2. Re: Chebyshev Inequality

let $\bar{X}$ be the sample mean. It has variance $\sigma^2 = \dfrac{\sigma_p^2}{n}$
and thus $\sigma = \dfrac{\sigma_p}{\sqrt{n}}$

Assume worst case and take $\sigma_p^2 = 5$

$\sigma=\sqrt{\dfrac{5}{n}}$

by Chebyshev

$P[|\bar{X}-\mu| \geq k \sigma] \leq \dfrac{1}{k^2}$

we want to ensure with 95% confidence that $|\bar{X}-\mu| \leq 1$ so

$1 = k \sigma$

$k = \dfrac{1}{\sigma} = \sqrt{\dfrac{n}{5}}$

and we need $\dfrac{1}{k^2} \leq 0.05 \Rightarrow k \approx 4.472$

$\sqrt{\dfrac{n}{5}} \approx 4.472$

$n = 100$