Answer:
50%
Explanation:
We have that the mean is 52000 miles (m) and the standard deviation is 2000 miles (sd), so we are asked:
P (x <52000)
P ((x - m) / sd <(52000 -52000) /2)
P (z <0)
If we look in the normal distribution table we have to:
P (x <52000) = 0.5
That is, the probability is of randomly selecting a car with a mileage less than 52,000 miles 50%