/*! 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 73 Emissions of nitrogen oxides, wh... [FREE SOLUTION] | 91Ó°ÊÓ

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Emissions of nitrogen oxides, which are major constituents of \(\mathrm{smog}\), can be modeled using a normal distribution. Let \(x\) denote the amount of this pollutant emitted by a randomly selected vehicle (in parts per billion). The distribution of \(x\) can be described by a normal distribution with \(\mu=1.6\) and \(\sigma=0.4\). Suppose that the EPA wants to offer some sort of incentive to get the worst polluters off the road. What emission levels constitute the worst \(10 \%\) of the vehicles?

Short Answer

Expert verified
The emission level that constitutes the worst \(10 \%\) of vehicles is around \(2.112\) parts per billion.

Step by step solution

01

Standardize the Normal Distribution

Convert the standard normal distribution with a mean of \(\mu=1.6\) and a standard deviation of \(\sigma=0.4\) to a standard normal distribution with a mean of 0 and a standard deviation of 1. This is done by using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(Z\) is the standardized normal random variable, \(X\) is the original random variable, \(\mu\) is the mean and \(\sigma\) is the standard deviation.
02

Find the Z Score for the 90th percentile

To identify the worst \(10\%\) of the vehicles, we need to find the \(90^{th}\) percentile of the distribution. We can do this by looking at a standard Z table or using a calculator that provides this functionality. The Z score for the \(90^{th}\) percentile is approximately \(1.28\). This means that \(90\%\) of the data in a standard normal distribution falls below the Z score of \(1.28\).
03

Determine the emission level

Now, using the Z-score found for the standard normal distribution, you can find the corresponding emission level in the original distribution by using the formula \(X = Z\sigma + \mu \). Substitute \(Z = 1.28\), \(\sigma = 0.4\) and \(\mu = 1.6\) into the formula to get the cut-off emission level. Hence, the cut-off emission level for the worst \(10\%\) vehicles would be \(X = 1.28(0.4) + 1.6 = 2.112\).

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Key Concepts

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

Statistical Modeling in Emission Levels
Statistical modeling is a powerful tool used in understanding real-world phenomena through mathematics. In our scenario, the emissions of nitrogen oxides by vehicles can be modeled using a normal distribution.
Normal distribution is a type of continuous probability distribution for a real-valued random variable. It is symmetric about the mean, meaning that it occurs frequently in both natural and social sciences to represent real-valued random variables whose distributions are not known.
  • Mean (\(\mu\)) indicates the central point of the distribution. In our example, it's 1.6 parts per billion for emissions.
  • Standard deviation (\(\sigma\)) measures the variability or spread of the distribution. Here, it’s 0.4 parts per billion.
Using statistical modeling helps interpret data about emissions, highlighting how frequently different levels of nitrogen oxides emissions occur, and assisting in making informed decisions, like removing the worst polluters.
Understanding Standardization in Normal Distribution
Standardization simplifies the comparison of different data by converting everything to a standard scale. When dealing with a normal distribution, standardization involves converting data to fit a standard normal distribution which has a mean of 0 and a standard deviation of 1.
This is achieved by using the formula: \[Z = \frac{X - \mu}{\sigma}\]. Here:
  • \(Z\): Standardized value
  • \(X\): Original value
  • \(\mu\): Mean of the distribution
  • \(\sigma\): Standard deviation
By translating the original data into the Z-score, complex calculations and comparisons are made easier. For instance, it allows us to find percentages and make inferences about data. In the car emissions example, understanding which vehicles fall into the worst 10% becomes manageable when we standardize the emissions data using the Z-score method.
Using Percentiles to Identify Extreme Values
Percentiles are a crucial concept when dealing with normal distributions as they indicate the value below which a given percentage of the data falls. They help in understanding the relative standing of a particular data point within a dataset.
The 90th percentile, for example, is the value below which 90% of the observations lie. It means that only 10% of the values are greater, marking them as extreme.
In our case using percentiles helps identify the worst emitters of nitrogen oxides. We find the Z-score corresponding to the 90th percentile, allowing us to determine the cutoff for the worst 10% of vehicular emissions. This helps in setting regulations or incentives, as it clearly points out which vehicles need to be targeted for reducing pollution.

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Most popular questions from this chapter

The Los Angeles Times (December 13,1992\()\) reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost \(80 \%\) did so. Suppose that for a particular route the actual percentage is exactly \(80 \%\), and consider randomly selecting six passengers. Then \(x\), the number among the selected six who rested or slept, is a binomial random variable with \(n=6\) and \(p=.8\). a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6)\), the probability that all six selected passengers rested or slept. c. Determine \(P(x \geq 4)\).

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?

Of all airline flight requests received by a certain discount ticket broker, \(70 \%\) are for domestic travel (D) and \(30 \%\) are for international flights (I). Let \(x\) be the number of requests among the next three requests received that are for domestic flights. Assuming independence of successive requests, determine the probability distribution of \(x\). (Hint: One possible outcome is DID, with the probability \((.7)(.3)(.7)=.147 .\) )

Consider babies born in the "normal" range of \(37-43\) weeks gestational age. The paper referenced in Example \(7.27\) ("Fetal Growth Parameters and Birth Weight. Their Relationship to Neonatal Body Composition." Ultrasound in Obstetrics and Gynecology [2009]: \(441-446\) ) suggests that a normal distribution with mean \(\mu=3500\) grams and standard deviation \(\sigma=\) 600 grams is a reasonable model for the probability distribution of the continuous numerical variable \(x=\) birth weight of a randomly selected full-term baby. a. What is the probability that the birth weight of a randomly selected full- term baby exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g} ?\) b. What is the probability that the birth weight of a randomly selected full- term baby is cither less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected full- term baby exceeds 7 pounds? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .\) ) d. How would you characterize the most extreme \(0.1 \%\) of all full-term baby birth weights? e. If \(x\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0)\), then \(y=a x\) also has a normal distribution. Use this formula to determine the distribution of full-term baby birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (c). How does this compare to your previous answer?

State whether each of the following random variables is discrete or continuous: a. The number of defective tires on a car b. The body temperature of a hospital patient c. The number of pages in a book d. The number of draws (with replacement) from a deck of cards until a heart is selected e. The lifetime of a lightbulb

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