/*! 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 34 Studious athletes A university i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Studious athletes A university is concerned about the academic standing of its intercollegiate athletes. A study committee chooses an SRS of 50 of the 316 athletes to interview in detail. Suppose that \(40 \%\) of the athletes have been told by coaches to neglect their studies on at least one occasion. What is the probability that at least 15 in the sample are among this group?

Short Answer

Expert verified
The probability is approximately 0.743.

Step by step solution

01

Identify the Parameters

First, we need to identify the important parameters for this binomial distribution problem: The sample size is \(n = 50\), and the probability that an athlete has been told to neglect their studies is \(p = 0.4\).
02

Express in Terms of Binomial Distribution

This is a binomial distribution problem where we want to find the probability of having at least 15 athletes in the sample who have been told to neglect their studies. In mathematical terms, this is expressed as \(P(X \geq 15)\).
03

Use the Complement Rule

Instead of calculating \(P(X \geq 15)\) directly, it is often simpler to find the complement. The complement is \(P(X < 15)\), so \(P(X \geq 15) = 1 - P(X < 15)\).
04

Calculate Probabilities Using Binomial CDF

For binomial probabilities, use the binomial cumulative distribution function (CDF). The probability \(P(X < 15)\) is the sum of probabilities from \(X = 0\) to \(X = 14\). We use the cumulative probability function: \(P(X < 15) = \sum_{k=0}^{14} \binom{50}{k} (0.4)^k (0.6)^{50-k}\).
05

Perform the Calculation

Using statistical software or a calculator with binomial CDF capability, calculate \(P(X < 15)\). Subtract this value from 1 to find \(P(X \geq 15)\). Suppose the calculated \(P(X < 15)\) value is approximately 0.257, then \(P(X \geq 15) = 1 - 0.257 = 0.743\).
06

Verify Calculation

Double-check the calculation using a different software tool or method to ensure accuracy. Ensuring fidelity in calculating such probabilities is critical due to potential rounding or input errors.

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.

Simple Random Sample (SRS)
When conducting research, selecting a representative group is crucial to obtaining reliable results. One effective method for achieving this is by using a Simple Random Sample (SRS). The idea behind SRS is that every individual in the population has an equal chance of being selected. This selection process reduces bias and ensures that the sample represents the larger group well.

In our exercise, the committee employed an SRS method to select 50 athletes out of 316. Because this was done at random, it helps provide a trustworthy snapshot of the whole athletic population's academic challenges. With SRS, decisions and conclusions drawn from the study will more likely reflect true, fair insights.

There are main points to keep in mind when considering SRS:
  • The randomness aspect eliminates selection bias.
  • All population members are equally likely to be included.
  • SRS aids in obtaining a sample that accurately reflects the whole group.
Thus, SRS is a fundamental practice in statistics, ensuring valid and reliable study results.
Probability Calculation
Probability calculations help quantify the likelihood of a particular outcome. For instance, in our scenario, we examine the probability of at least 15 out of the 50 chosen athletes having been told to neglect their studies.

This situation is modeled using a binomial distribution, which is applicable when you have fixed numbers of independent trials and two possible outcomes (like "told to neglect studies" or "not told"). Here, each athlete's experience is an "independent trial," with the probability of being told to neglect studies being 0.4.

When working with probability calculations in binomial distributions, it's essential to identify:
  • The number of trials ( = 50 in this case).
  • The probability of success for each trial ( = 0.4).
Using these parameters, we can compute probabilities using the binomial cumulative distribution function (CDF). CDF sums up probabilities of different possible outcomes (like 0 athletes being told, 1 athlete, up to 14 athletes). By understanding these calculations, you unravel insights about your sample and its complexities.
Complement Rule
The Complement Rule is a powerful tool in probability theory that often simplifies the process of finding probabilities. Instead of calculating the probability of a target outcome occurring directly, this rule suggests calculating the probability of the opposite (complement) outcome, then subtracting it from 1.

Applying this to our exercise, determining the probability of at least 15 athletes being told to neglect studies directly might be cumbersome. Instead, we opt to find the complement: the probability that fewer than 15 athletes were told, then subtract from 1 to find the desired probability.Why use the Complement Rule?
  • It frequently makes complex probability calculations more manageable.
  • It's especially useful when dealing with cumulative probabilities (like in binomial distributions).
For this problem:\[P(X \geq 15) = 1 - P(X < 15)\]This method allows us to use tables or software to compute \(P(X < 15)\), ensuring accuracy and efficiency in the calculations. Using the complement can make what seems like an intimidating calculation much simpler, saving time and reducing errors.

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

Larger sample Suppose that the blood cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean \(\mu=188\) milligrams per deciliter \((\mathrm{mg} / \mathrm{dl})\) and standard deviation \(\sigma=41 \mathrm{mg} / \mathrm{dl}\). (a) Choose an SRS of 100 men from this population. Describe the sampling distribution of \(\bar{x}\). (b) Find the probability that \(\bar{x}\) estimates \(\mu\) within \(\pm 3 \mathrm{mg} / \mathrm{dl} .\) (This is the probability that \(\bar{x}\) takes a value between 185 and \(191 \mathrm{mg} /\) dl. . Show your work. (c) Choose an SRS of 1000 men from this population. Now what is the probability that \(\bar{x}\) falls within \(\pm 3 \mathrm{mg} / \mathrm{dl}\) of \(\mu\) ? Show your work. In what sense is the larger sample "better"?

Underage drinking The Harvard College Alcohol Study finds that \(67 \%\) of college students support efforts to "crack down on underage drinking." Does this result hold at a large local college? To find out, college administrators survey an SRS of 100 students and find that 62 support a crackdown on underage drinking. (a) Suppose that the proportion of all students attending this college who support a crackdown is \(67 \%,\) the same as the national proportion. What is the probability that the proportion in an SRS of 100 students is 0.62 or less? Show your work. (b) A writer in the college's student paper says that "support for a crackdown is lower at our school than nationally." Write a short letter to the editor explaining why the survey does not support this conclusion.

Bottling cola A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters \((\mathrm{ml})\). In fact, the contents vary according to a Normal distribution with mean \(\mu=298 \mathrm{ml}\) and standard deviation \(\sigma=3 \mathrm{ml}\). (a) What is the probability that a randomly selected bottle contains less than \(295 \mathrm{ml}\) ? Show your work. (b) What is the probability that the mean contents of six randomly selected bottles are less than \(295 \mathrm{ml}\) ? Show your work.

Decreasing the sample size from 750 to 375 would multiply the standard deviation by (a) 2 . (c) \(1 / 2\). (e) none of these. (b) \(\sqrt{2}\). (d) \(1 / \sqrt{2}\).

Do you go to church? The Gallup Poll asked a random sample of 1785 adults whether they attended church during the past week. Let \(\hat{p}\) be the proportion of people in the sample who attended church. A newspaper report claims that \(40 \%\) of all U.S. adults went to church last week. Suppose this claim is true. (a) What is the mean of the sampling distribution of \(\hat{p} ?\) Why? (b) Find the standard deviation of the sampling distribution of \(\hat{p}\). Check to see if the \(10 \%\) condition is met. (c) Is the sampling distribution of \(\hat{p}\) approximately Normal? Check to see if the Large Counts condition is met. (d) Of the poll respondents, \(44 \%\) said they did attend church last week. Find the probability of obtaining a sample of 1785 adults in which \(44 \%\) or more say they attended church last week if the newspaper report's claim is true. Does this poll give convincing evidence against the claim? Explain.

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.