/*! 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 63 Auto emissions The amount of nit... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 63

Auto emissions The amount of nitrogen oxides (NOX) present in the exhaust of a particular type of car varies from car to car according to a Normal distribution with mean 1.4 grams per mile \((g / \mathrm{mi})\) and standard deviation \(0.3 \mathrm{~g} / \mathrm{mi}\). Two randomly selected cars of this type are tested. One has \(1.1 \mathrm{~g} / \mathrm{mi}\) of \(\mathrm{NOX}\) the other has \(1.9 \mathrm{~g} / \mathrm{mi}\). The test station attendant finds this difference in emissions between two similar cars surprising. If the NOX levels for two randomly chosen cars of this type are independent, find the probability that the difference is at least as large as the value the attendant observed.

Short Answer

Expert verified
The probability that the difference in emissions is at least 0.8 is approximately 0.0584.

Step by step solution

01

Understand the Distribution

The nitrogen oxides levels in the exhaust are usually distributed normally with a mean of \(\mu = 1.4 \) grams per mile and a standard deviation of \(\sigma = 0.3 \) grams per mile. This implies that for any car of this type, its NOx emissions follow the normal distribution \(X \sim N(1.4, 0.3^2)\).
02

Define the Difference between Car Emissions

Since the problem involves the difference in NOx emissions between two cars, define this difference as \(D = X_1 - X_2\), where \(X_1\) and \(X_2\) represent the NOx emissions of each car, respectively. The observed difference between the two given cars is \(D = 1.9 - 1.1 = 0.8\).
03

Find the Distribution of the Difference

Since \(X_1\) and \(X_2\) are independent, their difference \(D = X_1 - X_2\) also follows a normal distribution. The mean of \(D\) is \(\mu_D = \mu_{X_1} - \mu_{X_2} = 1.4 - 1.4 = 0\). The variance of \(D\) is \(\sigma_D^2 = \sigma_{X_1}^2 + \sigma_{X_2}^2 = 0.3^2 + 0.3^2 = 0.18\). Therefore, \(D \sim N(0, 0.18)\).
04

Convert the Observed Difference to a Z-Score

Convert the observed difference \(0.8\) to a Z-score using the distribution of \(D\). The Z-score is calculated as follows: \[ Z = \frac{D - \mu_D}{\sigma_D} = \frac{0.8 - 0}{\sqrt{0.18}} = \frac{0.8}{0.4243} \approx 1.886 \]
05

Find the Probability of the Observed Difference

To find the probability that the absolute difference is at least as large as 0.8, calculate \(P(|D| \geq 0.8)\). Since the normal distribution is symmetric: \[ P(|D| \geq 0.8) = 2 \, P(D \geq 0.8) = 2 (1 - P(D < 0.8)) = 2 (1 - \text{P}(Z < 1.886)) \] Using the standard normal distribution table, \(\text{P}(Z < 1.886) \approx 0.9708\). Thus, \[ P(|D| \geq 0.8) = 2(1 - 0.9708) \approx 0.0584 \].

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Key Concepts

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

Probability Calculation
Probability calculation is a fundamental aspect of statistics, particularly when working with data that follow a normal distribution.
In the context of the given exercise, the probability calculation involves determining how likely it is for the difference in emission levels from two cars to be as extreme or more extreme than what was observed.
This kind of probability is typically calculated using the cumulative distribution function (CDF) of a normal distribution. By transforming the observed difference into a Z-score, we can use the CDF to find the probability of observing that Z-score or more extreme values.
It's important to note that because normal distributions are symmetrical, we can look at the positive tail of the distribution and multiply by two to account for both tails, thus getting the probability of observing any extreme difference.
Calculating the probability in such scenarios not only helps in making statistical inferences but also supports hypothesis testing, allowing researchers to compare empirically observed data against a theoretical distribution.
Statistical Independence
Statistical independence is a key concept in probability and statistics, crucial for the analysis conducted in the exercise.
Two random variables are considered statistically independent if the occurrence of one event does not affect the probability of the other. In simpler terms, knowing the outcome of one does not change the score or outcome of the other.
In the context of the exhaust emissions problem, independence ensures that the emissions of NOx from one car do not influence those from another.
This assumption allows us to amalgamate individual variances to find the variance of the difference between the two measurements.
If the emissions were not independent, variance calculations would need extra terms accounting for covariance. Thus, assuming independence simplifies calculations and aligns with real-world expectations in this context.
Standard Deviation
Standard deviation is a measure that signifies the amount of variability or dispersion in a set of data values.
In the exercise concerning NOx emissions, the standard deviation of 0.3 grams per mile tells us how spread out the emissions measurements are around the mean of 1.4 grams per mile.

  • A smaller standard deviation would imply that most cars emit a NOx value close to the mean.

  • A larger standard deviation hints at a wider variety of emission levels among different cars.

When calculating differences in emissions between two cars, the standard deviation is crucial in determining the variability we can expect in those differences.
With two independent measurements, variances add up, meaning the standard deviation of the difference, in this case, is determined using the square root of the sum of individual variances.
Understanding standard deviation helps predict how much individual car emissions might deviate from the mean value and assess if observed differences are statistically significant.

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Most popular questions from this chapter

Smoking and social class (5.3) As the dangers of smoking have become more widely known, clear class differences in smoking have emerged. British government statistics classify adult men by occupation as "managerial and professional" \((43 \%\) of the population), "intermediate" \((34 \%),\) or "routine and manual" \((23 \%)\). A survey finds that \(20 \%\) of men in managerial and professional occupations smoke, \(29 \%\) of the intermediate group smoke, and \(38 \%\) in routine and manual occupations smoke. \({ }^{14}\) (a) Use a tree diagram to find the percent of all adult British men who smoke. (b) Find the percent of male smokers who have routine and manual occupations.

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