/*! 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 13 A psychologist determined that t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A psychologist determined that the number of sessions required to obtain the trust of a new patient is either \(1,2,\) or \(3 .\) Let \(x\) be a random variable indicating the number of sessions required to gain the patient's trust. The following probability function has been proposed. \\[ f(x)=\frac{x}{6} \quad \text { for } x=1,2, \text { or } 3 \\] a. Is this probability function valid? Explain. b. What is the probability that it takes exactly two sessions to gain the patient's trust? c. What is the probability that it takes at least two sessions to gain the patient's trust?

Short Answer

Expert verified
a. Yes, the function is valid. b. The probability is \(\frac{1}{3}\). c. The probability is \(\frac{5}{6}\).

Step by step solution

01

Verify Probability Function Validity

To check if the probability function is valid, we need to confirm two criteria: \(f(x)\) must be non-negative for all values of \(x\), and the sum of probabilities over all possible values must equal 1.Calculate \(f(x)\) for each value of \(x\):- \(f(1) = \frac{1}{6}\)- \(f(2) = \frac{2}{6} = \frac{1}{3}\)- \(f(3) = \frac{3}{6} = \frac{1}{2}\)Check if \(f(x)\) is non-negative: All values are positive.Sum the probabilities: \(\frac{1}{6} + \frac{1}{3} + \frac{1}{2} = \frac{1}{6} + \frac{2}{6} + \frac{3}{6} = \frac{6}{6} = 1\).Since both conditions are met, the probability function is valid.
02

Calculate Probability for Exactly Two Sessions

To find the probability that it takes exactly two sessions, use the given probability function:Calculate \(f(2)\):\[f(2) = \frac{2}{6} = \frac{1}{3}\]Thus, the probability is \(\frac{1}{3}\).
03

Calculate Probability for At Least Two Sessions

The probability of taking at least two sessions is the sum of the probabilities of taking two or three sessions:Calculate the sum:\[P(x \geq 2) = f(2) + f(3) = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}\]Thus, the probability is \(\frac{5}{6}\).

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.

Random Variables
Random variables are a fundamental concept in statistics and probability theory. They offer a way to represent uncertain quantities. For instance, in the exercise provided, the random variable \( x \) represents the number of sessions required for a psychologist to gain the trust of a new patient.

Random variables can be either discrete or continuous. Here, \( x \) is a discrete random variable as it takes on distinct values: 1, 2, or 3. These values correspond to the number of sessions, a countable quantity. The probability function assigned to the random variable, \( f(x) = \frac{x}{6} \), helps quantify the likelihood of each possible outcome.
  • A discrete random variable has a finite or countable number of possible values.
  • All probabilities must add up to 1, confirming that every outcome has been accounted for.
Understanding random variables is crucial in statistics education as they form the basis of probability distribution and other advanced topics. By modeling real-world scenarios with random variables, we gain insights into how likely different outcomes are.
Probability Distribution
Probability distributions describe how probabilities are assigned to values of a random variable. In the given exercise, the probability distribution is defined by the function \( f(x) = \frac{x}{6} \) for \( x = 1, 2, \) or \( 3 \). This function outlines how likely it is for each number of sessions (1, 2, or 3) to be needed for trust.

For a probability distribution to be valid, two conditions must be met:
  • The probability assigned to each outcome must be between 0 and 1.
  • The sum of the probabilities across all potential outcomes must equal 1.
These conditions ensure that each outcome's likelihood is realistically represented and that all possibilities are considered.
Probability distributions are a key component of statistics education, as they allow for calculation of various probabilities and help in predicting future events or analyzing past data. They serve as a foundation for many statistical methods and applications.
Statistics Education
Statistics education involves equipping learners with the skills and understanding of collecting, analyzing, and interpreting data. It empowers them to make informed decisions based on empirical evidence. In the context of probability functions and distributions, students learn how to model real-world scenarios using mathematical tools.

To excel in statistics, students must grasp several core concepts:
  • Understanding of random variables and their application to statistics problems.
  • Knowledge of probability distributions and their role in assigning probabilities.
  • Ability to verify the correctness of functions ensuring they conform to statistical principles.
By mastering these concepts, students can accurately analyze data sets, compute probabilities, and draw meaningful conclusions from their findings. This knowledge is critical not only for academics but also for real-life decision-making 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

Cars arrive at a car wash randomly and independently; the probability of an arrival is the same for any two time intervals of equal length. The mean arrival rate is 15 cars per hour. What is the probability that 20 or more cars will arrive during any given hour of operation?

Twelve of the top 20 finishers in the 2009 PGA Championship at Hazeltine National Golf Club in Chaska, Minnesota, used a Titleist brand golf ball (GolfBallTest website, November 12,2009 . Suppose these results are representative of the probability that a randomly selected PGA Tour player uses a Titleist brand golf ball. For a sample of 15 PGA Tour players, make the following calculations. a. Compute the probability that exactly 10 of the 15 PGA Tour players use a Titleist brand golf ball. b. Compute the probability that more than 10 of the 15 PGA Tour players use a Titleist brand golf ball. c. For a sample of 15 PGA Tour players, compute the expected number of players who use a Titleist brand golf ball. d. For a sample of 15 PGA Tour players, compute the variance and standard deviation of the number of players who use a Titleist brand golf ball.

In November the U.S. unemployment rate was \(8.7 \%\) (U.S. Department of Labor website, January 10,2010 ). The Census Bureau includes nine states in the Northeast region. Assume that the random variable of interest is the number of Northeastern states with an unemployment rate in November that was less than \(8.7 \%\). What values may this random variable have?

Consider a Poisson distribution with a mean of two occurrences per time period. a. Write the appropriate Poisson probability function. b. What is the expected number of occurrences in three time periods? c. Write the appropriate Poisson probability function to determine the probability of \(x\) occurrences in three time periods. d. Compute the probability of two occurrences in one time period. e. Compute the probability of six occurrences in three time periods. f. Compute the probability of five occurrences in two time periods.

The National Basketball Association (NBA) records a variety of statistics for each team. Two of these statistics are the percentage of field goals made by the team and the percentage of three-point shots made by the team. For a portion of the 2004 season, the shooting records of the 29 teams in the NBA showed that the probability of scoring two points by making a field goal was \(.44,\) and the probability of scoring three points by making a threepoint shot was .34 (NBA website, January 3,2004 ). a. What is the expected value of a two-point shot for these teams? b. What is the expected value of a three-point shot for these teams? c. If the probability of making a two-point shot is greater than the probability of making a three-point shot, why do coaches allow some players to shoot the three-point shot if they have the opportunity? Use expected value to explain your answer.

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.