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Chapter 7: Problem 49

In 2018 it was estimated that approximately \(45 \%\) of the American population watches the Super Bowl yearly. Suppose a sample of 120 Americans is randomly selected. After verifying the conditions for the Central Limit Theorem are met, find the probability that at the majority (more than \(50 \%\) ) watched the Super Bowl. (Source: vox.com)

Short Answer

Expert verified
20%. That is to say, the probability that more than 50% of a random sample of 120 Americans watched the Super Bowl is around 20%.

Step by step solution

01

Identify Given Information

The question provides the necessary information. It tells that estimated percentage of people who watch the super bowl is \(45%\) or \(0.45\) in decimal form. We also know that the sample size(n) is 120. We are interested in finding out the probability that more than \(50%\) people from this sample watch the super bowl which will be more than 60 people (since \(50%\) of 120 gives 60).
02

Calculate Mean and Standard Deviation for Binomial Distribution

Since, under the Central Limit Theorem, a binomial distribution with a large sample size and a probability near 0.5 approximates a normal distribution, we can calculate the mean and the standard deviation for a binomial distribution. The mean for a binomial distribution, by definition, is \(np\), and the standard deviation is \(\sqrt{npq}\), where \(q = 1 - p\). So, if \(n = 120\), \(p = 0.45\), then \(q = 1 - 0.45 = 0.55\). Thus, the mean \(\mu = np = 120 * 0.45 = 54\) and the standard deviation \(\sigma = \sqrt{npq} = \sqrt{120*0.45*0.55} \approx 6.95\).
03

Standardize to a Z-Score

To find the probability that more than 60 people will be watching the Super Bowl, convert that number to a Z-Score to apply it to the standardized normal distribution. The Z-Score for a value can be calculated with the formula \((value - mean) / standard\ deviation\). Here \(value = 60\), \(mean = 54\) and \(standard\ deviation = 6.95\). This gives us a Z-Score of \((60 - 54) / 6.95 \approx 0.86\). Note that we need the probability that more than 60 people will be watching the Super Bowl, so we're looking for the part of the distribution greater than this Z-Score.
04

Find the Probability

Once the Z-Score is found, use the standard normal distribution to find the desired probability. Because a z-table usually only gives the probability to the left of the observed Z-Score, and we want the probability to the right of the Z-Score, calculate \(1 - P(Z < 0.86)\). Using a standard Z-Table or calculator, we find that \(P(Z < 0.86) = 0.80\). Thus, \(1 - P(Z < 0.86) = 1 - 0.80 = 0.20\) or \(20\%\).

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Key Concepts

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

Probability
Probability is the measure of the likelihood that an event will occur. It's quantified as a number between 0 and 1, with 0 indicating that the event will not occur, and 1 indicating certainty that the event will occur.

When dealing with a real-world scenario, like estimating how many people watch the Super Bowl, we often use the concept of probability to make predictions about the outcome. In the given exercise, we are particularly interested in the probability of a sample of Americans where more than half watch the Super Bowl. By understanding probability, we can make informed guesses about the likelihood of certain events based on given data.
Binomial Distribution
A binomial distribution represents the number of successes in a fixed number of independent experiments, each asking a yes/no question, and each with its own boolean-valued outcome: a success or failure.

In this case, we're looking at a sample of 120 people and questioning whether each person watches the Super Bowl (success) or not (failure). The probability of success for each trial is 0.45. A binomial distribution is defined by two parameters: the number of trials (n) and the probability of success (p). The mean of a binomial distribution is calculated as \(np\), and the standard deviation as \(\sqrt{npq}\), where q is the probability of failure (1-p).
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution that has a mean of 0 and a standard deviation of 1. It is represented by the Z-distribution and is used to find probabilities and percentages for normal distributions after standardizing.

To relate a binomial distribution to this standard normal distribution, we employ the Central Limit Theorem, which states that the distribution of sample means approximates a normal distribution as the sample size becomes larger, irrespective of the shape of the population distribution. In this exercise, because we have a large sample size (n=120), we use the standard normal distribution to approximate the binomial distribution of Super Bowl watchers.
Z-Score
A Z-Score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is measured in terms of standard deviations from the mean. If a Z-Score is 0, it indicates that the data point's score is identical to the mean score.

A Z-Score is important because it allows for the standardization of scores on different scales, making it possible to calculate the probability of a score occurring within a normal distribution. This is exactly what we did in the exercise: we took the number of people who watched the Super Bowl (60), subtracted the mean of the distribution (54), and then divided by the standard deviation (approx. 6.95) to find the Z-Score, which we then used to find the desired probability.

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Most popular questions from this chapter

A 2017 Gallup poll reported that 658 out of 1028 U.S. adults believe that marijuana should be legalized. When Gallup first polled U.S. adults about this subject in 1969 , only \(12 \%\) supported legalization. Assume the conditions for using the CLT are met. a. Find and interpret a \(99 \%\) confidence interval for the proportion of U.S. adults in 2017 that believe marijuana should be legalized. b. Find and interpret a \(95 \%\) confidence interval for this population parameter. c. Find the margin of error for each of the confidence intervals found in parts a and b. d. Without computing it, how would the margin of error of a \(90 \%\) confidence interval compare with the margin of error for the \(95 \%\) and \(99 \%\) intervals? Construct the \(90 \%\) confidence interval to see if your prediction was correct.

From Formula \(7.2\), an estimate for margin of error for a \(95 \%\) confidence interval is \(m=2 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\mathrm{n}\) is the required sample size and \(\hat{p}\) is the sample proportion. Since we do not know a value for \(\hat{p}\), we use a conservative estimate of \(0.50\) for \(\hat{p}\). Replace \(\hat{p}\) with \(0.50\) in the formula and simplify.

While the majority of people who are color blind are male, the National Eye Institute reports that \(0.5 \%\) of women of with Northern European ancestry have the common form of red-green color blindness. Suppose a random sample of 100 women with Northern European ancestry is selected. Can we find the probability that less than \(0.3 \%\) of the sample is color blind? If so, find the probability. If not, explain why this probability cannot be calculated.

According to The Washington Post, \(72 \%\) of high school seniors have a driver's license. Suppose we take a random sample of 100 high school seniors and find the proportion who have a driver's license. a. What value should we expect for our sample proportion? b. What is the standard error? c. Use your answers to parts a and \(\mathrm{b}\) to complete this sentence: We expect _____\(\%\) to have their driver's license, give or take ____\(\%\). d. Suppose we increased the sample size from 100 to 500 . What effect would this have on the standard error? Recalculate the standard error to see if your prediction was correct.

In 2003 and 2017 Gallup asked Democratic voters about their views on the FBI. In \(2003,44 \%\) thought the \(\mathrm{FBI}\) did a good or excellent job. In \(2017,69 \%\) of Democratic voters felt this way. Assume these percentages are based on samples of 1200 Democratic voters. a. Can we conclude, on the basis of these two percentages alone, that the proportion of Democratic voters who think the FBI is doing a good or excellent job has increase from 2003 to \(2017 ?\) Why or why not? b. Check that the conditions for using a two-proportion confidence interval hold. You can assume that the sample is a random sample. c. Construct a \(95 \%\) confidence interval for the difference in the proportions of Democratic voters who believe the FBI is doing a good or excellent job, \(p_{1}-p_{2}\). Let \(p_{1}\) be the proportion of Democratic voters who felt this way in 2003 and \(p_{2}\) be the proportion of Democratic voters who felt this way in 2017 . d. Interpret the interval you constructed in part c. Has the proportion of Democratic voters who feel this way increased? Explain.
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