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Chapter 6: Problem 18

An article in the New york Times (March 2. 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

Short Answer

Expert verified
a. The relative frequency interpretation of the given probability would be 'Out of every 100 people that suffer a cardiac arrest in New York City, only 1 person survives.', b. From the given total of 2329 cardiac arrests in New York City, it is probable that about 23 people survived.

Step by step solution

01

Understand relative frequency interpretation

Relative frequency is a way of expressing probability. It's computed as the ratio of the number of times an outcome occurs to the total count of all outcomes. In this instance, \( P(\) survival \()=.01 \) implies that for every 100 instances of cardiac arrests, only 1 person survives.
02

Convert the probability into a relative frequency

The relative frequency interpretation of the given probability, \( P(\) survival \()=.01 \), would be 'Out of every 100 people that suffer a cardiac arrest in New York City, only 1 person survives.', or alternately, '1 out of 100 people survive a cardiac arrest in New York City.'
03

Understand '1 in 100 chance of survival'

'1 in 100 chance of survival' refers to the survival ratio or probability for a person who suffers a cardiac arrest in New York City. It is a way of expressing the survival rate.
04

Calculate the number of survivors from the given total

Given that for every 100 people that suffer cardiac arrests, only 1 person survives, then, out of the 2329 consecutive cardiac arrests, there might be approximately \(2329 \times .01 = 23.29\), which rounds down to about 23 people who survived. Since the number of people can't be a decimal, it is reasonable to round to the nearest whole number.

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Key Concepts

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

Relative Frequency
Relative frequency helps us interpret probability in a tangible way. It is defined as the ratio of the number of times a specific event occurs to the total number of trials or instances in any given experiment. This concept is crucial because it allows us to translate theoretical probabilities into practical, observed outcomes. For example, if you roll a die 100 times and the number "4" appears 15 times, the relative frequency of rolling a "4" would be 15/100 or 0.15.

In terms of the cardiac arrest scenario mentioned, the given probability of survival is stated as 0.01. This means that in every 100 cases of cardiac arrests, only 1 person is expected to survive. By understanding relative frequency, we can grasp the severity of the situation and appreciate why survival in such cases is heavily scrutinized and discussed.
Survival Rate
The survival rate is a measure of the proportion of survivors among a defined group of patients or a specific scenario over a set period. It's commonly expressed as a percentage, helping to convey the likelihood of living through a typically life-threatening event. In medical studies, survival rates are vital for understanding the effectiveness of treatments or the severity of diseases.

In the context of the cardiac arrest study in New York City, the survival rate was described as "1 in 100", translating to a 1% survival chance. This rate starkly highlights the challenges faced by emergency services in densely populated urban environments. Knowing this rate helps inform local efforts to improve emergency response times and educate the public on quick actions during cardiac-related emergencies.
  • Survival Rate = (Number of Survivors / Total Number of Events) 脳 100
  • It is critical in assessing healthcare interventions and policies.
Cardiac Arrest Study
Examining cardiac arrest statistics provides essential insights into the life-or-death nature of these events and the sobering survival rates associated with them. The specific study mentioned involved a comprehensive investigation of 2329 cardiac arrest incidents in New York City. Such studies are fundamental in understanding and improving medical responses in emergencies.

To calculate expected survivors from the study, we use the probability: of every 100 cardiac arrests, statistically, 1 person survives. Thus, in a total of 2329 cardiac arrests studied, approximately 23 people survived. This study sheds light on the factors that contribute to such low survival rates, including urban challenges like traffic congestion and delayed medical assistance.
  • Detailed studies enhance understanding of emergency medical response and its limitations.
  • Such data is used to propose changes to improve public health outcomes.

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Most popular questions from this chapter

The article "Anxiety Increases for Airline Passengers After Plane Crash" (San Luis Obispo Tribune. November 13,2001 ) reported that air passengers have a 1 in 11 million chance of dying in an airplane crash. This probability was then interpreted as "You could fly every day for 26,000 years before your number was up." Comment on why this probability interpretation is misleading.

The article "Chances Are You Know Someone with a Tattoo, and He's Not a Sailor" (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50 . The accompanying data are consistent with summary values given in the article. $$ \begin{array}{l|cc} & \text { At Least One Tattoo } & \text { No Tattoo } \\ \hline \text { Age 18-29 } & 18 & 32 \\ \text { Age 30-50 } & 6 & 44 \\ \hline \end{array} $$ Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities. a. \(P(\) tattoo \()\) b. \(P(\) tattoo \(\mid\) age \(18-29)\) c. \(P(\) tattoo \(\mid\) age \(30-50\) ) d. \(P(\) age \(18-29 \mid\) tattoo \()\)

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?

The manager of a music store has kept records of the number of CDs bought in a single transaction by customers who make a purchase at the store. The accompanying table gives six possible outcomes and the estimated probability associated with each of these outcomes for the chance experiment that consists of observing the number of CDs purchased by the next customer at the store. $$ \begin{aligned} &\begin{array}{l} \text { Number of CDs } \\ \text { purchased } \end{array} & 1 & 2 & 3 & 4 & 5 & 6 \text { or more } \\ &\begin{array}{c} \text { Estimated } \\ \text { probability } \end{array} & .45 & .25 & .10 & .10 & .07 & .03 \end{aligned} $$ a. What is the estimated probability that the next customer purchases three or fewer CDs? b. What is the estimated probability that the next customer purchases at most three CDs? How does this compare to the probability computed in Part (a)? c. What is the estimated probability that the next customer purchases five or more CDs? d. What is the estimated probability that the next customer purchases one or two CDs? e. What is the estimated probability that the next customer purchases more than two CDs? Show two different ways to compute this probability that use the probability rules of this section.

Medical 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 \(O_{1}=(\mathrm{C}, \mathrm{C})\), meaning that the first patient selected was covered and the second patient selected was also covered, \(O_{2}=(\mathrm{C}, \mathrm{N}), O_{3}=(\mathrm{N}, \mathrm{C})\), 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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