/*! 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 9 Susan is taking Western Civiliza... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Susan is taking Western Civilization this semester on a pass/fail basis. The department teaching the course has a history of passing \(77 \%\) of the students in Western Civilization each term. Let \(n=1\) \(2,3, \ldots\) represent the number of times a student takes western civilization until the first passing grade is received. (Assume the trials are independent.) (a) Write out a formula for the probability distribution of the random variable \(n\) (b) What is the probability that Susan passes on the first try \((n=1) ?\) (c) What is the probability that Susan first passes on the second try \((n=2) ?\) (d) What is the probability that Susan needs three or more tries to pass western civilization? (e) What is the expected number of attempts at western civilization Susan must make to have her (first) pass? Hint: Use \(\mu\) for the geometric distribution and round.

Short Answer

Expert verified
(a) \( P(n = k) = (1 - 0.77)^{k-1} \times 0.77 \); (b) 0.77; (c) 0.1771; (d) 0.0529; (e) 1.30 attempts.

Step by step solution

01

Understand the Problem

The problem involves a sequential process until a success is reached (Susan passing the course), represented by the random variable \( n \), indicating the trial number on which the first success occurs. This is a classic setup for a geometric distribution, where each trial is independent, and the probability of success in each trial is constant at 77%.
02

Formula for Geometric Distribution

For a geometric distribution, the probability that the first success occurs on the \(n\)-th trial is given by: \( P(n = k) = (1 - p)^{k-1} \times p \), where \( p \) is the probability of success (passing the course), and \( 1 - p \) is the probability of failure. In this case, \( p = 0.77 \).
03

Calculate Probability for First Try

For Susan to pass on her first try \((n=1)\), substitute \( n = 1 \) into the formula: \( P(n = 1) = (1 - 0.77)^{1-1} \times 0.77 = 0.77 \).
04

Calculate Probability for Second Try

To find the probability that Susan first passes on her second try \((n=2)\), substitute \( n = 2 \) into the formula: \( P(n = 2) = (1 - 0.77)^{2-1} \times 0.77 = 0.23 \times 0.77 = 0.1771 \).
05

Probability for Three or More Tries

The probability that Susan passes on the third or later try is the complement of passing within the first two tries: \( P(n \geq 3) = 1 - (P(n =1) + P(n=2)) = 1 - (0.77 + 0.1771) = 0.0529 \).
06

Expected Value for Geometric Distribution

The expected number of trials \( \mu \) until the first success for a geometric distribution is calculated as \( \mu = \frac{1}{p} \). Here, \( \mu = \frac{1}{0.77} \approx 1.30 \). Thus, Susan is expected to pass approximately after 1.3 attempts.

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 Distribution
The concept of a probability distribution is key in understanding events governed by randomness. It describes how probabilities are assigned to outcomes of a random variable, indicating the likelihood of occurrence of each outcome in a specific scenario. In our example of a geometric distribution, the probability distribution is focused on identifying when the first success occurs, given repeated independent trials. This is particularly useful in situations where events, like Susan passing her Western Civilization course, happen under consistent conditions. For the geometric distribution applied here, the probability that success occurs on the nth try is calculated with the formula:
  • \( P(n = k) = (1 - p)^{k-1} \times p \)
where:
  • \( p \) is the probability of passing, which is 0.77,
  • \(1 - p \) represents the probability of not passing on any particular attempt.
This distribution helps calculate exact probabilities for each potential number of trials Susan may go through before passing.
Expected Value
The expected value is a critical concept that provides the average outcome of a random variable over numerous trials. For a geometric distribution like Susan's passing attempts, the expected value tells us the average number of tries before achieving the first success.Expected value is given by:
  • \( \mu = \frac{1}{p} \)
For Susan, the probability of success \( p \) is 0.77, leading to an expected value:
  • \( \mu = \frac{1}{0.77} \approx 1.30 \)
This means on average, Susan needs about 1.3 attempts to pass her course. Note that you can't actually have 1.3 attempts in real life, but the number represents a statistical average, indicating that Susan is more likely to pass in one or two tries. Understanding the expected value here provides insight into how often Susan generally needs to attempt the course to succeed, reflecting the efficiency of the probability distribution.
Random Variable
A random variable is a mathematical representation of outcomes from a random process. It is a crucial element in probability and statistics, allowing us to quantify outcomes numerically.In Susan's case, the random variable \( n \) represents the number of times she attempts the Western Civilization course until she passes. Each attempt is a trial, and the outcome (passing or not passing) is what we're interested in capturing with this variable.The random variable in a geometric distribution like this one focuses on the trial number where the first success occurs. Since it tracks discrete attempts, it's a discrete random variable. The use of a random variable provides a structured way to evaluate the process, enabling us to apply probability models like the geometric distribution to understand and calculate the likelihood of different outcomes.Recognizing how a random variable operates in the context of attempts and outcomes can make the analysis of courses, exams, and similar scenarios more intuitive and manageable.

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

Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in USA Today. According to the survey, \(37 \%\) of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability that (a) out of five adults, none is concerned that employers are monitoring phone calls. (b) out of five adults, all are concerned that employers are monitoring phone calls. (c) out of five adults, exactly three are concerned that employers are monitoring phone calls.

Consider a binomial distribution with \(n=10\) trials and the probability of success on a single trial \(p=0.85\) (a) Is the distribution skewed left, skewed right, or symmetric? (b) Compute the expected number of successes in 10 trials. (c) Given the high probability of success \(p\) on a single trial, would you expect \(P(r \leq 3)\) to be very high or very low? Explain. (d) Given the high probability of success \(p\) on a single trial, would you expect \(P(r \geq 8)\) to be very high or very low? Explain.

Consider a binomial experiment with \(n=20\) trials and \(p=0.40\) (a) Find the expected value and the standard deviation of the distribution. (b) Would it be unusual to obtain fewer than 3 successes? Explain. Confirm your answer by looking at the binomial probability distribution table.

Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backward by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination (Source: Lies' Liesi't Lies\%' The Psychology of Deceit, by C. V. Ford, professor of psychiatry, University of Alabama). In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), \(85 \%\) of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of nine students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What is the probability that (a) all the students are able to pass the polygraph examination? (b) more than half the students are able to pass the polygraph examination? (c) no more than four of the students are able to pass the polygraph examination? (d) all the students fail the polygraph examination?

Which of the following are continuous variables, and which are discrete? (a) Number of traffic fatalities per year in the state of Florida (b) Distance a golf ball travels after being hit with a driver (c) Time required to drive from home to college on any given day (d) Number of ships in Pearl Harbor on any given day (e) Your weight before breakfast each morning

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.