\( 1 \leftarrow \) Carbon monoxide (CO) emissions for a certain kind of car vary with mean \( 2.022 \mathrm{~g} / \mathrm{mi} \) and standard deviation \( 0.7 \mathrm{~g} / \mathrm{mi} \). 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 2.1 and \( 2.3 \mathrm{~g} / \mathrm{mi} \). c) There is only a \( 10 \% \) chance that the fleet's mean CO level is greater than what value? a) Select the correct choice below and, if necessary, fill in the answer boxes within your choice. A. A Normal model with mean 2.022 and standard deviation .078 (Round to three decimal places as needed.) B. A Binomial model with trials and a probability of success of (Round to three decimal places as needed.) C. A Poisson model with mean number of successes (Round to three decimal places as needed.) D. There is no model that fits this distribution. b) P(2.1< \( \bar{y}<2.3)=1 \) (Round to three decimal places as needed.)

The distribution of \( \bar{y} \) follows a Normal model due to the Central Limit Theorem, which states that the sampling distribution of the sample mean will be normally distributed if the sample size is sufficiently large (which it is with \( n = 80 \)). The mean of \( \bar{y} \) is the same as the population mean, \( 2.022 \mathrm{~g} / \mathrm{mi} \), while the standard deviation is given by \( \sigma_{\bar{y}} = \frac{\sigma}{\sqrt{n}} = \frac{0.7}{\sqrt{80}} \approx 0.078 \mathrm{~g} / \mathrm{mi} \). To estimate the probability that \( \bar{y} \) is between \( 2.1 \) and \( 2.3 \), we need to standardize the values using the \( z \)-score formula. For \( 2.1 \), \( z = \frac{2.1 - 2.022}{0.078} \approx 1.000 \), and for \( 2.3 \), \( z = \frac{2.3 - 2.022}{0.078} \approx 3.577 \). Using the standard normal distribution, you can find the probabilities for these \( z \)-scores using a z-table or calculator, and then subtract the smaller probability from the larger one to find \( P(2.1 < \bar{y} < 2.3) \). This results in approximately \( 0.158 \) (You might want to double-check these numbers.) To find where there’s only a \( 10\% \) chance the fleet's mean CO level is greater, you can look for the z-score that corresponds to the \( 90\% \) percentile, which is approximately \( 1.2816 \). Using this, you would rearrange the formula \( \bar{y} = \mu + z \cdot \sigma_{\bar{y}} \) to solve for \( \bar{y} \). ∼\( \bar{y} = 2.022 + 1.2816 \cdot 0.078 \approx 2.1 \mathrm{~g} / \mathrm{mi} \), meaning there's a \( 10\% \) chance that the fleet's mean CO level is greater than this value.