I'm confused how to solve the following question:
Consider a biased coin, in which Pr(heads)=0.6. Assume 100 independent flips of this coin, and use Chebyshev's inequality to bound Pr(heads≤50).
We are given the modified formula of Chebyshev's inequality: $$ Pr(|heads – \frac{n}{2}| ≥ c\sqrt{n}) ≤ \frac{V(heads)}{(c\sqrt{n})^2} $$
Edit:
I got 6/25, could someone help me check my answer? –
$$ Pr(|X-60| ≥ a) ≤ \frac{np(1-p)}{a^2} $$
$$ Pr(x≤50) ≤ Pr(x≤50 *or* x≥70) < (.24*100)/100 $$
Best Answer
The Chebyshev inequality says that $$ P(|X-E(X)| \ge c) \le \frac{\mathrm{Var(X)}}{c^2}.$$
Since the number of heads is binomially distributed with $n=100$ and $p=0.6,$ you can use the formula $\mathrm{Var(X) = np(1-p)}$ to compute the variance. The mean $E(X) = np = 60.$
The probability that there are $\le 50$ heads is the probability that there at least $10$ fewer than the mean. This is only one tail (whereas the inequality written above is for two tails) but it will still hold for the one tail, it will just be an even worse overestimate. So you can plug $c=10$ and whatever you calculate for $\mathrm{Var}(X)$ into the formula and get a valid inequality.