
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 $\displaystyle ${15}/{16}$ $ of all bitstrings $\displaystyle $x \in \{0,1\}^n$ $ of length n have hamming weight that could satisfy the following relation $\displaystyle $ {n}/{2}  {4 \sqrt{n}}/{2} \leq wt(x) \leq {n}/{2} + {4 \sqrt{n}}/{2}$$. wt(x) is hamming wieght.
Thanks in advance
Regards,

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.