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Chapter 6: Problem 17

A recent survey by the American Accounting Association revealed \(23 \%\) of students graduating with a major in accounting select public accounting. Suppose we select a sample of 15 recent graduates. a. What is the probability two select public accounting? b. What is the probability five select public accounting? c. How many graduates would you expect to select public accounting?

Short Answer

Expert verified
a) 0.231, b) 0.237, c) 3.45 graduates.

Step by step solution

01

Understanding the Problem

We need to find the probability that a certain number of accounting graduates choose public accounting from a sample of 15. Additionally, we need to calculate the expected number of graduates selecting public accounting. Since this is a binomial probability problem, where each graduate can either select public accounting or not, we will use the binomial probability formula.
02

Defining the Parameters

Define the parameters for a binomial distribution: number of trials \( n = 15 \), probability of success (choosing public accounting) \( p = 0.23 \). We will calculate the probability of exactly 2 and 5 successes and also determine the expected number of graduates choosing public accounting.
03

Using the Binomial Probability Formula

The binomial probability formula is given by \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( \binom{n}{k} \) is the binomial coefficient.
04

Calculating Probability for Two Graduates

Using the formula for \( k = 2 \): \[ P(X = 2) = \binom{15}{2} (0.23)^2 (1-0.23)^{15-2} \]Calculate \( \binom{15}{2} = \frac{15 \times 14}{2 \times 1} = 105 \).Then, \( (0.23)^2 = 0.0529 \) and \( (0.77)^{13} \approx 0.047893 \).So, \[ P(X = 2) \approx 105 \times 0.0529 \times 0.047893 \approx 0.231 \]
05

Calculating Probability for Five Graduates

For \( k = 5 \):\[ P(X = 5) = \binom{15}{5} (0.23)^5 (1-0.23)^{15-5} \]Calculate \( \binom{15}{5} = 3003 \).Then, \( (0.23)^5 \approx 0.002985983 \) and \( (0.77)^{10} \approx 0.196874404 \).So, \[ P(X = 5) \approx 3003 \times 0.002985983 \times 0.196874404 \approx 0.237 \]
06

Finding the Expected Number of Graduates

The expected number of graduates selecting public accounting can be found using the formula for expected value of a binomial distribution, \( E(X) = n \times p \).Substitute the given values to find the expected number: \[ E(X) = 15 \times 0.23 = 3.45 \] graduates.

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Key Concepts

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

Probability
Probability is the likelihood of an event happening. In this context, it refers to how likely it is that a certain number of accounting graduates choose public accounting. The probability value is always between 0 and 1, where 0 means an impossible event, and 1 means a certain event. The probability of different outcomes can be calculated with various mathematical tools, such as probability formulas and equations.
In binomial experiments, such as the one with the accounting graduates, each trial is independent which means the outcome of one does not affect the next. For each graduate, there are two possible outcomes: they either choose public accounting or they don't.
  • When we calculate the probability of two graduates choosing public accounting, we are looking for a specific outcome from a larger sample, which involves mathematical calculations based on combinations and the probability of individual events.
Expected Value
The expected value is a predicted value of a variable and is often seen as the average outcome if the experiment were to be repeated many times. In a binomial distribution, like the graduates, the expected value helps to determine a central tendency.
The formula to find the expected value, which is denoted as \( E(X) \), is the product of the number of trials \( n \) and the probability of success \( p \). This formula is given by \( E(X) = n \times p \).
  • This means for our graduates scenario: if 15 students are graduating, 23% are expected to go into public accounting, so we'd expect around 3 or 4 students (when rounded) to choose this path.
  • This method gives an average expectation in the long run, helping to understand the overall behavior of the distribution.
Binomial Probability Formula
The binomial probability formula is essential for solving problems where there are two possible outcomes, like our graduates situation. The formula calculates the probability that a given number of successes, that is, students choosing public accounting, will occur out of a fixed number of trials.
The formula is \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \).
  • Here, \( \binom{n}{k} \) is the binomial coefficient, representing the number of ways to choose \( k \) successes from \( n \) trials.
  • \( p \) is the probability of success on an individual trial, and \( 1-p \) is the probability of failure.
Grab your calculator for these computations, as they involve multiplying large numbers.The binomial probability begins to give you a clearer picture of how likely certain outcomes are, helping with predictions based on initial parameters.
You can break down the calculation steps just like in the example to determine specific counts of success.

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Most popular questions from this chapter

For each of the following, indicate whether the random variable is discrete or continuous. a. The length of time to get a haircut. b. The number of cars a jogger passes each morning while running. c. The number of hits for a team in a high school girls' softball game. d. The number of patients treated at the South Strand Medical Center between 6 and 10 p.m. each night. e. The distance your car traveled on the last fill-up. f. The number of customers at the Oak Street Wendy's who used the drive- through facility. g. The distance between Gainesville, Florida, and all Florida cities with a population of at least 50,000 .

Acceptance sampling is a statistical method used to monitor the quality of purchased parts and components. To ensure the quality of incoming parts, a purchaser or manufacturer normally samples 20 parts and allows one defect. a. What is the likelihood of accepting a lot that is \(1 \%\) defective? b. If the quality of the incoming lot was actually \(2 \%,\) what is the likelihood of accepting it? c. If the quality of the incoming lot was actually \(5 \%,\) what is the likelihood of accepting it?

Recent statistics suggest that \(15 \%\) of those who visit a retail site on the Internet make a purchase. A retailer wished to verify this claim. To do so, she selected a sample of 16 "hits" to her site and found that 4 had actually made a purchase. a. What is the likelihood of exactly four purchases? b. How many purchases should she expect? c. What is the likelihood that four or more "hits" result in a purchase?

The U.S. Postal Service reports \(95 \%\) of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to different locations. a. What is the probability that all six arrive within 2 days? b. What is the probability that exactly five arrive within 2 days? c. Find the mean number of letters that will arrive within 2 days. d. Compute the variance and standard deviation of the number that will arrive within 2 days.

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly one non-work-related e-mail between 4 p.m. and 5 p.m. yesterday? b. What is the probability she received five or more non-work-related e-mails during the same period? c. What is the probability she did not receive any non-work-related e-mails during the period?
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