/*! 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 1 An article in the New York Times... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An article in the New York Times 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 \()=0.01\) for people who suffer cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and difficulty finding victims in large buildings. Similar studies in smaller cities showed higher survival rates.) a. Give a relative frequency interpretation of the given probability. b. The basis for the New York Times article was a research study of 2,329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2,329 cardiac arrest sufferers do you think survived? Explain.

Short Answer

Expert verified
The relative frequency interpretation of the given probability suggests that in a very large number of cardiac arrests in New York City, we would expect about 1% of them to survive. When justified with the data from the study, it suggests that from the 2,329 cardiac arrests, around 23 people would have survived.

Step by step solution

01

Understanding Relative Frequency Interpretation

The relative frequency interpretation of a probability refers to an estimate of the chance that a particular event will occur. It is based on observed data and calculated as the ratio of the number of observed occurrences of the event to the total number of trials. In this problem, a relative frequency interpretation of the given probability, \(P(\) survival \()=0.01\) , suggests that if we were to observe a very large number of cardiac arrests in New York City, we would expect about 1% of them to survive.
02

Calculation of The Number of Survivors

Next, to justify the 1 in 100 chance of survival statement, and figure out the number of survivors out of the 2,329 cardiac arrest sufferers, multiply the total number of sufferers, which is 2,329, by the survival probability of 0.01. This accounts for the given probability.
03

Justification and Explanation

The statement that 1 in 100 people survive cardiac arrest in New York City is justified by the fact that, of the 2,329 consecutive cardiac arrests in the city, approximately 2,329 * 0.01 = 23 people are expected to have survived. The numbers are provided by a study, justifying the relative frequency interpretation of the survival probability.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Relative Frequency Interpretation of Probability
When we talk about the relative frequency interpretation of probability, we are referring to how often an event occurs relative to the number of trials or opportunities for it to happen. It's like taking a tally of how many times something has happened and relating it to the total count of attempts or observations.

Consider a baseball player with a batting average of .300. This means that, relative to frequency, the player hits safely in 30% of his at bats over a season. It's calculated by dividing the number of hits by the total at bats. Similarly, the reported probability of survival from cardiac arrest (\( P(\text{survival}) = 0.01 \) is a statistical estimate indicating that, historically, for every 100 cases of cardiac arrest in New York City, about one person survives.

The interpretation is essentially a prediction based on past data. If New York City continues to have the same conditions, then this probability tells us that the pattern of survival is expected to stay consistent - roughly one survivor for every 100 instances of cardiac arrest. The value of using relative frequency is that it provides a practical way to understand and communicate the likelihood of events, especially in high-stakes situations like emergency medical response.
Understanding Probability Notation
Probability notation is a shorthand way to express the likelihood of an event. It allows statisticians, mathematicians, and students alike to quickly understand and calculate the odds of something happening. In our cardiac arrest example, the probability notation \( P(\text{survival}) = 0.01 \) is read as 'the probability of survival equals 0.01', or in plain terms, there's a 1% chance of surviving.

This notation is essential because it provides a common language for discussing uncertain outcomes. It can be broken down into parts: the 'P' stands for 'probability', and the event of interest is enclosed in parentheses. The equal sign then links this to the numerical likelihood - in this case, 0.01.

Being comfortable with reading and writing probabilities in this way is crucial for understanding studies, reports, and analyses across a variety of fields, not just survival statistics. It scales from everyday probabilities (like the chance of rain) to those with serious societal implications (like survival rates for illnesses).
Survival Rate Statistics
Survival rate statistics are vital numbers used in evaluating the effectiveness of healthcare systems, and understanding the risks associated with various conditions. These rates tell us the proportion of people who continue to live for a certain period of time after being diagnosed with a disease or experiencing a critical health event like a cardiac arrest.

In the context of our exercise, the survival rate for cardiac arrest in New York City is given as 1 in 100, derived from a study of 2,329 cases. If we were to convert this into a percentage, it shows a survival rate of 1%. To find the number of individuals who survived from the study, we would use the following calculation: \( 2,329 \times 0.01 = 23.29 \) - which, when rounded, suggests approximately 23 survivors.

Understanding these statistics helps public health officials to identify areas for improvement, such as faster emergency response times. It can also provide the public with valuable information on the likelihood of recovery from certain incidents, aiding in preventive measures and emergency preparedness.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. a. Use the given probability information to set up a "hypothetical 1000 " table. b. Use the table from Part (a) to find the following probabilities: i. the probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. ii. the probability that a randomly selected customer purchases an extended warranty for neither the washer nor the dryer.

An airline reports that for a particular flight operating daily between Phoenix and Atlanta, the probability of an on-time arrival is \(0.86 .\) Give a relative frequency interpretation of this probability.

The paper "Predictors of Complementary Therapy Use Among Asthma Patients: Results of a Primary Care Survey" (Health and Social Care in the Community [2008]: \(155-164)\) described a study in which each person in a large sample of asthma patients responded to two questions: Question 1: Do conventional asthma medications usually help your symptoms? Question 2: Do you use complementary therapies (such as herbs, acupuncture, aroma therapy) in the treatment of your asthma? Suppose that this sample is representative of asthma patients. Consider the following events: \(E=\) event that the patient uses complementary therapies \(F=\) event that the patient reports conventional medications usually help The data from the sample were used to estimate the following probabilities: $$P(E)=0.146 \quad P(F)=0.879 \quad P(E \cap F)=0.122$$ a. Use the given probability information to set up a "hypothetical 1000 " table with columns corresponding to \(E\) and \(n o t E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. The probability that an asthma patient responds that conventional medications do not help and that patient uses complementary therapies. ii. The probability that an asthma patient responds that conventional medications do not help and that patient does not use complementary therapies. iii. The probability that an asthma patient responds that conventional medications usually help or the patient uses complementary therapies. c. Are the events \(E\) and \(F\) independent? Explain.

A large cable company reports that \(42 \%\) of its customers subscribe to its Internet service, \(32 \%\) subscribe to its phone service, and \(51 \%\) subscribe to its Internet service or its phone service (or both). a. Use the given probability information to set up a "hypothetical \(1000 "\) table. b. Use the table to find the following: i. the probability that a randomly selected customer subscribes to both the Internet service and the phone service. ii. the probability that a randomly selected customer subscribes to exactly one of the two services.

The same issue of The Chronicle for Higher Education referenced in Exercise 5.17 also reported the following information for degrees awarded to Hispanic students by U.S. colleges in the \(2008-2009\) academic year: A total of 274,515 degrees were awarded to Hispanic students. \- 97,921 of these degrees were Associate degrees. \- 129,526 of these degrees were Bachelor's degrees. \- The remaining degrees were either graduate or professional degrees. What is the probability that a randomly selected Hispanic student who received a degree in \(2008-2009\) a. received an associate degree? b. received a graduate or professional degree? c. did not receive a bachelor's degree?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.