How to find "n" by using weak law of large number?

Question

Use the weak law of large numbers to find a number $n$ such that if one tosses a (fair) coin $n$ times the proportion of "heads" will lie in the range $[0.49, 0.51]$ with probability at least $0.95$.

I really have no idea on how to start working on this question as the value of variance is not provided. Can anyone help me for it?

Answer 1

The weak law of large numbers only states that the expectation value $\hat{p}$ of your probability of observing head converges to the true value $p=0.5$ for $n\rightarrow \infty$. In order to calculate confidence intervals you need something more. The repeated toss of fair coins follows a binomial distribution, which is very well approximated by a normal distribution according to the central limit theorem. You then get the 95% confidence interval

$$ \hat{p}\pm 1.96 \sqrt{\hat{p}(1-\hat{p})/n} $$

In your case $\hat{p} = 0.5$ (according to the weak law of large numbers) and we obtain the required number of tosses $n\approx 10^4$.

Answer 2

Using Cebishev's inequality you have

$$\mathbb{P}[|\overline {X}_n-\mu|\leq \epsilon]\geq 1-\frac{V(\overline {X}_n)}{\epsilon ^2}$$

You have only to substitute and set the probability at 0.95

The variance of the coin flip is 0.25 thus the variance $V(\overline {X}_n)=\frac{0.25}{n}$