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According to the college registrar's office, \(40 \%\) of students enrolled in an introductory statistics class this semester are freshmen, \(25 \%\) are sophomores, \(15 \%\) are juniors, and \(20 \%\) are seniors. You want to determine the probability that in a random sample of five students enrolled in introductory statistics this semester, exactly two are freshmen. (a) Describe a trial. Can we model a trial as having only two outcomes? If so, what is success? What is failure? What is the probability of success? (b) We are sampling without replacement. If only 30 students are enrolled in introductory statistics this semester, is it appropriate to model 5 trials as independent, with the same probability of success on each trial? Explain. What other probability distribution would be more appropriate in this setting?

Short Answer

Expert verified
A trial is successful if a freshman is selected, with a success probability of 0.40. Since sampling is without replacement from 30 students, use the hypergeometric distribution.

Step by step solution

01

Define the Trial and Possible Outcomes

A trial consists of selecting one student at random from those enrolled in the introductory statistics class and determining whether the student is a freshman or not. We can model the trial as having two outcomes: "success" if the student is a freshman and "failure" if the student is not a freshman.
02

Determine the Probability of Success

Given the problem, the probability that a randomly selected student is a freshman (probability of success) is 40%. Thus, we have \( P(\text{success}) = 0.40 \) and \( P(\text{failure}) = 1 - 0.40 = 0.60 \).
03

Consider Sampling Without Replacement

If we sample without replacement and there are only 30 students, it's not suitable to assume trials are independent with the same probability of success because the sample is 5 out of 30, affecting the probability with each draw.
04

Suggest Suitable Probability Distribution

Due to the changing probabilities from sampling without replacement, it is more appropriate to use the hypergeometric distribution rather than the binomial distribution. The hypergeometric distribution accounts for trials dependent on each other with finite populations.

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Key Concepts

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

Understanding the Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent and identical trials, each with the same probability of success. Imagine flipping a coin - head could be a success, and tail a failure. If you flip a coin ten times, you might want to know the probability of getting exactly six heads.

In this example, each flip (or trial) is independent of the others, meaning that the result of one flip doesn't affect the outcome of another. Furthermore, the probability of landing heads (success) remains the same for each flip. This makes our ten coin flips a perfect candidate for a binomial distribution model.

When dealing with cases like students in a statistics class, it means each student could either be a freshman (success) or not a freshman (failure). If we assume there's a large enough population, selecting students can also mimic the independent trials seen in coin flipping. However, remember, real-life problems often have constraints that should be considered, like changes in probability with each selection.
Exploring the Hypergeometric Distribution
The hypergeometric distribution becomes important when your scenario involves sampling without replacement, leading to dependent trials. This distribution applies when you draw items without putting them back, just like picking colored balls from a bag and not returning them before the next pick.

Let's say you have a small class of thirty students, where some are freshmen. If you sample five students (without replacement), the probability of selecting a freshman changes with each pick. As you choose more students, the remaining population shifts, affecting your chances of drawing another freshman. This dependency means that each choice isn't independent from the last, differentiating it from a binomial scenario.

The hypergeometric distribution appropriately models such a scenario by considering the finite population's changing structure, helping predict probabilities as you remove individuals from the pool.
Concept of Sampling Without Replacement
Sampling without replacement is a method where each sample drawn from a population isn't returned before the next draw. This approach contrasts with sampling with replacement, where each sampled item is returned, keeping the population size constant.

Imagine a bowl of marbles: if you pick one and don't replace it, the number of marbles decreases, altering the odds for the subsequent draw. This scenario makes the trials interdependent, as previous selections influence future probabilities.

In statistical exercises, like determining class composition, sampling without replacement means the probability of choosing a freshman changes depending on the prior selections. This method is realistic for small populations, like the thirty-student class example, where each draw directly impacts future outcomes. Such dependency requires considering the hypergeometric distribution for accurate probability calculations.

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Most popular questions from this chapter

What does the random variable for a binomial experiment of \(n\) trials measure?

Harper's Index reported that the number of (Orange County, California) convicted drunk drivers whose sentence included a tour of the morgue was 569 , of which only 1 became a repeat offender. (a) Suppose that of 1000 newly convicted drunk drivers, all were required to take a tour of the morgue. Let us assume that the probability of a repeat offender is still \(p=1 / 569 .\) Explain why the Poisson approximation to the binomial would be a good choice for \(r=\) number of repeat offenders out of 1000 convicted drunk drivers who toured the morgue. What is \(\lambda\) to the nearest tenth? (b) What is the probability that \(r=0\) ? (c) What is the probability that \(r>1\) ? (d) What is the probability that \(r>2\) ? (e) What is the probability that \(r>3\) ?

Henry Petroski is a professor of civil engineering at Duke University. In his book To Engineer Is Human: The Role of Failure in Successful Design, Professor Petroski says that up to \(95 \%\) of all structural failures, including those of bridges, airplanes, and other commonplace products of technology, are believed to be the result of crack growth. In most cases, the cracks grow slowly. It is only when the cracks reach intolerable proportions and still go undetected that catastrophe can occur. In a cement retaining wall, occasional hairline cracks are normal and nothing to worry about. If these cracks are spread out and not too close together, the wall is considered safe. However, if a number of cracks group together in a small region, there may be real trouble. Suppose a given cement retaining wall is considered safe if hairline cracks are evenly spread out and occur on the average of \(4.2\) cracks per 30 -foot section of wall. (a) Explain why a Poisson probability distribution would be a good choice for the random variable \(r=\) number of hairline cracks for a given length of retaining wall. (b) In a 50 -foot section of safe wall, what is the probability of three (evenly spread out) hairline cracks? What is the probability of three or more (evenly spread out) hairline cracks? (c) Answer part (b) for a 20-foot section of wall. (d) Answer part (b) for a 2 -foot section of wall. Round \(\lambda\) to the nearest tenth. (e) Consider your answers to parts (b), (c), and (d). If you had three hairline cracks evenly spread out over a 50 -foot section of wall, should this be cause for concern? The probability is low. Could this mean that you are lucky to have so few cracks? On a 20 -foot section of wall [part (c)], the probability of three cracks is higher. Does this mean that this distribution of cracks is closer to what we should expect? For part (d), the probability is very small. Could this mean you are not so lucky and have something to worry about? Explain your answers.

Consider two binomial distributions, with \(n\) trials each. The first distribution has a higher probability of success on each trial than the second. How does the expected value of the first distribution compare to that of the second?

Which of the following are continuous variables, and which are discrete? (a) Speed of an airplane (b) Age of a college professor chosen at random (c) Number of books in the college bookstore (d) Weight of a football player chosen at random (e) Number of lightning strikes in Rocky Mountain National Park on a given day

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