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We are given that the gas mileage data for 1231 vehicles follows a normal distribution with a mean of 22.8 mpg and a standard deviation of 4.8 mpg. The goal is to find the percentage of vehicles that have a gas mileage better than the 2019 Volkswagen Beetle, which has a gas mileage of 29 mpg.

Here are the steps to solve this problem:

1. **Standardize the value of 29 mpg**:
   We use the z-score formula:
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   where:
   - \( x = 29 \) mpg (Volkswagen Beetle's gas mileage),
   - \( \mu = 22.8 \) mpg (mean gas mileage),
   - \( \sigma = 4.8 \) mpg (standard deviation).

2. **Find the z-score**: We will compute the z-score for 29 mpg.

3. **Find the cumulative probability**: Using the standard normal distribution table or a calculator, we will determine the cumulative probability corresponding to the z-score.

4. **Calculate the percentage**: Since we want the percentage of vehicles with a gas mileage higher than the Beetle, we subtract the cumulative probability from 1 and multiply by 100 to get the percentage.

Let me calculate the z-score and percentage now.The z-score for the Volkswagen Beetle's gas mileage of 29 mpg is approximately 1.29.

Next, we will find the cumulative probability associated with this z-score using the standard normal distribution and then calculate the percentage of vehicles with better gas mileage than the Beetle.Approximately **9.82%** of vehicles have better gas mileage than the 2019 Volkswagen Beetle.

Would you like more details on the calculations or any clarifications? Here are some related questions to expand your understanding:

1. What is the significance of a z-score in a normal distribution?
2. How would the percentage change if the Beetle's mileage were 25 mpg?
3. Can you explain why we subtract the cumulative probability from 1?
4. How is the standard deviation important in determining the z-score?
5. What other real-world scenarios can we apply this type of analysis to?

**Tip:** When dealing with normally distributed data, converting values to z-scores simplifies comparisons across different datasets.

The 2019 Volkswagen Beetle with a four-cylinder 2.0-L engine and automatic transmission has combined gas mileage of 29 mpg. Find the percentage of 2019 vehicles that have better gas mileage than the Beetle's gas mileage. The data is given as a normal distribution with a mean of 22.8 mpg and a standard deviation of 4.8 mpg.

Learn how to calculate the percentage of vehicles that have better gas mileage than the 2019 Volkswagen Beetle using normal distribution. This problem involves the calculation of z-scores, cumulative probabilities, and percentiles. Given: mean = 22.8 mpg, standard deviation = 4.8 mpg, and Beetle gas mileage = 29 mpg.

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