Carbon monoxide (CO) emissions for a certain kind of car vary with mean \( 3.786 \mathrm{~g} / \mathrm{mi} \) and standard deviation \( 0.8 \mathrm{~g} / \mathrm{mi} \). A company has 80 of these cars in its fleet. Let \( \overline{\mathrm{y}} \) 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 3.9 and \( 4 \mathrm{~g} / \mathrm{mi} \). c) There is only a \( 10 \% \) chance that the fleet's mean CO level is greater than what value? A. A Poisson model with \( \square \) mean number of successes (Round to three decimal places as needed.) B. A Normal model with mean \( \square \) and standard deviation (Round to three decimal places as needed.) C. A Binomial model with \( \square \) trials and a probability of success of \( \square \) (Round to three decimal places as needed.) D. There is no model that fits this distribution. b) \( P(3.9<\bar{y}<4)=\square \) (Round to three decimal places as needed.) c) \( P(\bar{y}> \) A \( \square)=0.1 \) (Round to one decimal place as needed.)

To tackle the problem, let’s break it down step by step. ### a) Approximate Model for the Distribution of \( \overline{y} \) The mean CO level \( \overline{y} \) for the company's fleet can be approximated using the Central Limit Theorem (CLT), which states that with a sufficiently large sample size (in this case, 80 cars), the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the population distribution. Therefore, since we have a sample size of 80, the distribution of \( \overline{y} \) is approximately a normal distribution with a mean of \( 3.786 \mathrm{~g} / \mathrm{mi} \) and a standard deviation calculated as the population standard deviation divided by the square root of the sample size: \[ \sigma_{\overline{y}} = \frac{\sigma}{\sqrt{n}} = \frac{0.8}{\sqrt{80}} \approx 0.0894 \mathrm{~g} / \mathrm{mi}. \] ### b) Estimate the Probability that \( \overline{y} \) is Between 3.9 and 4 g/mi To find the probability \( P(3.9 < \overline{y} < 4) \), we can standardize the values using the z-score formula: \[ z = \frac{\overline{y} - \mu}{\sigma_{\overline{y}}}. \] Calculating the z-scores: 1. For \( \overline{y} = 3.9 \): \[ z_1 = \frac{3.9 - 3.786}{0.0894} \approx 1.273. \] 2. For \( \overline{y} = 4 \): \[ z_2 = \frac{4 - 3.786}{0.0894} \approx 2.379. \] Next, we find the area under the curve for these z-scores using the standard normal distribution table. - The area to the left of \( z_1 \approx 1.273 \) is about \( 0.898 \). - The area to the left of \( z_2 \approx 2.379 \) is about \( 0.991 \). The probability \( P(3.9 < \overline{y} < 4) \) is then: \[ P(3.9 < \overline{y} < 4) = P(Z < 2.379) - P(Z < 1.273) = 0.991 - 0.898 \approx 0.093. \] ### c) Value Where There is a 10% Chance the Fleet's Mean CO Level is Greater Than This Value To find the value such that there is a \( 10\% \) chance that \( \overline{y} \) is greater than it, we need to find the z-score that corresponds to the \( 90\% \) percentile (100% - 10%). The z-score for the 90th percentile is approximately \( 1.2816 \). Using the z-score formula to solve for \( \overline{y} \): \[ \overline{y} = \mu + z \cdot \sigma_{\overline{y}}, \] where: - \( \mu = 3.786 \) - \( z \approx 1.2816 \) - \( \sigma_{\overline{y}} \approx 0.0894 \) Calculating it, we get: \[ \overline{y} = 3.786 + (1.2816 \cdot 0.0894) \approx 3.868. \] Thus, there is only a 10% chance that the fleet's mean CO level is greater than approximately \( 3.9 \mathrm{~g}/\mathrm{mi} \) when rounded to one decimal place. ### Summary of Answers a) A Normal model with mean \( 3.786 \) and standard deviation \( 0.089 \). b) \( P(3.9 < \bar{y} < 4) \approx 0.093 \). c) \( P(\bar{y} > 3.868) = 0.1 \)