Final answer:
Using the standard normal distribution and a Z-score calculation, approximately 13.25% of vehicles have better gas mileage than the 2014 Volkswagen Beetle with a mileage of 28 mpg. The closest answer option is 10%.
Step-by-step explanation:
To determine what percent of all vehicles have better gas mileage than the 2014 Volkswagen Beetle with a combined gas mileage of 28 mpg, we can use the standard normal distribution. The mean gas mileage for the vehicles is 22.2 mpg, and the standard deviation is 5.2 mpg. First, we calculate the Z-score for the Beetle's mpg, which is the number of standard deviations away from the mean:
Z = (X - μ) / σ
Where:
- X is the gas mileage of the Beetle (28 mpg)
- μ (mu) is the mean gas mileage (22.2 mpg)
- σ (sigma) is the standard deviation (5.2 mpg)
Plugging in the values we get:
Z = (28 - 22.2) / 5.2 = 1.1154
We can now use a standard normal distribution table or a calculator to find the area to the right of the Z-score. This area represents the proportion of vehicles in the dataset with better gas mileage than the Beetle. The area to the right of the Z-score of 1.1154 is approximately 0.1325, or 13.25%. That means that about 86.75% of vehicles have worse gas mileage than the Beetle, and thus approximately 13.25% have better gas mileage.
Looking at the options provided, none of them precisely match 13.25%, so we need to choose the closest one. Option A) 10% is the closest to our calculated percentage.