Answer:
15.386% of the 2016 card will have better gas mileage than the 2016 beetle.
Explanation:
Hello!
The study variable is:
X: combined gas mileage of city and highway of a 2016 vehicle.
X~N(μ;σ²)
μ= 23.0 mpg
σ= 4.9 mpg
The 2016 Volkswagen Beetle has a combined has a mileage of 28 mpg and you need to find what percentage of 2016 vehicles have a better gas mileage than the Bettle's, symbolically:
P(X > 28)
Since the variable has a normal distribution you can use the standard normal to find out the asked percentage:
P(Z >
) = P(Z > 1,02) =
1 - P(Z ≤ 1.02)= 1 - 0.84614 = 0.15386
15.386% of the 2016 card will have better gas mileage than the 2016 beetle.
I hope it helps!