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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 $P(|S_n| \ge x) \ge \ldots\,{}$?

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  • $\begingroup$ Chebyshev's inequality applies to all distributions on the real line, not just to binomial distributions. But you've specified a binomial distribution. There may be inequalities of the sort you seek that would apply to binomial distributions but not to other distributions. It's unclear whether such an inequality would answer your question. $\endgroup$ – Michael Hardy Feb 7 '15 at 1:00
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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 $P(|S_n| \leq x)$.

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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.

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