So I am pondering the following question: Suppose $A$ is an event which has the probability of 0.4 of occurring. How many trials must be performed to assert with probability 0.9 that the relative frequency of A differs from 0.4 by no more than 0.1.
Here is what I tried. Let
$$ S_n = \xi_1 + \cdots + \xi_n $$ where each $\xi_i$ is the random variable which gives 1 if $A$ occurs and 0 otherwise. Then we are trying to determine the smallest $n$ for which $$ P(|S_n/n - 0.4| > 0.1) \leq 0.1. $$
By Chebyshev's inequality we have $$ P(|S_n/n - 0.4| > 0.1) \leq \frac{(0.4 \cdot 0.6)}{n (0.1)^2} $$ and we want to make the right hand side smaller than 0.1. This yields the inequality $$ 10^3 \cdot (0.4 \cdot 0.6) = 240 \leq n $$ which seems to say that we need $n$ to be at least about 240.
However, the book I am using claims the answer is about 65, which I am not sure where that is coming from. Is it the case that Chebyshev's inequality here is too crude and there are better ways of getting $n$?
Thanks!
edit: previously I misapplied Chebyshev's inequality getting a different lower bound.
Chebyshev's inequality is for upper bound, which is not very exact. But you can use the central limit theorem. Basically the equation is
$$2\cdot \Phi\left(\frac{\frac{S_n}{n}-\mu}{\sqrt{Var\left(\frac{S_n}{n} \right)}} \right)-1=0.9$$
$$2\cdot \Phi\left(\frac{0.1}{\sqrt{\frac{0.4\cdot 0.6}{n}}} \right)-1=0.9$$
$2\cdot \Phi(x_0)-1$ is the area bounded by the interval from $-x_0$ to $x_0$ of the standard normal distribution.