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Chapter 6: Problem 38

The USA Today article referenced in Exercise \(6.37\) also gave information on seat belt usage by age, which is summarized in the following table of counts: $$ \begin{array}{lcc} & \begin{array}{c} \text { Does Not Use } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} & \begin{array}{c} \text { Uses } \\ \text { Seat Belt } \\ \text { Regularly } \end{array} \\ \hline 18-24 & 59 & 41 \\ 25-34 & 73 & 27 \\ 35-44 & 74 & 26 \\ 45-54 & 70 & 30 \\ 55-64 & 70 & 30 \\ 65 \text { and older } & 82 & 18 \\ \hline \end{array} $$ Consider the following events: \(S=\) event that a randomly selected individual uses a seat belt regularly, \(A_{1}=\) event that a randomly selected individual is in age group \(18-24\), and \(A_{6}=\) event that a randomly selected individual is in age group 65 and older. a. Convert the counts to proportions and then use them to compute the following probabilities: i. \(P\left(A_{1}\right)\) ii. \(P\left(A_{1} \cap S\right)\) iii. \(P\left(A_{1} \mid S\right)\) iv. \(P\left(\right.\) not \(\left.A_{1}\right) \quad\) v. \(P\left(S \mid A_{1}\right) \quad\) vi. \(P\left(S \mid A_{6}\right)\) b. Using the probabilities \(P\left(S \mid A_{1}\right)\) and \(P\left(S \mid A_{6}\right)\) computed in Part (a), comment on how \(18-24\) -year-olds and seniors differ with respect to seat belt usage.

Short Answer

Expert verified
a. i. \(P(A_1)\) is approximately \(0.06\), ii. \(P(A_1 \cap S)\) is approximately \(0.03\), iii. \(P(A_1 | S)\) is approximately \(0.21\), iv. \(P(\) not \(A_1)\) is approximately \(0.94\), v. \(P(S | A_1)\) is approximately \(0.41\), vi. \(P(S | A_6)\) is approximately \(0.18\). b. The \(18-24\) age group has a higher probability of using a seat belt regularly than individuals who are \(65\) and older, indicating that younger individuals are more likely to use seat belts regularly than older individuals.

Step by step solution

01

Calculate total number of individuals and proportions

First, sum up all the counts given in the table to get the total number of individuals surveyed. Then, calculate the proportion of individuals in each age group and those who use or do not use seat belts regularly.
02

Calculate \(P(A_1)\)

\(P(A_1)\) is the probability that a randomly selected individual is in the age group \(18-24\). This can be calculated by dividing the total number of individuals in the \(18-24\) age group by the total number of individuals surveyed.
03

Calculate \(P(A_1 \cap S)\)

\(P(A_1 \cap S)\) is the probability that a randomly selected individual is in the age group \(18-24\) and uses a seat belt regularly. This can be calculated by dividing the number of individuals in the \(18-24\) age group who use a seat belt regularly by the total number of individuals surveyed.
04

Calculate \(P(A_1 | S)\)

\(P(A_1 | S)\) is the conditional probability that the individual is in the \(18-24\) age group given they use a seat belt regularly. This is calculated using the formula for conditional probability: \(P(A_1 | S) = P(A_1 \cap S)/P(S)\), where \(P(S)\) is probability that a randomly selected individual uses seat belt regularly and can be calculated by summing the proportions of those individuals who use seat belt regularly in all age groups.
05

Calculate \(P(S | A_1)\) and \(P(S | A_6)\)

\(P(S | A_1)\) and \(P(S | A_6)\) are the conditional probabilities that the individual uses a seat belt regularly given they are in the \(18-24\) age group and 65 and older age group respectively. These can be calculated using the formula: \(P(S | A_i) = P(A_i \cap S)/P(A_i)\), where \(i=1,6\) and \(P(A_i)\) is the probability that a randomly selected individual is in the \(i^{th}\) age group.

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Key Concepts

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

Conditional Probability
Understanding conditional probability is crucial for analyzing situations where the probability of an event is influenced by the occurrence of another event. In the context of seat belt usage, we investigate conditional probability as the likelihood of someone using a seat belt given that they are from a specific age group. This is denoted as P(S | A1) or P(S | A6), where S represents the event of using a seat belt regularly, and A1 and A6 represent being in the age groups 18-24 and 65 and older, respectively.

Conditional probability is computed by dividing the probability of the intersection of the two events (P(A1 ∩ S) or P(A6 ∩ S)) by the probability of the condition or given event (P(A1) or P(A6)). The general formula for conditional probability is P(B | A) = P(A ∩ B) / P(A).

For example, to calculate P(S | A1), which is the probability of using a seat belt given that an individual is 18-24 years old, we divide the number of 18-24 year olds who use a seat belt by the total number of individuals in that age group. An analogous approach is taken for calculating P(S | A6). This concept helps us understand how behavior, such as seat belt usage, might vary across different age groups.
Probability Computation
Probability computation involves using mathematical formulas to determine the likelihood of various events. In our exercise, we must compute the basic probabilities such as P(A1), which is the probability of selecting an individual aged 18-24, and more complex probabilities like P(A1 ∩ S), the probability of selecting an individual aged 18-24 who uses a seat belt regularly. These calculations are foundational in probability theory and are done by dividing the count of the targeted group by the total population size.

The complexity of probability computations increases when we need to find conditional probabilities or the likelihood of composite events. These require awareness of the relationships between the events in question. For example, to compute P(A1 | S), it is necessary to understand that the probability of being in a certain age group given regular seat belt usage can be different from the probability of being in that age group overall.
Statistical Data Analysis
In statistical data analysis, we analyze data in order to make inferences or understand patterns. In the exercise at hand, we are analyzing data related to age groups and seat belt usage. To begin, we convert raw data counts into proportions or probabilities, which provide a clearer picture of how often certain events are occurring.

With the calculated probabilities in hand, such as P(S | A1) and P(S | A6), we can conduct further analysis. For instance, by comparing these probabilities, we gain insights into how seat belt usage differs among different age groups, which is essential for targeted safety campaigns or policies. Statistical data analysis can be as simple as computing descriptive statistics like these probabilities, or it may involve more complex inferential statistics, allowing researchers to derive conclusions about larger populations based on sample data.

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Most popular questions from this chapter

The Australian newspaper The Mercury (May 30 , 1995) reported that, based on a survey of 600 reformed and current smokers, \(11.3 \%\) of those who had attempted to quit smoking in the previous 2 years had used a nicotine aid (such as a nicotine patch). It also reported that \(62 \%\) of those who quit smoking without a nicotine aid be gan smoking again within 2 weeks and \(60 \%\) of those who used a nicotine aid began smoking again within 2 weeks. If a smoker who is trying to quit smoking is selected at random, are the events selected smoker who is trying to quit uses a nicotine aid and selected smoker who has attempted to quit begins smoking again within 2 weeks inde pendent or dependent events? Justify your answer using the given information.

\(6.4\) A tennis shop sells five different brands of rackets, each of which comes in either a midsize version or an oversize version. Consider the chance experiment in which brand and size are noted for the next racket purchased. One possible outcome is Head midsize, and another is Prince oversize. Possible outcomes correspond to cells in the following table: $$ \begin{array}{|l|l|l|l|l|l|} \hline & \text { Head } & \text { Prince } & \text { Slazenger } & \text { Wimbledon } & \text { Wilson } \\ \hline \text { Midsize } & & & & & \\ \hline \text { Oversize } & & & & & \\ \hline \end{array} $$ a. Let \(A\) denote the event that an oversize racket is purchased. List the outcomes in \(A\). b. Let \(B\) denote the event that the name of the brand purchased begins with a W. List the outcomes in \(B\). c. List the outcomes in the event \(n o t \bar{B}\). d. Head, Prince, and Wilson are U.S. companies. Let \(C\) denote the event that the racket purchased is made by a U.S. company. List the outcomes in the event \(B\) or \(C\). e. List outcomes in \(B\) and \(C\). f. Display the possible outcomes on a tree diagram, with a first-generation branch for each brand.

An individual is presented with three different glasses of cola, labeled C, D, and P. He is asked to taste all three and then list them in order of preference. Suppose that the same cola has actually been put into all three glasses. a. What are the simple events in this chance experiment, and what probability would you assign to each one? b. What is the probability that \(\mathrm{C}\) is ranked first? c. What is the probability that \(\mathrm{C}\) is ranked first \(a n d \mathrm{D}\) is ranked last?

The article "Birth Beats Long Odds for Leap Year Mom, Baby" (San Luis Obispo Tribune, March 2, 1996) reported that a leap year baby (someone born on February 29 ) became a leap year mom when she gave birth to a baby on February \(29,1996 .\) The article stated that a hospital spokesperson said that the probability of a leap year baby giving birth on her birthday was one in \(2.1\) million (approximately .00000047). a. In computing the given probability, the hospital spokesperson used the fact that a leap day occurs only once in 1461 days. Write a few sentences explaining how the hospital spokesperson computed the stated probability. b. To compute the stated probability, the hospital spokesperson had to assume that the birth was equally likely to occur on any of the 1461 days in a four- year period. Do you think that this is a reasonable assumption? Explain. c. Based on your answer to Part (b), do you think that the probability given by the hospital spokesperson is too small, about right, or too large? Explain.

Insurance status - covered (C) or not covered (N) \- is determined for each individual arriving for treatment at a hospital's emergency room. Consider the chance experiment in which this determination is made for two randomly selected patients. The simple events are \(O_{1}=(\mathrm{C}, \mathrm{C})\) \(O_{2}=(\mathrm{C}, \mathrm{N}), O_{3}=(\mathrm{N}, \mathrm{C})\), and \(O_{4}=(\mathrm{N}, \mathrm{N}) .\) Suppose that probabilities are \(P\left(O_{1}\right)=.81, P\left(O_{2}\right)=.09, P\left(O_{3}\right)=.09\), and \(P\left(O_{4}\right)=.01\). a. What outcomes are contained in \(A\), the event that at most one patient is covered, and what is \(P(A)\) ? b. What outcomes are contained in \(B\), the event that the two patients have the same status with respect to coverage, and what is \(P(B)\) ?
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