/*! 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 58 Suppose that \(90 \%\) of all re... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(90 \%\) of all registered California voters favor banning the release of information from exit polls in presidential elections until after the polls in California close. A random sample of 25 registered California voters is to be selected. a. What is the probability that more than 20 favor the ban? b. What is the probability that at least 20 favor the ban? c. What are the mean value and standard deviation of the number of voters in the sample who favor the ban? d. If fewer than 20 voters in the sample favor the ban, is this at odds with the assertion that (at least) \(90 \%\) of California registered voters favors the ban? (Hint: Consider \(P(x<20)\) when \(p=.9 .)\)

Short Answer

Expert verified
a) The probability that more than 20 voters favor the ban is approximately 0.564. b) The probability that at least 20 voters favor the ban is approximately 0.866. c) The mean number of voters who favor the ban is 22.5 and the standard deviation is 1.5. d) Yes, fewer than 20 voters in the sample favoring the ban is at odds with the assertion that 90% favor the ban. The probability of getting fewer than 20 in a sample of 25, given that the true population proportion is 0.9, is quite small (approximately 0.01365).

Step by step solution

01

Determine the probability of more than 20 voters in favor of the ban

As each voter independently favors the ban with probability 0.9, the number who favor the ban follows a binomial distribution. The probability that more than 20 favor the ban is found by subtracting the cumulative probability up to 20 from 1. Using the formula for binomial probability: \( P(x=k) = C(n, k)*(p^k)*(1-p)^{n-k} \) and the fact that \( C(n, k) \) is the binomial coefficient, we find the probabilities for k = 0 to 20 and subtract the sum of these from 1.
02

Determine the probability of at least 20 voters in favor of the ban

To find the probability that at least 20 favor the ban, we calculate the cumulative probability up to and including 20. This includes the probabilities for k = 20, 21, 22, 23, 24, and 25.
03

Compute the mean value and standard deviation

The mean value for a binomial distribution is given by \( np \), and the standard deviation by \( sqrt(np(1-p)) \). Substituting n=25 and p=0.9 gives the mean and standard deviation.
04

Compare the output of a sample to a population assertion

In our case, fewer than 20 voters in the sample favoring the ban might be considered as evidence against the assertion that 90% of registered voters favor the ban. We can quantify how strong this evidence is by calculating \( P(x<20) \).

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 Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena. The central object is the probability of events occurring within a defined set of outcomes. In our context, it examines the likelihood of a certain number of California registered voters favoring the ban on exit poll information release.

To evaluate this, we use the concept of random variables, which are functions that assign a numerical value to each outcome in a sample space. The binomial distribution is a type of random variable that emerges when we consider the number of successes in a sequence of independent trials, here representing the voters' responses. Understanding the probability theory aids in predicting outcomes and making informed decisions based on the likelihood of various events.
Standard Deviation
Standard deviation is a statistic that measures the amount of variation or dispersion of a set of values. In simpler terms, it tells us how much the values (in this case, the number of voters who favor the ban) deviate from the average value (the mean).

A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. In the context of our exercise, the standard deviation helps us understand the variability in the number of voters who favor the ban. This is particularly useful when assessing whether the sample of 25 voters is representative of the population of all California voters.
Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of trials of a binary experiment. In our example, a binary experiment is whether a voter favors the ban or not, with only two possible outcomes: success (favoring the ban) or failure (not favoring the ban).

The probability distribution is characterized by two parameters: the number of trials () and the probability of success in each trial (p). The binomial distribution is used to calculate different probabilities, such as the likelihood that exactly ) voters favor the ban, at least ) voters are in favor, or fewer than ) voters favor it, which are critical inputs in our decision-making process.
Cumulative Probability
Cumulative probability refers to the probability that a random variable is less than or equal to a specified value. It is essentially the sum of probabilities for all outcomes up to and including the desired outcome. In the realm of the binomial distribution, it means accounting for all the probabilities of obtaining a number of successes from zero up to the specified number.

For example, to find the probability of at least 20 voters favoring the ban, we sum the probabilities of having 20, 21, 22, 23, 24, or 25 voters in favor. This cumulative approach gives us a complete picture of the likelihood of reaching a certain threshold and is critical in scenarios where we care about the probability of achieving a minimum number of successes, as is the case in the textbook exercise.

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

Symptom validity tests (SVTs) are sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for PostTraumatic Stress Disorder: Application of the Binomial Distribution" (journal of Anxiety Disorders [2008 ]: \(1297-1302\) ) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the MENT test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is \(.7\) (compared to \(.96\) for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=.70\) to compute various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: i. \(\quad P(x=42)\) ii. \(\quad P(x<42)\) iii. \(P(x \leq 42)\) c. Explain why the probabilities computed in Part (b) are not all equal. d. The authors computed the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=.96\), they found \(p(x \leq 42)=.000000000013\) Explain why the authors computed this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.

Flashlight bulbs manufactured by a certain company are sometimes defective. a. If \(5 \%\) of all such bulbs are defective, could the techniques of this section be used to approximate the probability that at least five of the bulbs in a random sample of size 50 are defective? If so, calculate this probability; if not, explain why not. b. Reconsider the question posed in Part (a) for the probability that at least 20 bulbs in a random sample of size 500 are defective.

Suppose that \(x\) has a binomial distribution with \(n=\) 50 and \(p=.6\), so that \(\mu=n p=30\) and \(\sigma=\sqrt{n p}(1-p)\) \(=3.4641 .\) Calculate the following probabilities using the normal approximation with the continuity correction: a. \(P(x=30)\) b. \(P(x=25)\) c. \(P(x \leq 25)\) d. \(\quad P(25 \leq x \leq 40)\) e. \(P(25

Consider two binomial experiments. a. The first binomial experiment consists of six trials. How many outcomes have exactly one success, and what are these outcomes? b. The second binomial experiment consists of 20 trials. How many outcomes have exactly 10 successes? exactly 15 successes? exactly five successes?

The number of vehicles leaving a turnpike at a certain exit during a particular time period has approximately a normal distribution with mean value 500 and standard deviation \(75 .\) What is the probability that the number of cars exiting during this period is a. At least 650 ? b. Strictly between 400 and 550 ? (Strictly means that the values 400 and 550 are not included.) c. Between 400 and 550 (inclusive)?

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.