# Chebyshev inequality

• September 18th 2012, 08:12 AM
robmas
Chebyshev inequality
Hi all,
I would appreciate if someone could help me with this question.

How I can ,using Chebyshev inequality, prove that a least ${15}/{16}$ of all bitstrings $x \in \{0,1\}^n$ of length n have hamming weight that could satisfy the following relation ${n}/{2} - {4 \sqrt{n}}/{2} \leq wt(x) \leq {n}/{2} + {4 \sqrt{n}}/{2}$. wt(x) is hamming wieght.

Regards,
• September 18th 2012, 03:06 PM
awkward
Re: Chebyshev inequality
I think you have to make the assumption that all the bitstrings are equally likely. With that assumption, the hamming weight has a Binomial(n, p) distribution with p = 1/2. From this you can get the mean and standard deviation and apply Chebyshev's inequality.