Chebyshev Application

Question

If we have a sample mean of $\overline{X} = (X_1 + X_2+\ldots+ X_n)/n$ and mean $m$ and standard deviation $s$, how large should the sample size $n$ be so that with probability $.99$ the error $|\overline{X} - m|$ is less than 2 standard deviations? (This should be done according to Chebyshev's inequality)

Answer 1

Assuming that $X_i$'s are independent you have $var(\bar{X})=\frac{s^2}{n}$, and $E \bar{X}=m$.

Then, by Chebyshev's inequality you have

$$P(|\bar{X}-m| >2)\le \frac{s^2}{4n}=0.01$$

Now you get that $n=25s^2$ does the job...