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3.19 Drawing box plots. After an introductory statistics course, \(80 \%\) of students can successfully construct box plots. Of those who can construct box plots, \(86 \%\) passed, while only \(65 \%\) of those students who could not construct box plots passed. (a) Construct a tree diagram of this scenario. (b) Calculate the probability that a student is able to construct a box plot if it is known that he passed.

Short Answer

Expert verified
The probability that a student can construct a box plot, given they passed, is approximately 0.841.

Step by step solution

01

Define the Events

Let's define the following events: Let \( C \) denote the event a student can construct a box plot, and \( P \) the event a student passed. We have the following probabilities: \( P(C) = 0.80 \), \( P(P|C) = 0.86 \), and \( P(P|C^c) = 0.65 \) where \( C^c \) denotes the event a student cannot construct a box plot.
02

Complete the Probability Tree Diagram

First, draw the starting point for 'Can Construct' and 'Cannot Construct'. Attach branches from each start point with conditional probabilities: For the 'Can Construct' branch, the probability of passing is \(0.86\) and not passing is \(0.14\). For the 'Cannot Construct' branch, the probability of passing is \(0.65\) and not passing is \(0.35\).
03

Calculate Overall Probability of Passing

Use the law of total probability: \( P(P) = P(P|C)P(C) + P(P|C^c)P(C^c) \). Substitute the values: \( P(P) = (0.86)(0.80) + (0.65)(0.20) = 0.688 + 0.13 = 0.818 \).
04

Calculate Conditional Probability

Use Bayes’ Theorem to find \( P(C | P) \): \[ P(C | P) = \frac{P(P | C)P(C)}{P(P)} = \frac{0.86 \times 0.80}{0.818} \] Calculate \( P(C | P) = \frac{0.688}{0.818} \approx 0.841 \).

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Key Concepts

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

Tree Diagram
A tree diagram is a fantastic tool for visualizing and analyzing probabilities in a step-by-step manner. It's like a map that guides you through different possible scenarios and their respective probabilities. In this exercise, we started by considering two main events:
  • Can Construct a box plot
  • Cannot Construct a box plot
From these two starting points, we draw branches representing the outcomes of passing or failing the course.
  • For students who can construct a box plot, there are two branches: one for passing with a probability of 0.86 and one for failing with a probability of 0.14.
  • Similarly, for students who cannot construct a box plot, the branches represent passing with a probability of 0.65 and failing with a probability of 0.35.
By using a tree diagram, you can clearly see how each possible outcome contributes to the overall probability scenario. It's like following a path through a forest, with each branch leading you closer to your destination.
Conditional Probability
Conditional probability is a fundamental concept in statistics, representing the probability of an event occurring given that another event has already occurred. In this context, we want to know the probability that a student can construct a box plot, provided that they passed the course.
To tackle this, we use the formula for conditional probability:
\[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]where:
  • P(A | B) is the probability of event A given that event B has occurred.
  • P(A ∩ B) is the probability of both events A and B occurring.
  • P(B) is the probability of event B.
In our example, we demonstrate this by using Bayes’ Theorem, which allows us to calculate the probability of a student being able to construct a box plot given that they passed the course. The formula eventually becomes:
  • \[ P(C | P) = \frac{P(P | C)P(C)}{P(P)} \]
By substituting the known values, you can compute the probability of interest, illustrating how conditional probabilities can provide insights into complex scenarios.
Law of Total Probability
The law of total probability is a crucial tool in probability theory, especially when you need to find the probability of an event based on various scenarios or partitions. It allows us to account for all possible ways that an event can occur, by considering each partition and its contribution to the total probability.
In our scenario, we use this law to calculate the overall probability that a student has passed the course. We considered two partitions:
  • The student can construct a box plot
  • The student cannot construct a box plot
Using the law of total probability, we calculated
\[ P(P) = P(P|C)P(C) + P(P|C^c)P(C^c) \]
This formula suggests that to find the likelihood of passing (P), we need to consider both the probability of passing given that the student can construct a box plot, and the probability of passing given they cannot, each weighted by their respective probabilities.
The calculations walk us through leveraging each part's contribution, culminating in an understanding of how students' ability status collectively affects the overall passing rate. By tying all possibilities together, the law of total probability gives us a complete picture of the probability landscape.

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Most popular questions from this chapter

\(\mathrm{P}(\mathrm{A})=0.3, \mathrm{P}(\mathrm{B})=0.7\) (a) Can you compute \(\mathrm{P}(\mathrm{A}\) and \(\mathrm{B})\) if you only know \(\mathrm{P}(\mathrm{A})\) and \(\mathrm{P}(\mathrm{B}) ?\) (b) Assuming that events \(A\) and \(B\) arise from independent random processes, i. what is \(\mathrm{P}(\mathrm{A}\) and \(\mathrm{B}) ?\) ii. what is \(\mathrm{P}(\mathrm{A}\) or \(\mathrm{B}) ?\) iii. what is \(\mathrm{P}(\mathrm{A} \mid \mathrm{B}) ?\) (c) If we are given that \(\mathrm{P}(\mathrm{A}\) and \(\mathrm{B})=0.1,\) are the random variables giving rise to events \(\mathrm{A}\) and \(\mathrm{B}\) independent? (d) If we are given that \(\mathrm{P}(\mathrm{A}\) and \(\mathrm{B})=0.1\), what is \(\mathrm{P}(\mathrm{A} \mid \mathrm{B}) ?\)

The birthday problem. Suppose we pick three people at random. For each of the following questions, ignore the special case where someone might be born on February 29th, and assume that births are evenly distributed throughout the year. (a) What is the probability that the first two people share a birthday? (b) What is the probability that at least two people share a birthday?

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The Behavioral Risk Factor Surveillance System (BRFSS) is an annual telephone survey designed to identify risk factors in the adult population and report emerging health trends. The following table displays the distribution of health status of respondents to this survey (excellent, very good, good, fair, poor) and whether or not they have health insurance. $$\begin{array}{rrrrrrr} & {\text { Health Status }} & \\ & \text { Excellent } & \text { Very good } & \text { Good } & \text { Fair } & \text { Poor } & \text { Total } \\ \hline \text { No } & 0.0230 & 0.0364 & 0.0427 & 0.0192 & 0.0050 & 0.1262 \\ \text { Yes } & 0.2099 & 0.3123 & 0.2410 & 0.0817 & 0.0289 & 0.8738 \\ \hline \text { Total } & 0.2329 & 0.3486 & 0.2838 & 0.1009 & 0.0338 & 1.0000 \end{array}$$ (a) Are being in excellent health and having health coverage mutually exclusive? (b) What is the probability that a randomly chosen individual has excellent health? (c) What is the probability that a randomly chosen individual has excellent health given that he has health coverage? (d) What is the probability that a randomly chosen individual has excellent health given that he doesn't have health coverage? (e) Do having excellent health and having health coverage appear to be independent?

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