Finding $n$ using Chebyshev's inequality The height of a person is a random variable with variance $\leq 5$ square inches. According to Mr. Chebyshev, how many people do we need to sample to ensure that the sample mean is at most $1$ inch away from the distribution mean with probability $\geq 95$%?  
According to Chebyshev's inequality, 

$$P(|X-\mu| \leq \alpha) \geq \sigma^2/\alpha^2.$$

In my case I have $\alpha=1$. 
But I am not able to understand how to apply this formula into the question.  
 A: Chebyshev's inequality is of the form
$\newcommand{\Var}{\operatorname{Var}}$
$\newcommand{\E}{\operatorname{E}}$
$\def\Xbar{\,\overline{\!X}}$
$$P(|Y-\E(Y)| \geq \alpha) \leq \dfrac{\Var(Y)}{\alpha^2},$$
where $Y$ is a random variable.
(Please note that the $\geq$ and $\leq$ are reversed as to how you typed them.)
But as stated in your question, you are not to apply Chebyshev to a single random variable but rather a sample mean (i.e., a mean of random samples)
$$Y=\Xbar=\frac{X_1+\dots+X_n}{n}$$
where $X$ is a random variable (the height of a person) and the $X_1$, $\ldots$, $X_n$ are just $n$ numbers generated by $X$ at distinct times. It so happens that the $X_1$, $\ldots$, $X_n$ can be treated as independent, identically distributed random variables, and by the properties of expectation and variance we have
\begin{align*}
\E(\Xbar) &= \E\left(\frac{X_1+\dots+X_n}{n}\right)\\
&=\frac{\E(X_1)+\dots+\E(X_n)}{n}\\
&=\frac{n\E(X_1)}{n}\\
&=\E(X);\\
\Var(\Xbar) &= \Var\left(\frac{X_1+\dots+X_n}{n}\right)\\
&=\frac{\Var(X_1)+\dots+\Var(X_n)}{n^2}\\
&=\frac{n\Var(X_1)}{n^2}\\
&=\tfrac{1}{n} \Var(X).
\end{align*}
Then the Chebyshev's inequality for $Y=\Xbar$ takes the form
$$P(|\Xbar-\E(X)| \geq \alpha) \leq \dfrac{\Var(X)}{n\ \alpha^2},$$
and you can plug in your values.
But since you want the probability of the "$\leq$" case, the inequality yelds
$$P(|\Xbar-\E(X)| < \alpha) = 1 - P(|\Xbar-\E(X)| \geq \alpha) \geq 1 -\dfrac{\Var(X)}{n\ \alpha^2}.$$
P.S.: It is a little bit odd that your desired probability is $\geq 95$% rather than just $=95$%, since that would yield
$$0.95 \leq P(|\Xbar-\E(X)| < \alpha) \geq 1 -\dfrac{\Var(X)}{n\ \alpha^2}.$$
Then, there is no way to actually relate the leftmost side with the rightmost side.
P.S. $\boldsymbol 2$: Never mind the P.S. (which was a confussion all on my own); this answer suggests relating all the quantities as
$$P(|\Xbar-\E(X)| < \alpha) \geq 1 -\dfrac{\Var(X)}{n\ \alpha^2} \geq 1 -\dfrac{5}{n\ \alpha^2} \geq 0.95$$
since $1 -\Var(X)/n\alpha^2$ is an increasing function on $\alpha$.
