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 \( \overline{\mathrm{y}} \) ? Explain. b) Estimate the probability that \( \overline{\mathrm{y}} \) 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 \( \square \) and standard deviation (Round to three decimal places as needed.) B. A Binomial model with \( \square \) trials and a probability of success of (Round to three decimal places as needed.) C. A Poisson model with \( \square \) mean number of successes (Round to three decimal places as needed.) D. There is no model that fits this distribution.

**a)** Since the individual car CO emissions have mean \(\mu = 2.022\) and standard deviation \(\sigma = 0.7\), by the Central Limit Theorem the sampling distribution of the fleet mean \(\bar{y}\) (with \(n=80\) cars) is approximately normal. Its mean is \[ \mu_{\bar{y}} = 2.022 \] and its standard deviation is \[ \sigma_{\bar{y}} = \frac{0.7}{\sqrt{80}}. \] Thus, the appropriate model is **A. A Normal model with mean \(2.022\) and standard deviation \(\frac{0.7}{\sqrt{80}}\).** --- **b)** We want to estimate \[ P(2.1 \leq \bar{y} \leq 2.3). \] 1. First, compute the standard deviation of \(\bar{y}\): \[ \sigma_{\bar{y}} = \frac{0.7}{\sqrt{80}} \approx \frac{0.7}{8.944} \approx 0.0783. \] 2. Next, convert the endpoints to \(z\)-scores: - For \(\bar{y}=2.1\): \[ z = \frac{2.1 - 2.022}{0.0783} \approx \frac{0.078}{0.0783} \approx 1.0. \] - For \(\bar{y}=2.3\): \[ z = \frac{2.3 - 2.022}{0.0783} \approx \frac{0.278}{0.0783} \approx 3.55. \] 3. Using the standard normal distribution: \[ P(2.1 \leq \bar{y} \leq 2.3) \approx P(1.0 \leq Z \leq 3.55). \] From standard normal tables, \[ P(Z \leq 1.0) \approx 0.8413 \quad \text{and} \quad P(Z \leq 3.55) \approx 0.9998. \] Thus, \[ P(1.0 \leq Z \leq 3.55) \approx 0.9998 - 0.8413 = 0.1585. \] So, the probability is approximately \(0.1585\) (or 15.85%). --- **c)** We need to find the value \(t\) such that there is only a 10% chance the fleet mean exceeds \(t\); that is: \[ P(\bar{y} > t) = 0.10. \] This is equivalent to: \[ P(\bar{y} \leq t) = 0.90. \] 1. Find the corresponding \(z\)-score for \(0.90\) in the standard normal distribution. The \(90^{\text{th}}\) percentile is approximately: \[ z_{0.90} \approx 1.282. \] 2. Convert this \(z\)-score back to \(t\): \[ t = \mu_{\bar{y}} + z_{0.90} \cdot \sigma_{\bar{y}}. \] Substituting the values: \[ t \approx 2.022 + 1.282 \times 0.0783 \approx 2.022 + 0.1003 \approx 2.1223. \] Thus, there is only a 10% chance that the fleet's mean CO level exceeds approximately \(2.122 \, \mathrm{g/mi}\).