68-95-99.7 Rule and Normal Distribution Question. I need help solving this question. It is on my final exam review. I was given the answer but i do not know how to solve it. 
1) Not everyone pays the same price for the same model of a car. Suppose the price of a car is normally distributed. The mean price of a particular model of a new car is 22,000 and the standard deviation is 750. Use both the 68-95-99.7 Rule and Normal Distribution to find the percentages of buyers who paid more than 22,750? 
ANS: P(x>22,750) = 0.16
 Normal Distribution
 P(x>22,750) = 0.1587


In addition to my comment above, you could standardize by obtaining the Z-Score and find the probability that way(where Mean = 1, Standard Deviation = 0)
$$Z = \frac{X-\mu }{\sigma }$$
$$Z =\frac{22750-22000}{750}$$
$$Z=1$$
$$P(Z\geq1) = .1587$$
$$and$$
$$NormalCDF(22750,1000000,22000,750) = .1587$$
