What is the width of the Gaussian distribution that is generated from performing $N$ trials of coin tossing?
Example: In a trial of 1000 tosses of a coin, $P(H)=0.5$ and $P'(H)=0.5$, where $H$ refers to heads. Intuitively, if this trial is repeated $N$ times, then the most common occurrence would be $500$ heads. Although, theoretically, heads will also occur $501$, $502$, $503$, $504$... times, albeit less frequently than for $500$ heads. I am assuming this would follow a normal distribution.
In which case, what would the $1\,\sigma$ width of this normal distribution be? That is to say, given $1000$ tosses of a coin, what values would be expected 68% of the time?
Or, alternatively, what is $\sigma$, given that:
$$ \mathrm{Expected\ value} = 500 \pm \sigma? $$
It's not, $\sqrt{N}$, is it?
To be more precise I take 68,27%.
Now you can approximate the binomial distribution by the normal distribution.
$P(X\leq x)=\Phi\left( \frac{x-\mu}{\sqrt{n\cdot p \cdot q}} \right)=0.15856$
$\Phi(\cdot )$ if the function of the normal distribution.
We need the left side of the outside of the interval. Because of the symmetry we can divide it by 2. It is $\frac{100-68,27}{2}=0.15856$ Inserting the values:
$P(X\leq x)=\Phi\left( \frac{x-250}{\sqrt{500\cdot 0.5 \cdot 0.5}} \right)=0.15856$
Taking the inverse funkcion:
$ \frac{x-250}{\sqrt{500\cdot 0.5 \cdot 0.5}}= \Phi ^{-1}\left(0.15856\right)$
$ \frac{x-250}{\sqrt{500\cdot 0.5 \cdot 0.5}}= -1$
Now you can solve the equation for x.The result have to be round up to get the lower bound of the inverval. Because of the symmetry of the interval it is easy to calculate the upper bound.