proportion of the voters/ Central limit theorem

Question

I want to compute the proportion of the voters p. Therefore I consider random variables $X_k$ for $k=1,...,n$: $$ X_k=\left\{\begin{array}{ll} 1, party \ is \ elected: "p" \\ 0, party \ is \ not \ elected: "1-p" \end{array}\right. $$ I can approximate $p$ by $p'_n= \frac{1}{n} \sum_{k=1}^n X_k $ Now I want to calculate the number of voters n, so that $$ P(|p'_n-p| \leq 0,02) \geq 0,95$$ holds As a hint I should use $p(1-p)\leq \frac{1}{4}$ How can I get to this inequality and how can I compute this using central limit theorem?

Answer 1

The idea is that the central limit theorem tells us that $p_n'$ can be approximated like a normal random variable with mean $p$ and some variance.

Can you compute what the variance is?

Let $\sigma_n^2$ denote the variance---and thus $\sigma_n$ the standard deviation; then the variable $$\frac{p_n' - p}{\sigma_n}$$ is close to a standard normal.

Knowing that $(p_n' - p)/\sigma_n$ is roughly a standard normal, how can you examine the probability $P(|p_n' - p| \leq .02)$?