Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How often do you watch movies at home? A USA Today Snapshot found that about 7 in 10 adults say they watch movies at home at least once a week. \(^{5}\) Suppose a random sample of \(n=50\) adults are polled and asked if they had watched a movie at home this week. Let us assume that \(p=.7\) is, in fact, correct. What are the probabilities for the following events? a. Fewer than 30 individuals watched a movie at home this week? b. More than 42 individuals watched a movie at home this week? c. Fewer than 10 individuals did not watch a movie at home this week?

Short Answer

Expert verified
Question: In a random sample of 50 adults, the probability of an adult watching a movie at home at least once a week is 0.7. Calculate the probabilities of the following events: a. Fewer than 30 individuals watched a movie at home this week. b. More than 42 individuals watched a movie at home this week. c. Fewer than 10 individuals did not watch a movie at home this week. Solution: a. The probability that fewer than 30 individuals watched a movie at home this week is approximately __(insert calculated value)__. b. The probability that more than 42 individuals watched a movie at home this week is approximately __(insert calculated value)__. c. The probability that fewer than 10 individuals did not watch a movie at home this week is approximately __(insert calculated value)__.

Step by step solution

01

Determine the binomial distribution function formula

For binomial distribution problems, we use the binomial distribution function formula: \(P(X=k) = {n \choose k} \cdot p^k \cdot (1-p)^{n-k}\) where: - \(n\) is the number of trials - \(k\) is the number of successes - \(p\) is the probability of success - \({n \choose k} = \frac{n!}{k!(n-k)!}\), which is the combination formula. We will use this formula to find the probabilities for the given events.
02

Calculate the probability for event a

Event a: Fewer than 30 individuals watched a movie at home this week. To find this probability, we will need to sum the probabilities of 0 to 29 individuals watching a movie, using the formula from step 1 with \(k\) ranging from 0 to 29. \(P(X<30) = \sum_{k=0}^{29} {n \choose k} \cdot p^k \cdot (1-p)^{n-k}\) Replace n with 50 and p with 0.7, and compute the sum to get the probability.
03

Calculate the probability for event b

Event b: More than 42 individuals watched a movie at home this week. To find this probability, we will need to sum the probabilities of 43 to 50 individuals watching a movie, using the formula from step 1 with \(k\) ranging from 43 to 50. \(P(X>42) = \sum_{k=43}^{50} {n \choose k} \cdot p^k \cdot (1-p)^{n-k}\) Replace n with 50 and p with 0.7, and compute the sum to get the probability.
04

Calculate the probability for event c

Event c: Fewer than 10 individuals did not watch a movie at home this week. Since 50 - 10 = 40, this is equivalent to at least 40 individuals watching a movie, so we need to sum the probabilities of 40 to 50 individuals watching a movie, using the formula from step 1 with \(k\) ranging from 40 to 50. \(P(X\geq40) = \sum_{k=40}^{50} {n \choose k} \cdot p^k \cdot (1-p)^{n-k}\) Replace n with 50 and p with 0.7, and compute the sum to get the probability. After calculating the probabilities for each event, the final answers will be: a. The probability that fewer than 30 individuals watched a movie at home this week is approximately \(__\). b. The probability that more than 42 individuals watched a movie at home this week is approximately \(__\). c. The probability that fewer than 10 individuals did not watch a movie at home this week is approximately \(__\).

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 Vaia!

Key Concepts

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

Understanding Probability
Probability is the measure of the likelihood of an event occurring. In the context of our problem, we are interested in finding the probability of a certain number of adults watching a movie at home. Probability values range from 0 to 1, where 0 indicates an impossible event and 1 indicates a certain event.

To find probabilities in scenarios involving several outcomes, we can use formulas and distribution models. One such model is the binomial distribution which is relevant when we have a fixed number of trials, like asking 50 adults in a survey. Each trial is independent, and there are only two outcomes (e.g., watching a movie or not watching a movie).
  • Probability is used to quantify uncertainty and assess the potential outcomes of different events.
  • It helps in predicting how likely it is for a specific event to occur within a fixed number of trials.
Exploring probability in everyday contexts helps you understand patterns and predictions, which are useful skills for many real-world applications.
The Role of Random Sampling
Random sampling is a method used to select a group of individuals from a larger population, ensuring that each individual has an equal chance of being chosen. This technique helps reduce bias, allowing for more accurate and reliable inferences about a population.

In the movie-watching scenario, by selecting a random sample of 50 adults, we assume that each sample has a similar distribution to the overall population. This is why the random sampling method is essential—it helps to ensure that the data collected reflects broader trends.
  • Random sampling ensures that our conclusions are not skewed by selecting only a specific type of participant.
  • It's a cornerstone of statistical analysis, helping to confirm that findings are representative of the wider population.
Effective random sampling means that the probabilities we calculate are a good approximation of the population's actual behaviors.
Applying the Binomial Probability Formula
The binomial probability formula allows us to calculate the likelihood of achieving exactly "k" successes in "n" trials, where the probability of success on each trial is "p." In this context, a "success" is defined as an adult watching a movie.

The formula is:\[ P(X=k) = {n \choose k} \cdot p^k \cdot (1-p)^{n-k} \]Here:
  • "\(n\)" represents the total number of trials (e.g., 50 adults participating in the survey).
  • "\(k\)" is the number of successful outcomes (e.g., the number of adults who watch a movie).
  • "\(p\)" is the probability of a single trial resulting in success (e.g., 0.7 or 70%).
The formula also includes combinations, which help in determining how many ways "k" successes can occur in "n" trials. This is calculated using:\[{n \choose k} = \frac{n!}{k!(n-k)!} \]This approach is invaluable for understanding binomial distribution, enabling you to solve problems involving different numbers of successes in a set number of trials.

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

Consider a binomial experiment with \(n=20\) and \(p=.4 .\) Calculate \(P(x \geq 10)\) using each of these methods: a. Table 1 in Appendix I b. The normal approximation to the binomial probability distribution

A researcher notes that senior corporation executives are not very accurate forecasters of their own annual earnings. He states that his studies of a large number of company executive forecasts "showed that the average estimate missed the mark by \(15 \%\)." a. Suppose the distribution of these forecast errors has a mean of \(15 \%\) and a standard deviation of \(10 \%\). Is it likely that the distribution of forecast errors is approximately normal? b. Suppose the probability is .5 that a corporate executive's forecast error exceeds \(15 \% .\) If you were to sample the forecasts of 100 corporate executives, what is the probability that more than 60 would be in error by more than \(15 \% ?\)

Is a tall president better than a short one? Do Americans tend to vote for the taller of the two candidates in a presidential selection? In 33 of our presidential elections between 1856 and \(2006,\) 17 of the winners were taller than their opponents. Assume that Americans are not biased by a candidate's height and that the winner is just as likely to be taller or shorter than his opponent. Is the observed number of taller winners in the U.S. presidential elections unusual? a. Find the approximate probability of finding 17 or more of the 33 pairs in which the taller candidate wins. b. Based on your answer to part a, can you conclude that Americans might consider a candidate's height when casting their ballot?

Find the following probabilities for the standard normal random variable: a. \(P(.3

The discharge of suspended solids from a phosphate mine is normally distributed, with a mean daily discharge of 27 milligrams per liter \((\mathrm{mg} / \mathrm{l})\) and a standard deviation of \(14 \mathrm{mg} / \mathrm{l}\). What proportion of days will the daily discharge exceed \(50 \mathrm{mg} / \mathrm{l} ?\)

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.

Sign-up for free