Approximation of the binomial distribution Let $S_n=\dfrac{B_n - np}{\sqrt{n\cdot p\cdot (1-p)}}$ be a random variable which has the standardized binomial distribution. From Chebyshev's inequality I know that $$P(|S_n| \ge x) \le \frac{1}{x^2}$$ Is there also a good approximation of the form $P(|S_n| \ge x) \ge \ldots\,{}$?
 A: Start by expressing the binomial distribution as
$$
P(B_n = k) = \binom{n}{k}p^k(1-p)^{n-k}
$$
so that
$$ 
P(|S_n| \leq  x) = \sum_{i=np-x\sqrt{np(1-p)}}^{np-x\sqrt{np(1-p)}}\binom{n}{k}p^k(1-p)^{n-k}
$$
and express 
$$
\binom{n}{k} = \frac{n!}{k!(n-k)!}
$$
Then use the Stirling approximation with the first 2 non-trivial terms:
$$
\sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}-\frac{1}{360n^3}} < m! < \sqrt{2\pi n} \left(\frac{n}{e}\right)^n e^{\frac{1}{12n}}
$$
This yields really tight bounds on $P(|S_n| \leq  x)$.
A: For large values of $n$, $S_n$ has a CDF that is well-approximated by the CDF
of a standard normal random variable. Thus, one approximation to a
lower bound is
$$P\{|S_n| \geq x\} \geq  \sqrt{\frac{2}{\pi}}\left (\frac{1}{x} - \frac{1}{x^3}\right )\exp(-x^2/2)~~ \text{for}~~ x > 0.$$
The corresponding upper bound is
$$P\{|S_n| \geq x\} \leq  \sqrt{\frac{2}{\pi}}\frac{\exp(-x^2/2)}{x}
~~ \text{for}~~ x > 0.$$
See, for example, this answer for
details on how to arrive at these bounds.
