Answer
P(X > 34) = 0.2843. This implies that about 28 out of 100 samples will have a mean mileage greater than 34 mpg.
Work Step by Step
Since the distribution is normal, we can use the z score:
$z = \frac{34-32}{3.5} = 0.57$
P(X > 34) = P(z > 0.57) = 0.2843
