/*! 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 53 Carbon monoxide (CO) emissions f... [FREE SOLUTION] | 91Ó°ÊÓ

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Carbon monoxide (CO) emissions for a certain kind of car vary with mean \(2.9 \mathrm{g} / \mathrm{mi}\) and standard deviation 0.4 \(\mathrm{g} / \mathrm{mi}\). A company has 80 of these cars in its fleet. Let \(\bar{y}\) represent the mean CO level for the company's fleet. a) What's the approximate model for the distribution of \(\bar{y} ?\) Explain. b) Estimate the probability that \(\bar{y}\) is between 3.0 and \(3.1 \mathrm{g} / \mathrm{mi}\) c) There is only a \(5 \%\) chance that the fleet's mean CO level is greater than what value?

Short Answer

Expert verified
a) \(\bar{y}\) approximately follows a normal distribution with mean 2.9 and SD 0.0447 g/mi. b) \(P(3.0 < \bar{y} < 3.1) \approx 0.0125\). c) 95th percentile is \(2.9736\) g/mi.

Step by step solution

01

Understand the Problem

First, identify what we're dealing with in the problem. We have a fleet of 80 cars, each with carbon monoxide emissions following a distribution with mean \(2.9\) g/mi and standard deviation \(0.4\) g/mi. We'll calculate and interpret the mean and standard deviation of a sample mean distribution for \(\bar{y}\), the average emissions for these cars.
02

Approximate Model for \(\bar{y}\)

The distribution of \(\bar{y}\) can be modeled using the Central Limit Theorem (CLT), which states that the sampling distribution of the sample mean \(\bar{y}\) is approximately normal if the sample size is large enough (typically \(n \geq 30\)). Here, the number of cars \(n = 80\) is sufficient. The mean of this distribution is \(\mu_{\bar{y}} = \mu = 2.9\) g/mi. The standard deviation is given by the formula for the standard error: \(\sigma_{\bar{y}} = \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{80}} \approx 0.0447\) g/mi.
03

Calculate Probability for \(\bar{y}\) between 3.0 and 3.1

To find the probability that \(\bar{y}\) is between 3.0 and 3.1 g/mi, we use the properties of the normal distribution. First, calculate the z-scores for 3.0 and 3.1 g/mi using the formula: \(z = \frac{\bar{y} - \mu_{\bar{y}}}{\sigma_{\bar{y}}}\). For 3.0, \(z = \frac{3.0 - 2.9}{0.0447} \approx 2.24\). For 3.1, \(z = \frac{3.1 - 2.9}{0.0447} \approx 4.47\). Then, find the probability using a standard normal distribution table or calculator. The probability of \(z\) being between 2.24 and 4.47 is the difference between \(P(Z < 4.47)\) and \(P(Z < 2.24)\). This gives approximately 0.0125 as the probability.
04

Calculate 5% Threshold for \(\bar{y}\)

To find the threshold where there's only a 5% chance of being greater, we need the 95th percentile of the sampling distribution. Using a normal distribution table, find the z-score that corresponds to 95%, which is approximately \(z \approx 1.645\). Use the z-score formula to solve for \(\bar{y}\): \(\bar{y} = \mu_{\bar{y}} + z \cdot \sigma_{\bar{y}} = 2.9 + 1.645 \times 0.0447 \approx 2.9736\) g/mi. Therefore, there's a 5% chance that \(\bar{y}\) is greater than 2.9736 g/mi.

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Key Concepts

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

Sampling Distribution
Imagine you have a large group of items, and you're curious about their average quality. Instead of checking every single one, you take a few samples at a time. This is the essence of the sampling distribution, which refers to the distribution of a sample statistic, such as the sample mean, from a large number of repeated samples.
The Central Limit Theorem (CLT) is key here; it tells us that, regardless of the original distribution of our data, the sampling distribution of the sample mean will be approximately normal if the sample size is large enough.
  • In the exercise, we have 80 cars forming our sample, which is indeed large enough.
  • The mean of this sampling distribution is the same as the population mean, which is 2.9 g/mi for CO emissions.
  • To measure the spread, we use the standard error, calculated as the standard deviation of the population divided by the square root of the sample size.
This tells us how the sample means will vary around the true mean. In our case, it's about 0.0447 g/mi.
Normal Distribution
The normal distribution is like a bell curve, representing how values of a dataset are distributed. Many natural phenomena follow this pattern, and the normal distribution is symmetric about its mean.
For most data following this distribution:
  • About 68% lies within one standard deviation of the mean.
  • 95% within two standard deviations.
  • And 99.7% within three standard deviations.
In the context of the exercise, the Central Limit Theorem assures us that the distribution of average CO emissions in the company's fleet is normal, even if the individual emissions are not. Since the sample size is sufficiently large, we can predict probabilities and thresholds accurately using this distribution. This brings a level of predictability and certainty to our analyses.
Mean and Standard Deviation
The mean and standard deviation are the heartbeats of any dataset analysis. They tell us a lot about the overall distribution and variance of the data.
The mean is the average; you add up all the data points and divide by the number of points:
  • For the CO emissions, the mean or average is 2.9 g/mi.
The standard deviation indicates how spread out the values are around the mean. A small standard deviation indicates that the values are closely concentrated around the mean, while a larger one indicates a wider spread.
  • In this exercise, the standard deviation is 0.4 g/mi.
The standard error shrinks this spread in the context of sample means. This happens as we divide the population standard deviation by the square root of our sample size. Much like squeezing the bell curve width, it provides a clearer view of the average tendencies of the sample.
Z-scores
Z-scores are like the universal translators for data values into a standardized scale. They tell us how many standard deviations a particular value is from the mean, making it easier to find probabilities across different data sets.
To calculate a z-score for a value, you use the formula: \[ z = \frac{(x - \text{mean})}{\text{standard deviation}} \]In the context of the exercise:
  • For calculating whether the mean CO level is between 3.0 and 3.1 g/mi, we transform these values into z-scores using the sample mean and standard error instead of the population mean and standard deviation.
  • As calculated, the z-scores are 2.24 for 3.0 g/mi and 4.47 for 3.1 g/mi.
  • These z-scores help in determining how likely these sample mean values occur under our normalized sampling distribution.
Z-scores open the door to using the standard normal distribution table, which helps in estimating probabilities and determining threshold values more efficiently.

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