Use Chebychev's Inequality to find a lower bound. Let $X$ be the number of heads one would obtain in $140$ flips of a fair coin.
Use Chebychev's Inequality to find a lower bound on the probability $P(60 < X < 80)$.
Okay so Chebychev's Inequality is $P(|X - E(X)| > kσ) \le 1/k^2$ for $ k > 0$, where $σ^2$ is the variance of $X$.
I'm not sure how to fill this in or anything. My probabilty test is tomorrow so help is much appreciated! Descriptive answers would be awesome.
 A: Before the solution, a minor comment. 
The Chebyshev Inequality is not quite quoted correctly. It should be 
$$\Pr(|X-\mu|\ge k\sigma)\le \frac{1}{k^2}.\tag{$1$}$$
For continuous distributions there is no need to distinguish between $\le$ and $\lt$. Here we are working with a discrete distribution. 
A standard calculation shows that in our case $\mu=np=70$ and $\sigma^2=np(1-p)=35$.  We want a lower bound on  $\Pr(60\lt  X\lt 80)$. The complementary event is $|X-70|\ge 10$. We first find an upper bound for $\Pr(|X-70|\ge 10)$.  
Compare with Inequality $(1)$ quoted above. In our case we have  $k\sigma=10$, and therefore 
$$k=\frac{10}{\sigma}, \quad\text{so}\quad\frac{1}{k^2}=\frac{\sigma^2}{100}=\frac{35}{100}.$$
It follows that $\Pr(|X-70|\ge 10)$ is $\le \frac{35}{100}$. Thus 
$$\Pr(60\lt  X\lt 80)\ge 1-\frac{35}{100}=\frac{65}{100}.$$
That is the lower bound given by the Chebyshev Inequality.
Remark: It is not a very good lower bound. You might want to use software such as the free-to-use Wolfram Alpha to calculate the exact probability. It's not Chebyshev's fault. An inequality that works for every distribution that has a mean and variance, including some pretty weird ones, cannot be expected to compete against estimates based on more information. 
A: Hint: You should recognize this experiment as a series of Bernoulli trials, and so its probability distribution is given by the binomial distribution! Compute the mean and variance and fill in the blanks and you are basically done! 
A: Note that $X$ is binomial with $n=140, p = 0.5$ so $\mathbb{E}[X] = np = 70$, $\sigma^2 = Var(X) = np(1-p) = 35,$ giving $\sigma = \sqrt{35} $ .
Now,
$\begin{split}
\mathbb{P}[60 < X < 80]
  &= \mathbb{P}[-10 < X - \mathbb{E}[X] < 10] \\
  &= \mathbb{P}[|X - \mathbb{E}[X]| < 10] \\
  &= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10}{\sqrt{35}} \sigma]\\
  &= \mathbb{P}[|X - \mathbb{E}[X]| < \frac{10\sigma}{\sqrt{35}}]
\end{split}
$
By Chebyshev's Inequality,
$\mathbb{P}[X \not \in (60,80)]
  = \mathbb{P}[|X - \mathbb{E}[X]| > \frac{10\sigma}{\sqrt{35}}]
  \leq (\frac{\sqrt{35}}{10})^2 = 35/100 = 7/20$.
Hence,
$\mathbb{P}[60 < X < 80]
   = 1 - \mathbb{P}[X \not \in (60,80)]
   \geq 1-7/20 = 13/20.
$
