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 Normal model with mean 2.022 and standard deviation .078 (Round to three decimal places as needed.) A Binomial model with \( \quad \) 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< \( \bar{y}<2.3)=.159 \) (Round to three decimal places as needed.) c) P( \( \bar{y}>\square)=0.1 \) (Round to one decimal place as needed.)

The sampling distribution of the mean \( \bar{y} \) for a sample size of 80 can be approximated by a normal distribution due to the Central Limit Theorem. This theorem states that, regardless of the shape of the population distribution, the distribution of the sample means will approach a normal distribution as the sample size increases. Since the population mean is \( 2.022 \, \text{g/mi} \) and the standard deviation is \( 0.7 \, \text{g/mi} \), the standard deviation of the sampling distribution (standard error) is \( \frac{0.7}{\sqrt{80}} \approx 0.078 \). To estimate the probability that \( \bar{y} \) falls between \( 2.1 \) and \( 2.3 \, \text{g/mi} \), we need to standardize these values first using the z-score formula: \( z = \frac{\bar{y} - \mu}{\sigma/\sqrt{n}} \). For \( \bar{y} = 2.1 \), the z-score is approximately \( \frac{2.1 - 2.022}{0.078} \approx 1.0 \), and for \( \bar{y} = 2.3 \), the z-score is about \( \frac{2.3 - 2.022}{0.078} \approx 3.56 \). Using standard normal distribution tables or calculators, the probability between these z-scores is \( P(1.0 < z < 3.56) \approx 0.159 \) (rounded to three decimal places). To find the value for which there is a 10% chance that the fleet's mean CO level \( \bar{y} \) is greater than it, we identify the z-value corresponding to the 90th percentile (since 10% is in the upper tail). The z-score associated with the 90th percentile is approximately \( 1.28 \). Thus, we use the formula \( \bar{y} = \mu + z \cdot \sigma/\sqrt{n} \) to solve for \( \bar{y} \): \[ 2.022 + 1.28 \times 0.078 \approx 2.1 \, \text{g/mi} \] (rounded to one decimal place). So, to summarize: a) A Normal model with mean \( 2.022 \) and standard deviation \( 0.078 \). b) \( P(2.1 < \bar{y} < 2.3) = 0.159 \) (rounded to three decimal places). c) \( P(\bar{y} > 2.1) = 0.1 \) (rounded to one decimal place).