# Math Help - Normal approximation: Method

1. ## Normal approximation: Method

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.

2. Originally Posted by Yan
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 $10000$ tosses of a fair coin the mean number of heads is $\mu=5000$, and the standard deviation of the number of heads is $\sigma=\sqrt{10000\times 0.25}$

Also the number of heads is approximatly normally distributed $\sim N(\mu,\sigma^2)$.

Now a symetrical interval that contains $2/3$ of a normal distribution is $\mu \pm 0.965 \sigma$ (you get this from looking up a tail probability of $0.16667$ backwards so to speak in a table of the standard normal distribution).

Now finish this yourself.

CB