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.
Best Answer
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.