a fair coin is tossed 10,000 times. Find a number m such that the chance of the number of heads being between 5000-m and 5000+m is approximately 2/3.
In $\displaystyle 10000$ tosses of a fair coin the mean number of heads is $\displaystyle \mu=5000$, and the standard deviation of the number of heads is $\displaystyle \sigma=\sqrt{10000\times 0.25}$
Also the number of heads is approximatly normally distributed $\displaystyle \sim N(\mu,\sigma^2)$.
Now a symetrical interval that contains $\displaystyle 2/3$ of a normal distribution is $\displaystyle \mu \pm 0.965 \sigma$ (you get this from looking up a tail probability of $\displaystyle 0.16667$ backwards so to speak in a table of the standard normal distribution).
Now finish this yourself.
CB