Suppose that $X$ is a random variable with the unknown mean $\mu$ and variance $\sigma$ 100. How many observations of $X$ must be taken? Problem: Suppose that $X$ is a random variable with the unknown mean $\mu$ and variance 100. How many observations of $X$ must be taken so that the probability $X$ within 2 units of $\mu$ is 0.99 ?
This is how I did it.
Let us consider, $E[X_1]$ = $\mu$, so $E[\frac{1}{n}\Sigma(X_i)]$ = $\frac{1}{n}$ $E[\Sigma(X_i)]$ = $\mu$, and $Var[X_i]$ = $\sigma^{2}$ = 100. Now, $Var[\overline{X}]$ = $Var[\frac{1}{n} \Sigma(X_i)]$ = $\frac{1}{n^{2}}$ $\Sigma(X_i)$ = $\frac{100}{n}$
Now, using Chebyshev's inequality, $P[ |\overline{X}-\mu| \leq a]$ $\geq$ $ 1 - \frac{Var(\overline{X})}{a^{2}}$. Here, $a = 2$, therefore plugging in that value gives $P[ |\overline{X}-\mu| \leq 2]$ $\geq$ $ 1 - \frac{Var(\overline{X})}{2^{2}}$, So,  $1 - \frac{100}{4n}$ = 0.99 (Given), therefore calculating for n, we get n = 2500. Therefore, the total number of observations, $n$ = 2500.
I checked this with my prof, and  he told me that its wrong. Can someone help me on this? I don't know where my mistake is and how I calculated it wrong. Appreciate your help and support.    
 A: As far as I can tell you correctly applied Chebychev to answer "how many independent samples do you need for $\bar{X}$ to be within $2$ units of $\mu$ with probability $0.99$." (Is this the question you are asking? Is there a typo in the question statement in your question? "Probability $X$ is within..." does not make sense.)
Your work correctly shows that with $n \ge 2500$ independent samples, $P(|\bar{X}-\mu| \le 2) \ge 0.99$.
It is unclear why your professor said it is wrong. It is possible he/she is looking for a smaller $n$. Chebychev's inequality guarantees that $n \ge 2500$ is enough, but because it is an inequality, it might be that you don't need that many samples.
As suggested in the comments, for large $n$, the sample mean $\bar{X}$ is approximately normal (with mean $\mu$ and variance $\sigma^2/n$) and thus $Z:=\frac{\bar{X}-\mu}{\sigma^2/n}$ approximately standard normal. Then
$$P(|\bar{X}-\mu| \le 2) = P(|Z| \le 2n/\sigma^2)$$
is approximately the area under the standard normal density between $-2n/\sigma^2$ and $2n/\sigma^2$. You can show (by looking up in a table or using a computer) that the area from $-2.576$ to $2.576$ under the standard normal density is $0.99$, so setting $2n/\sigma^2 = 2.576$ yields $n = 2.576 \sigma^2/2 \approx 129$. This result is valid only if the normal approximation is good here.
