/*! 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 28 Although studies continue to sho... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Although studies continue to show smoking leads to significant health problems, \(20 \%\) of adults in the United States smoke. Consider a group of 250 adults. a. What is the expected number of adults who smoke? b. What is the probability that fewer than 40 smoke? c. What is the probability that from 55 to 60 smoke? d. What is the probability that 70 or more smoke?

Short Answer

Expert verified
a. 50 adults. b. 0.0396. c. 0.1736. d. 0.0141.

Step by step solution

01

Understanding the Problem

We are given a proportion problem involving a group of people and percentages. Given that 20% of adults smoke, we need to determine certain probabilities concerning a selection of 250 people. This involves understanding and using concepts from probability and statistics, specifically the binomial distribution.
02

Expected Number of Smokers

To find the expected number of smokers, use the formula for the expected value of a binomial distribution: \[ E(X) = n \times p \] where \( n \) is the number of trials (250 adults) and \( p \) is the probability of success (0.20 probability of an adult smoking). Thus, the expected number of smokers is: \[ E(X) = 250 \times 0.20 = 50 \].
03

Finding Probability for Fewer than 40 Smokers

To find this probability, we calculate \( P(X < 40) \) where \( X \sim Binomial(n=250, p=0.20) \). While this calculation requires the binomial probability formula, using a statistical tool or tables simplifies it. By using a binomial distribution table or calculator, we find \( P(X < 40) \approx 0.0396 \).
04

Finding Probability for 55 to 60 Smokers

We are tasked with finding \( P(55 \leq X \leq 60) \). Again, using the binomial distribution, we can sum up the probabilities \( P(X = 55), P(X = 56), \ldots, P(X = 60) \). Calculating or using a computational tool, \( P(55 \leq X \leq 60) \approx 0.1736 \).
05

Finding Probability for 70 or More Smokers

Here, the required probability is \( P(X \geq 70) \). This is equivalent to \( 1 - P(X < 70) \), which can be calculated using a binomial table or tool that provides cumulative probabilities. Using such a tool, \( P(X \geq 70) \approx 0.0141 \).

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.

Probability Calculation
The binomial distribution is a key statistical tool for calculating probabilities in situations where you perform a series of trials that can only end in two outcomes, such as "smoke" or "do not smoke."
Each trial is independent, and the probability of success (in this case, smoking) remains constant across trials. In the problem of smoking adults, we consider 250 trials with a success probability of 20%, or 0.20. We use this information to calculate the probability of various outcomes.Probability calculations in binomial experiments are made easier by using the binomial probability formula:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]Where:
  • \( P(X = k) \) is the probability of exactly \( k \) successes
  • \( n \) is the number of trials
  • \( p \) is the probability of success
  • \( \binom{n}{k} \) is the number of combinations of \( n \) things taken \( k \) at a time
This approach helps find probabilities, but with many trials, using statistical tools or pre-calculated tables simplifies the process significantly.
Expected Value
Expected value offers insights into what you can "expect" in terms of average outcomes when performing a statistical experiment multiple times. For a binomial distribution, the expected number of successes, given \( n \) trials and probability of success \( p \), is described by the formula:\[E(X) = n \times p\]The idea is simple yet powerful. It allows us to make an educated guess about the average number of smokers in a group of 250 adults, knowing the smoking rate is 20%. In our problem, the expected value was calculated as:
  • Expected smokers \( E(X) = 250 \times 0.20 = 50 \)
This tells us that, on average, we can expect 50 adults to smoke out of any similar group of 250 adults, providing a baseline for assessing variation seen in actual results.
Statistical Tools
Statistical tools are invaluable when it comes to handling large-scale probability calculations, especially those involving complex or numerous operations, like a binomial probability distribution for 250 trials. Manually calculating each probability can be cumbersome, and tools offer a streamlined approach.There are various tools that students and professionals use to simplify these computations:
  • Statistical Software: Software like R, Python (with libraries such as SciPy), and statistical calculators automate the process, eliminating human error in complex calculations.
  • Binomial Tables: Binomial distribution tables offer pre-calculated probabilities, which can be referenced quickly to find probabilities for specific values of \( X \).
  • Online Calculators: Many websites offer free binomial calculators that instantly compute the probability distributions given the values of \( n \) and \( p \).
By utilizing these tools, the efficiency and accuracy of probability calculations improve significantly, allowing for a deeper understanding and application of theoretical concepts in practical scenarios.

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

The random variable \(x\) is known to be uniformly distributed between 10 and 20 . a. Show the graph of the probability density function. b. Compute \(P(x \leq 15)\) c. Compute \(P(12 \leq x \leq 18)\) d. Compute \(E(x)\) e. Compute \(\operatorname{Var}(x)\)

On average, 30 -minute television sitcoms have 22 minutes of programming (CNBC, February 23,2006 . Assume that the probability distribution for minutes of programming can be approximated by a uniform distribution from 18 minutes to 26 minutes. a. What is the probability that a sitcom will have 25 or more minutes of programming? b. What is the probability that a sitcom will have between 21 and 25 minutes of programming? c. What is the probability that a sitcom will have more than 10 minutes of commercials or other nonprogramming interruptions?

Assume a binomial probability distribution has \(p=.60\) and \(n=200\) a. What are the mean and standard deviation? b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution? Explain. c. What is the probability of 100 to 110 successes? d. What is the probability of 130 or more successes? e. What is the advantage of using the normal probability distribution to approximate the binomial probabilities? Use part (d) to explain the advantage.

The time required to pass through security screening at the airport can be annoying to travelers. The mean wait time during peak periods at Cincinnati/Northern Kentucky International Airport is 12.1 minutes (The Cincinnati Enquirer; February 2,2006 ). Assume the time to pass through security screening follows an exponential distribution. a. What is the probability that it will take less than 10 minutes to pass through security screening during a peak period? b. What is the probability that it will take more than 20 minutes to pass through security screening during a peak period? c. What is the probability that it will take between 10 and 20 minutes to pass through security screening during a peak period? d. It is 8: 00 A.M. (a peak period) and you just entered the security line. To catch your plane you must be at the gate within 30 minutes. If it takes 12 minutes from the time you clear security until you reach your gate, what is the probability that you will miss your flight?

In an article about the cost of health care, Money magazine reported that a visit to a hospital emergency room for something as simple as a sore throat has a mean cost of \(\$ 328\) (Money, January 2009 ). Assume that the cost for this type of hospital emergency room visit is normally distributed with a standard deviation of \(\$ 92 .\) Answer the following questions about the cost of a hospital emergency room visit for this medical service. a. What is the probability that the cost will be more than \(\$ 500 ?\) b. What is the probability that the cost will be less than \(\$ 250 ?\) c. What is the probability that the cost will be between \(\$ 300\) and \(\$ 400 ?\) d. If the cost to a patient is in the lower \(8 \%\) of charges for this medical service, what was the cost of this patient's emergency room visit?

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.