Chebyshev’s inequality and variance I have the following exercice:


  Use Chebyshev’s inequality to find an upper bound on the probability that the number of tails that come up when a fair coin is tossed n times deviates from the mean by more than $5\sqrt{n}$.


My question is: how should I approach this kind of problems? I know I will need the variance to use Chebyshev's inequality, and it seems fair to say that the expected value of the number of tails that show up when a fair coin is tossed $n$ times will be $n/2$. But this doesn't allow me to calculate $\mathbb{E}[X^2]$, and I'm really lost at how I should find them with a bound such as n tosses.

Any input is much appreciated!
Hint: Each coin flip is an independent Bernoulli random variable $X_i$ of probability $1/2$ and variance $(1-1/2)(1/2)=1/4$. Since $X_i$ are independent the variance of $X = \sum_{i=1}^n X_i$ (n coin flips) is 
\begin{equation}
Var(X) = \sum Var(X_i) = n/4
\end{equation}
Hint: Each coin flip is an independent Bernoulli random variable $X_i$ of probability $1/2$ and variance $(1-1/2)(1/2)=1/4$. Since $X_i$ are independent the variance of $X = \sum_{i=1}^n X_i$ (n coin flips) is 
\begin{equation}
Var(X) = \sum Var(X_i) = n/4
\end{equation}