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Chapter 7: Problem 62

How many people in a car? A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban inter- change. Suppose that this count has mean 1.5 and standard deviation 0.75 in the population of all cars that enter at this interchange during rush hours. (a) Could the exact distribution of the count be Normal? Why or why not? (b) Traffic engineers estimate that the capacity of the interchange is 700 cars per hour. Find the probability that 700 cars will carry more than 1075 people. Show your work. (Hint: Restate this event in terms of the mean number of people \(\overline{x}\) per car.)

Short Answer

Expert verified
(a) No, the exact distribution cannot be Normal. (b) Probability is approximately 12.51%.

Step by step solution

01

Determine the Distribution Type

The problem states a count of people in cars, which is a discrete measurement. The normal distribution, meanwhile, is a continuous distribution. Since the normal distribution cannot exactly model discrete data, especially if the condition for normality on discrete data is not ideal, the distribution of the count cannot be exact Normal.
02

Define the Event

We are interested in the event that 700 cars will carry more than 1075 people. This event can be stated using the sample mean, \( \overline{x} \), which needs to be greater than \( \frac{1075}{700} \approx 1.536 \) people per car.
03

Apply Central Limit Theorem

Since 700 is a large number, we can apply the Central Limit Theorem (CLT), which states that the sampling distribution of the sample mean \( \overline{x} \) will be approximately Normal with mean \( \mu = 1.5 \) and standard deviation \( \frac{\sigma}{\sqrt{n}} = \frac{0.75}{\sqrt{700}} \).
04

Calculate the Probability

To find the probability that \( \overline{x} > 1.536 \), we first standardize this normal variable: \( z = \frac{1.536 - 1.5}{0.75/\sqrt{700}} \). Calculate \( z \), then use the standard normal distribution table to find the probability that \( Z > z \).
05

Compute the Z-Score

\[ z = \frac{1.536 - 1.5}{0.75/\sqrt{700}} = \frac{0.036}{\frac{0.75}{\sqrt{700}}} \approx 1.15 \]
06

Find the Probability from Z-Score

Using the standard normal distribution table, find \( P(Z > 1.15) \). This corresponds to approximately 0.1251, meaning there is a 12.51% probability that 700 cars will carry more than 1075 people.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics, often referred to as a Gaussian distribution. It is important because many variables in nature, such as heights, weights, and test scores, follow this pattern when collected under sufficient conditions. In a normal distribution, we have:
  • A symmetrical bell-shaped curve centered around the mean.
  • The majority of the data falls within three standard deviations from the mean.
  • It is defined by two parameters: the mean ( \( \mu \) ) and the standard deviation ( \( \sigma \) ).
However, the original exercise highlighted that because we're counting people, a discrete measure, the exact distribution cannot be "normal." The normal distribution is continuous, and thus it is not a perfect fit for discrete data unless conditions for normality are well met, typically with a large enough sample size that allows for approximation.
Sampling Distribution
The concept of sampling distribution is critical when we deal with samples instead of populations. A sampling distribution is the probability distribution of a given statistic based on many random samples from a population.Here's what you need to know:
  • The mean of the sampling distribution ( \( \overline{x} \) ) is equal to the mean of the population ( \( \mu \) ).
  • The standard deviation of the sampling distribution, known as the standard error, is \( \frac{\sigma}{\sqrt{n}} \) , where \( n \) is the sample size.
  • The larger the sample size, the closer the sample mean is to the population mean.
In our exercise, applying the Central Limit Theorem (CLT) allows us to assume the sampling distribution of the mean number of people per car is approximately normal, which simplifies calculating probabilities.
Z-Score Calculation
The Z-score is a measure that indicates how many standard deviations an element is from the mean. Z-scores are essential because they enable us to compare different data points across normal distributions or determine probabilities:
  • A Z-score of 0 indicates the value is exactly at the mean.
  • Positive Z-scores signify values above the mean, and negative ones below.
  • To calculate a Z-score, use the formula: \( z = \frac{X - \mu}{\sigma} \) , where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
In the exercise, the Z-score tells us how unusual it is to have a sample mean of 1.536 or more for the number of people per car. By getting a Z-score of 1.15, we concluded there is a 12.51% chance the sample mean exceeds this threshold using the standard normal distribution table.
Rush Hour Traffic Study
Studies like the rush hour traffic study in San Francisco provide invaluable insights into urban traffic patterns. Understanding this can help city planners and engineers make informed decisions to ease congestion and improve public safety. Here's why this type of study is necessary:
  • It helps in estimating demand for road capacity and improving infrastructure.
  • By knowing the average number of people per car, they can implement policies to encourage carpooling.
  • These studies can further guide design in placing carpool lanes or incentivizing public transport.
Specifically, our exercise involved calculating the probability of 700 cars carrying more than 1075 people. Such calculations help in resource allocation during peak transit times for smoother traffic management. Moreover, it's an example of practical applications of statistics to solve real-world issues.

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Most popular questions from this chapter

Do you drink the cereal milk? AUSA Today Poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Of the respondents, 67\(\%\) said that they drink it. Suppose that 70\(\%\) of U.S. adults actually drink the cereal milk. Let \(\hat{p}\) be the proportion of people in the sample who drink the cereal milk. (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 Normal condition is met. (d) Find the probability of obtaining a sample of 1012 adults in which 67\(\%\) or fewer say they drink the cereal milk. Do you have any doubts about the result of this poll?

Who owns a Harley? Harley-Davidson motorcycles make up 14\(\%\) of all the motorcycles registered in the United States. You plan to interview an SRS of 500 motorcycle owners. How likely is your sample to contain 20\(\%\) or more who own Harleys? Follow the four-step process.

Multiple choice: Select the best answer for Exercises 43 to \(46,\) which refer to the following setting. The magazine Sports Illustrated asked a random sample of 750 Division I college athletes, "Do you believe performance- enhancing drugs are a problem in college sports?" Suppose that 30\(\%\) of all Division I athletes think that these drugs are a problem. Let \(\hat{p}\) be the sample proportion who say that these drugs are a problem. The standard deviation of the sampling distribution is about \(\begin{array}{ll}{\text { (a) } 0.0006 .} & {\text { (c) } 0.017 .} & {\text { (e) none of these. }} \\ {\text { (b) } 0.033 .} & {\text { (d) } 1}\end{array}\)

Exercises 69 to 72 refer to the following setting. In the language of government statistics, you are "in the labor force" if you are available for work and either working or actively seeking work. The unemployment rate is the proportion of the labor force (not of the entire population) who are unemployed. Here are data from the Current Population Survey for the civilian population aged 25 years and over in a recent year. The table entries are counts in thousands of people. Unemployment \((5.3)\) If you know that a randomly chosen person 25 years of age or older is a college graduate, what is the probability that he or she is in the labor force? Show your work.

If we take a simple random sample of size \(n=500\) from a population of size \(5,000,000,\) the variability of our estimate will be (a) much less than the variability for a sample of size \(n=500\) from a population of size \(50,000,000\) . (b) slightly less than the variability for a sample of size \(n=500\) from a population of size \(50,000,000\) . (c) about the same as the variability for a sample of size \(n=500\) from a population of size \(50,000,000\) . (d) slightly greater than the variability for a sample of size \(n=500\) from a population of size \(50,000,000\) . (e) much greater than the variability for a sample of size \(n=500\) from a population of size \(50,000,000\) .
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