/*! 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 33 Pollutants in auto exhausts, con... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Pollutants in auto exhausts, continued. The level of nitrogen oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life \((150,000\) miles of driving) of cars of a particular model varies Normally with mean \(80 \mathrm{mg} / \mathrm{mi}\) and standard deviation \(4 \mathrm{mg} / \mathrm{mi}\). A company has 25 cars of this model in its fleet. What is the level \(L\) such that the probability that the average \(\mathrm{NOX}+\mathrm{NMOG}\) level \(\mathrm{x}^{-} 1\) for the fleet is greater than \(L\) is only \(0.01\) ? (Hint: This requires a backward Normal calculation. See page 91 in Chapter 3 if you need to review.)

Short Answer

Expert verified
L is approximately 81.864 mg/mi.

Step by step solution

01

Identify the Known Values

We know that the NOx + NMOG levels are Normally distributed with a mean (\(\mu\)) of 80 mg/mi and a standard deviation (\(\sigma\)) of 4 mg/mi. The fleet consists of 25 cars, so this is our sample size (\(n\)).
02

Calculate the Standard Error

The standard error (SE) of the sample mean is given by the formula \( SE = \frac{\sigma}{\sqrt{n}} \). For our problem, this becomes \( SE = \frac{4}{\sqrt{25}} = \frac{4}{5} = 0.8 \).
03

Determine Corresponding Z-Score

We are looking for a Z-score such that the probability of the mean being greater than it is 0.01. For a standard Normal distribution, this Z-score can be found by: \( P(Z < k) = 0.99 \). Using Z-table or a calculator, we find that \( k = 2.33 \).
04

Calculate the Critical Level L

Now we use the Z-score formula \( Z = \frac{(X - \mu)}{SE} \) and solve for \(X\) such that \(Z = 2.33\):\[ X = \mu + Z \times SE = 80 + 2.33 \times 0.8 \].Performing the calculation gives: \( X = 80 + 1.864 = 81.864 \). Thus, the level \( L \) is approximately 81.864 mg/mi.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Standard Deviation
Standard Deviation is a measure that indicates the amount of variability, or dispersion, in a set of values. It tells us how much the individual data points in a dataset deviate, on average, from the mean of the dataset. In simpler terms, it helps us understand how spread out the values are. If the standard deviation is small, it means that most of the data points are close to the mean. If the standard deviation is large, the data points are spread out over a larger range of values.To find the standard deviation in the context of a Normally distributed variable, we calculate the square root of the variance. The variance is the average of the squared differences from the Mean. For the given problem, where the mean \(\mu\) is 80 mg/mi, the standard deviation \(\sigma\) is 4 mg/mi. This suggests that most car emissions are within 4 mg/mi of this average.
Sample Size
Sample Size is simply the number of observations or data points we are considering in a study or survey. It is crucial in statistical analysis because it can significantly affect the reliability of our results. In our exercise, the sample size is 25, which refers to the number of cars being analyzed in the fleet. The larger the sample size, typically, the more accurate and less variable our estimates about the larger population will be. The sample size influences the calculation of the standard error, which adjusts the standard deviation to reflect the size of the sample. A larger sample size usually results in a smaller standard error, making our estimates more precise.
Z-Score
The Z-Score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is expressed in terms of standard deviations. A Z-score of 1.0, for example, indicates that the value is one standard deviation away from the mean. Z-scores are useful for understanding how unusual or typical a particular data point is within the distribution. In the context of our problem, we are looking for a Z-score that corresponds to a probability of 0.01 above it. By reversing the problem using standard normal tables, or online calculators, we determine a Z-score of 2.33. This Z-score forms the basis for finding the critical level L.
Standard Error
Standard Error measures the accuracy with which a sample represents a population. It is the standard deviation of the sample mean distribution, helping to understand how far the sample mean of the data is likely to be from the true population mean.The Standard Error (SE) is calculated by dividing the standard deviation by the square root of the sample size. In our case, \( SE = \frac{4}{\sqrt{25}} = 0.8 \). The smaller the standard error, the closer the sample mean is likely to be to the population mean, which implies that we have a more reliable estimate. Thus, understanding and calculating standard error is crucial in making sound conclusions about the data.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A Sample of Young Men. A government sample survey plans to measure the total cholesterol level of an SRS of men aged \(20-34\). The researchers will report the mean \(x^{-} x\) from their sample as an estimate of the mean total cholesterol level \(\mu\) in this population. (a) Explain to somene who knows no statistics what it means to say that \(x^{-} \bar{x}\) is an "unbiased" estimator of \(\mu\). (b) The sample result \(x^{-} \bar{x}\) is an unbiased estimator of the population truth \(\mu\) no matter what size SRS the study uses. Explain to someone who knows no statistics why a large sample gives more trustworthy results than a small sample.

Sampling Distribution versus Population Distribution. The 2015 American Time Use Survey contains data on how many minutes of sleep per night each of 10,900 survey participants estimated they get. 3 The times follow the Normal distribution with mean \(529.9\) minutes and standard deviation \(135.6\) minutes. An SRS of 100 of the participants has a mean time of \(x^{-}=514.4 \bar{x}=514.4\) minutes. A second SRS of size 100 has mean \(x^{-}=539.3 r=539.3\) minutes. After many SRSs, the many values of the sample mean \(x^{-} x\) follow the Normal distribution with mean \(529.9\) minutes and standard deviation \(13.56\) minutes. (a) What is the population? What values does the population distribution describe? What is this distribution? (b) What values does the sampling distribution of \(x^{-} x\) describe? What is the sampling distribution?

Testing glass. How well materials conduct heat matters when designing houses. As a test of a new measurement process, 10 measurements are made on pieces of glass known to have conductivity 1 . The average of the 10 measurements is 1.07. For each of the boldface numbers, indicate whether it is a parameter or a statistic. Explain your answer.

What Does the Central Limit Theorem Say? Asked what the central limit theorem says, a student replies, "As you take larger and larger samples from a population, the histogram of the sample values looks more and more Nonal." Is the student right? Explain your answer.

Playing the numbers: A gambler gets chance outcomes. The law of large numbers tells us what happens in the long run. Like many games of chance, the numbers racket has outcomes that vary considerably-one three-digit number wins \(\$ 600\) and all others win nothing - that gamblers never reach "the long run." Even after many bets, their average winnings may not be close to the mean. For the numbers racket, the mean payout for single bets is \(\$ 0.60\) (60 cents) and the standard deviation of payouts is about \(\$ 18.96\). If Joe plays 350 days a year for 40 years, he makes 14,000 bets. (a) What are the mean and standard deviation of the average payout \(\mathrm{x}^{-} \bar{x}\) that Joe receives from his 14,000 bets? (b) The central limit theorem says that his average payout is approximately Nomal with the mean and standard deviation you found in part (a). What is the approximate probability that Joe's average payout per bet is between \(\$ 0.50\) and \(\$ 0.70\) ? You see that Joe's average may not be very close to the mean \(\$ 0.60\) even after 14,000 bets.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.