How do I approximate this binomial problem correctly? If I have an experiment that has $1000$ trials, and $10$% of the time there is an error, what is the approximate probability that I will have $125$ failures? I figured out that $\mu =100$ and $\sigma =9.4868$. I'm trying to approximate with a normal distribution, since with this many trials, and the fact that each experiment is independent the total number of failrues should be normally distributed. So I set up my problem as so $$P(X\le 125)=P\left(z\le \frac{125-100}{\frac{9.4868}{\sqrt{1000}}}\right)=\Phi\left(\frac{125-100}{\frac{9.4868}{\sqrt{1000}}}\right)$$ But there's no way that with probability $1$ there are at most $125$ errors. So what am I missing?
 A: The DeMoivre-Laplace theorem says that if $X$ is a binomial random variable with parameters $(n,p)$, then 
$$P\{a < X < b\} \approx \Phi\left(\frac{b-np}{\sqrt{np(1-p)}}\right) - \Phi\left(\frac{a-np}{\sqrt{np(1-p)}}\right)$$
and so, for your problem,
$$P\{X \leq 125\} \approx \Phi\left(\frac{125-100}{\sqrt{1000\cdot 0.1\cdot 0.9}}\right)
= \Phi\left(\frac{25}{\sqrt{90}}\right)$$
This is a form of the central limit theorem applied to binomial random variables.
For a more standard application of the central limit theorem, note that if
$Y_1, Y_2, \ldots, Y_{n}$ are independent random variables with mean
$\mu$ and variance $\sigma^2$, then the central limit theorem says
that the CDF of $(Y_1+Y_2+\cdots+Y_{n} - n\mu)/\sigma\sqrt{n}$
is approximated by the CDF of a standard normal random variable $Z$.
If the $Y_i$ are Bernoulli random variables
with parameter $p$ (and hence mean $p$ and variance $p(1-p)$, standard deviation
$\sigma = \sqrt{p(1-p)}\, $), then 
$$\frac{(Y_1+Y_2+\cdots+Y_{1000})-1000p}{\sigma\sqrt{1000}}
= \frac{(Y_1+Y_2+\cdots+Y_{1000})-100}{\sqrt{0.1\cdot 0.9}\sqrt{1000}}
= \frac{X-100}{\sqrt{90}}$$
has approximately the same CDF as a standard normal random variable $Z$. Hence,
$$P\{X \leq 125\} \approx  P\left\{Z \leq \frac{25}{\sqrt{90}}\right\} 
= \Phi\left(\frac{25}{\sqrt{90}}\right)$$
just as we got from the Demoivre-Laplace theorem.
A: You are computing the value for going over $83$ standard deviations above the mean. The chance of going above that are approximately
$$
\frac1{\sqrt{2\pi}}\frac1{83}e^{-83^2/2}
$$
which is very small; $\sim2.5\times10^{-1494}$. The chance of getting less than 83 standard deviations is therefore approximately $1$.
However, as Dilip Sarwate points out, we should use the standard deviation of $9.4868$ and not $\frac{9.4868}{\sqrt{1000}}$. That puts us at $2.6352$ standard deviations above the mean for $0.99580$ probability. 
