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

 A: Since $22,750-22,000=750$, and $750$ is the standard deviation, the cutoff of $22,750$ is one standard deviation above the mean. Thus, you want to know what percentage of the data is more than one standard deviation above the mean
The $68$-$95$-$99.7$ rule says that about $68\%$ of the data in a normally distributed data set lie within one standard deviation of the mean. That leaves $100\%-68\%=32\%$ of the data more than one standard deviation away from the mean. The normal distribution is symmetric about the mean, so half of that $32\%$ is more than one standard deviation below the mean, and half of it is more than one standard deviation above the mean. Half of $32\%$ is $16\%$, so the $68$-$95$-$99.7$ rule says that about $16\%$ of the data are above $22,750$.
To get a more accurate value you’ll need to use the normal distribution itself. This requires either a calculator with the appropriate function or access to a table of the normal distribution. These tables come in several forms. The one on this web page gives the area under the normal curve to the left of a cutoff value. In your case the cutoff value is $1.00$, one standard deviation above the mean, and the table says that $0.8413$, or $84.13\%$, of the area under the curve is to the left of that value. That leaves $100.00\%-84.13\%=15.87\%$ of the area above the cutoff, meaning that about $15.87\%$ of the buyers will have paid more than $\$22,570$.
A: 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)
$$Z = \frac{X-\mu }{\sigma }$$
$$Z =\frac{22750-22000}{750}$$
$$Z=1$$
$$P(Z\geq1) = .1587$$
$$and$$
$$NormalCDF(22750,1000000,22000,750) = .1587$$
A: Trivial problem if you remember what the rule means: 68% is the proportion of buyers who pay within 1 standard deviation of the mean. 32% pay more or less, so 16% pay more. 22750 is 1 standard deviation above the mean.
