Mean & Standard deviation given probability of range in normal distrubition On SAT,scores range from 2000 to 2400, with two thirds of the scores falling in the range of 2200 to 2300. If we further assume that test scores are normally distributed in this range from 2000 to 2400, determine the mean and standard deviation.
I believe there is a mistake in the statement of your problem.
By symmetry, and for things to work out as intended '2200' has to be '2100'.

The Empirical Rule says that 68% of observations fall within one standard deviation of the mean. What you request can be done
only approximately, so we will use 68% instead of $2/3.$
Then the population mean needs to be halfway between 2100 and 2300; $\mu = 2200.$ Also, that implies that $\sigma = 100.$

Then using the CDF function pnorm in R, we can find the probability of a score less than 2000 and greater than 4000
for a test with scores distributed $\mathsf{Norm}(\mu=2200, \sigma=100).$ About 2.27% of students will have scores in the
'tails' of the distribution.

pnorm(2000, 2200, 100)
[1] 0.02275013
1 - pnorm(2400, 2200, 100)
[1] 0.02275013


The figure below shows the normal distribution of SAT scores. Vertical red lines mark values in the discussion above.
No matter how you fit a normal distribution to this data, it will always put positive probability outside of 2000 to 2400 range. So it is not possible to fix the range to a bounded set and assume that the underlying distribution is normal. 
However, we can approximate a normal distribution assuming that the probability of scores being less than 2000 is equal to the probability of scores being above 2400. 

Now, let $\mu$ denote the mean and $\sigma^2$ denote the variance. Then, 
$$\Phi(\frac{2300-\mu}{\sigma})-\Phi(\frac{2200-\mu}{\sigma})=\frac{2}{3}$$
 and 
$$1-\Phi(\frac{2400-\mu}{\sigma})=\Phi(\frac{2000-\mu}{\sigma})$$

where $\Phi(.)$ is the CDF of standard normal. Solving these two equations give $\mu
$ and $\sigma$.