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At a large university, in the never-ending quest for a satisfactory textbook, the Statistics Department has tried a different text during each of the last three quarters. During the fall quarter, 500 students used the text by Professor Mean; during the winter quarter, 300 students used the text by Professor Median; and during the spring quarter, \(200 \mathrm{stu}-\) dents used the text by Professor Mode. A survey at the end of each quarter showed that 200 students were satisfied with Mean's book, 150 were satisfied with Median's book, and 160 were satisfied with Mode's book. If a student who took statistics during one of these quarters is selected at random and admits to having been satisfied with the text, is the student most likely to have used the book by Mean, Median, or Mode? Who is the least likely author? [Hint: Draw a tree diagram or use Bayes' theorem.]

Short Answer

Expert verified
Student is most likely satisfied with Mean's book and least likely with Median's.

Step by step solution

01

Define the Problem Variables

We need to find which author is most and least likely to have satisfied a student, given they were satisfied with the textbook. Let's define the probabilities. Let \(M\) be the event the student used Mean's book, \(Md\) the event the student used Median's book, and \(Mo\) the event the student used Mode's book. The event \(S\) represents a student being satisfied.
02

Determine Prior Probabilities

The prior probabilities for choosing a student from each book are based on the number of students:\[P(M) = \frac{500}{1000}, \quad P(Md) = \frac{300}{1000}, \quad P(Mo) = \frac{200}{1000}\]These simplify to:\[P(M) = 0.5, \quad P(Md) = 0.3, \quad P(Mo) = 0.2\]
03

Determine Conditional Probabilities

These are the probabilities of being satisfied given a specific book:\[P(S|M) = \frac{200}{500} = 0.4, \quad P(S|Md) = \frac{150}{300} = 0.5, \quad P(S|Mo) = \frac{160}{200} = 0.8\]
04

Apply Bayes' Theorem

To find the probability that a satisfied student used each book, apply Bayes' theorem:\[P(M|S) = \frac{P(S|M) \cdot P(M)}{P(S)} \]For Mode and Median as well:\[P(Md|S) = \frac{P(S|Md) \cdot P(Md)}{P(S)} \]\[P(Mo|S) = \frac{P(S|Mo) \cdot P(Mo)}{P(S)} \]
05

Calculate Overall Satisfaction Probability

The total probability of being satisfied is given by:\[P(S) = P(S|M) \cdot P(M) + P(S|Md) \cdot P(Md) + P(S|Mo) \cdot P(Mo)\]\[ = 0.4\times 0.5 + 0.5\times 0.3 + 0.8\times 0.2 = 0.58\]
06

Calculate Posterior Probabilities

Using the results from Bayes' theorem:\[P(M|S) = \frac{0.4 \times 0.5}{0.58} = \frac{0.2}{0.58} = 0.3448\]\[P(Md|S) = \frac{0.5 \times 0.3}{0.58} = \frac{0.15}{0.58} = 0.2586\]\[P(Mo|S) = \frac{0.8 \times 0.2}{0.58} = \frac{0.16}{0.58} = 0.2759\]
07

Interpret the Results

The probability the satisfied student used Mean's book is about 34.48%, Median's book is about 25.86%, and Mode's book is about 27.59%. Therefore, the student is most likely to be satisfied with Mean and least likely with Median.

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Key Concepts

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

Conditional Probability: Understanding Dependencies
Conditional probability helps us calculate the likelihood of an event given that another event has already occurred. Imagine you want to figure out how likely it is for a student to have used Professor Mode's book given they were satisfied with it. This utilizes the concept of conditional probability.

Conditional probability is expressed as \( P(A|B) \), which reads as "the probability of A given B." In simpler terms, it's finding the probability of A happening under the condition that B has happened.

Here's how it works in the context of our problem:
  • We first identify the probability that a student might find a specific textbook satisfying.
  • Then, we calculate how likely it is that a satisfied student had used a particular textbook.
Understanding these dependencies with conditional probability helps us make informed predictions based on existing information.
Probability Theory: The Foundation of Calculations
Probability theory forms the backbone of these calculations, enabling us to work with uncertain events systematically. Probability in this context measures the likelihood of each textbook being chosen by a random student.

To apply probability theory effectively, we first determine prior probabilities, which involves dividing the number of students using each textbook by the total number of students. For instance:
  • The prior probability of using Mean's book is \( P(M) = 0.5 \).
  • The prior probability of using Median's book is \( P(Md) = 0.3 \).
  • The prior probability of using Mode's book is \( P(Mo) = 0.2 \).
These numbers are crucial as they form the initial step towards understanding the likeliness of events even before any condition is applied. Probability theory helps convert the abstract ideas of chance into quantifiable numbers we can analyze.
Statistical Analysis: Making Sense of Data
Statistical analysis allows us to make sense of data and interpret results effectively. In this problem, statistical analysis helps us find which book students were most or least likely satisfied with.

Using Bayes' Theorem, we conduct statistical analysis to determine the posterior probabilities, which provide the likelihood of each book being satisfactory once the satisfaction status is known.
  • The calculated posterior probability for Mean's book is about 34.48%.
  • The posterior probability for Median's book is approximately 25.86%.
  • The posterior probability for Mode's book is around 27.59%.
Statistical analysis is not just about crunching numbers; it's about understanding the story they tell. In this case, it's clear that while all books provided satisfaction, Mean's book is found to have the highest likelihood of being used by a satisfied student. Through this process, we gain insights that go beyond numbers, illustrating the power of statistical analysis in interpreting real-world scenarios.

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Most popular questions from this chapter

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

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