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Chapter 7: Problem 41

According to a 2017 article in 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. Find the probability that more than \(75 \%\) of the sample has a driver's license. Begin by verifying that the conditions for the Central Limit Theorem for Sample Proportions have been met.

Short Answer

Expert verified
Based on Central Limit Theorem for sample proportions, the probability that more than 75% of a random sample of 100 high school seniors has a driver's license is approximately 25.78%.

Step by step solution

01

Conditions for Central Limit Theorem

To verify the conditions for the Central Limit Theorem for Sample Proportions, we check two conditions: \n\n1. The sample size should be less than 10% of the population. As high school seniors considerably number in the millions, a sample size of 100 is less than 10%. \n\n2. The sample size (\(n \times p\)) and (\(n \times (1 - p)\)) should be greater than or equal to 5. If we assume \(p = 0.72\), then \(n \times p = 100 \times 0.72 = 72\) and \(n \times (1 - p) = 100 \times 0.28 = 28\), which are both greater than 5. Therefore, the conditions are met.
02

Calculate the Mean and Standard Deviation

The mean (\(渭\)) of a sample proportion is equal to the population proportion \(p\), so \(渭 = p = 0.72\). \n\nThe standard deviation (\(蟽\)) of a sample proportion is calculated as \(\sqrt{p \times (1 - p) / n}\). Hence \(蟽 = \sqrt{0.72 \times 0.28 / 100} = 0.0458\) approximately.
03

Determine the Z-Score and Find the Probability

The z-score can be found using the formula \(Z = (X - 渭)/蟽\), where \(X\) is the sample proportion expressed as a decimal (0.75). Hence \(Z = (0.75 - 0.72) / 0.0458 = 0.6557\) approximately. \n\nTherefore, the probability that more than 75% of the sample has a driver's license is \(P(Z > 0.65) = 1 - P(Z < 0.65)\). Using the standard normal distribution table, \(P(Z < 0.65) = 0.7422\) approximately, so the required probability is \(1 - 0.7422 = 0.2578\) or 25.78%.

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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 how likely an event is to occur. It ranges from 0 to 1, where 0 means the event cannot happen and 1 indicates certainty. In the context of our example, probability helps us understand the chance that more than 75% of high school seniors in a sample of 100 have a driver's license.

To calculate this, we can use the concept of the Central Limit Theorem for Sample Proportions, which applies probability theory to sampling distributions. It allows us to find out where a given sample proportion falls within the standard normal distribution and calculate the probability of obtaining a value higher than 75%.
Sample Proportion
Sample proportion is a statistic that estimates the proportion of a certain characteristic within a subset of a larger population. For instance, in our high school seniors' driver's license scenario, we're looking at the proportion of students in a sample of 100 that have a driver's license.

Understanding sample proportion is critical because it helps researchers make inferences about the population as a whole. In the exercise, using the sample, we want to determine the likelihood that the proportion exceeds 75%, which is where the calculation of sample proportions comes into play.
Z-Score
A z-score is a statistical measure 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 represents the score is identical to the mean. A positive z-score indicates the number is above the mean, and a negative z-score signifies it's below the mean.

In the given exercise, we used a z-score to determine how far away our sample proportion (75%) is from the population proportion (72%). After finding the z-score, we then look to the standard normal distribution to find out the probability associated with that z-score.
Standard Normal Distribution
The standard normal distribution is a specific type of normal distribution that is centered around zero and has a standard deviation of one. It's an important concept in statistics because it allows us to compare different data sets with different means and standard deviations on a common scale.

In reference to the sample proportion problem we're involved with, after obtaining the z-score, it corresponds to a point on the standard normal distribution curve. This position on the curve then tells us the probability of a z-score (sample proportion) being above or below a certain value, leading us to discover the likelihood of more than 75% of the sample having a driver's license.
Statistics Education
Statistics education is essential for almost every field as it equips learners with the tools to collect, analyze, and make conclusions from data. In the example provided, understanding the Central Limit Theorem for Sample Proportions is critical for interpreting sample data.

Good statistics education breaks complex concepts into easier-to-understand parts and explains the real-world application of these terms, much like explaining the relevance of probability in inferences about population characteristics, or the importance of the z-score in evaluating sample data against population norms. This way, students can better appreciate the value of statistics in everyday decision-making and research.

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Most popular questions from this chapter

According to a 2017 Pew Research Center report on voting issues, \(59 \%\) of Americans feel that the everything should be done to make it easy for every citizen to vote. Suppose a random sample of 200 Americans is selected. We are interested in finding the probability that the proportion of the sample who feel with way is greater than \(55 \%\). a. Without doing any calculations, determine whether this probability will be greater than \(50 \%\) or less than \(50 \%\). Explain your reasoning. b. Calculate the probability that the sample proportion is \(55 \%\) or more.

A Harris poll asked a sample of U.S. adults if they agreed with the statement "Artificial intelligence will widen the gap between the rich and poor in the U.S." Of those aged 18 to \(35,69 \%\) agreed with the statement. Of those aged 36 to \(50,60 \%\) agreed with the statement. A \(95 \%\) confidence interval for \(p_{1}-p_{2}\) (where \(p_{1}\) is the proportion of those aged \(18-35\) who agreed and \(p_{2}\) is the proportion of those aged \(36-50\) who agreed) is \((0.034,0.146) .\) Does the interval contain \(0 ?\) What does this tell us about the proportion of adults in these age groups who agree with the statement?

Ignaz Semmelweiss \((1818-1865)\) was the doctor who first encouraged other doctors to wash their hands with disinfectant before touching patients. Before the new procedure was established, the rate of infection at Dr. Semmelweiss's hospital was about \(10 \%\). Afterward the rate dropped to about \(1 \%\). Assuming the population proportion of infections was \(10 \%\), find the probability that the sample proportion will be \(1 \%\) or less, assuming a sample size of 200 . Start by checking the conditions required for the Central Limit Theorem to apply.

uppose you want to estimate the mean grade point average (GPA) of all students at your school. You set up a table in the library asking for volunteers to tell you their GPAs. Do you think you would get a representative sample? Why or why not?

Assume your class has 30 students and you want a random sample of 10 of them. Describe how to randomly select 10 people from your class using the random number table.
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