .0006. The corresponding proportion for recent immigrants (thought to be a high-risk group) was .0075. Suppose that a Santa Clara County kindergartner is selected at random. Are the events selected student is a recent immigrant and selected student has \(T B\) independent or dependent events? Justify your answer using the given information.

Short Answer

Expert verified
With the given data, it's impossible to definitively determine if the events 'selected student is a recent immigrant' and 'selected student has TB' are independent or dependent.

Step by step solution

01

Formulate the given data

We have two events here:\nEvent A: Selected student is a recent immigrant.\nEvent B: Selected student has TB.\nThe proportion of all tested kindergartners who were found to have TB is 0.0006, we denote it as P(B).\nThe corresponding proportion for recent immigrants (thought to be a high-risk group) was 0.0075, we represent it as P(A and B). Let's assume P(A) denotes the probability that a randomly selected kindergartener is a recent immigrant. Unfortunately, we do not have this data point explicitly stated. Therefore, without this data point, we can't definitively classify the events as independent or dependent.
02

Attempt to classify the event based on provided data

We can tell that if P(A and B) = P(A) * P(B), then the events A and B are independent. However, we do not have the value of P(A), so we can't make this calculation. Additionally, we can't determine whether P(A|B) = P(A), i.e., the probability of event A given event B has occurred equals the probability of event A, because we also lack the values, which are necessary to calculate.
03

Concluding the analysis

Since we don't have all the necessary data points (in this case, the proportion of kindergartners who are recent immigrants or P(A)), we can't definitively determine whether the events are independent or dependent. With the given data, we simply can't make a justified decision.

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Key Concepts

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

Independent Events
In probability, events are considered independent when the occurrence of one event does not affect the occurrence of another. This means that knowledge about one event gives us no information about the other event.
Imagine flipping a coin and rolling a die at the same time. The outcome of the die roll does not impact the outcome of the coin flip. Each event stands alone, adhering to the formula:
  • Two events, A and B, are independent if: \[ P(A \text{ and } B) = P(A) \times P(B) \]
Since both events do not influence each other, their probability is merely the product of their individual probabilities. In the problem regarding tuberculosis (TB) among kindergartners, not having the proportion of recent immigrants among the total population makes it difficult to prove this independence.
Dependent Events
Dependent events are situations where the outcome or occurrence of the first event affects the outcome or occurrence of the second event. Think about drawing cards from a deck: if you draw a card and do not replace it, the probabilities of future draws are altered.
In simpler terms, dependent events are linked; the occurrence of one means there has been a change in the likelihood of the other:
  • Two events, A and B, are dependent if: \[ P(A \text{ and } B) eq P(A) \times P(B) \]
In the TB problem, without knowing the proportion of recent immigrants (P(A)), we can't clearly say if the events "student is a recent immigrant" and "student has TB" are dependent. However, usually, we suspect dependence when there’s a significant increase in probability, like the higher TB risk in immigrant children.
Conditional Probability
Conditional probability is the likelihood of an event happening given that another event has already occurred. This probability is expressed as \( P(A|B) \), meaning the probability of A occurring given B has happened. It allows us to adjust our probability estimates based on new information.
The formula for conditional probability is:
  • \[ P(A|B) = \frac{P(A \text{ and } B)}{P(B)} \]
This adjustment is crucial when events are dependent upon one another. In the TB problem, calculating \( P(A|B) \) would help in understanding the TB risk among immigrants if we had the proportion of immigrants. Unfortunately, the absence of this data point means we cannot compute this and fully understand the relationship between the events.
Statistical Analysis
Statistical analysis involves collecting, reviewing, and interpreting data to draw conclusions. It’s essential in understanding patterns and relationships between different variables. In probability, it helps assess the independence or dependence of events by analyzing data trends.
To determine if events like in the TB scenario are independent or dependent, one conducts statistical analyses using appropriate data points, such as the proportion of recent immigrants, to calculate necessary probabilities.
Without all data pieces, the analysis may lack the depth needed to reach solid conclusions as was the case with the given exercise. Thus, statistical proficiency allows one to critically evaluate available data and make informed decisions, even while acknowledging limitations.

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Most popular questions from this chapter

At a large university, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used a book by Professor Mean; during the winter quarter, 300 students used a book by Professor Median; and during the spring quarter, 200 students used a book by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with the text in the fall quarter, 150 in the winter quarter, and 160 in the spring quarter. a. If a student who took statistics during one of these three quarters is selected at random, what is the probability that the student was satisfied with the textbook? b. If a randomly selected student reports being satisfied with the book, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? (Hint: Use Bayes' rule to compute three probabilities.)

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.2\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\) ). c. What is the probability that the firm is successful in at least one of the two bids?

In a school machine shop, \(60 \%\) of all machine breakdowns occur on lathes and \(15 \%\) occur on drill presses. Let \(E\) denote the event that the next machine breakdown is on a lathe, and let \(F\) denote the event that a drill press is the next machine to break down. With \(P(E)=.60\) and \(P(F)=.15\), calculate: a. \(P\left(E^{C}\right)\) b. \(P(E \cup F)\) c. \(P\left(E^{C} \cap F^{C}\right)\)

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let not \(E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(\left.E_{1}\right)\) ? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

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