How do I find the variance using the central limit theorem and knowing the estimator?

Question

In a business, a particular article is mass-produced. A certain percentage of production is considered to be defective. We assume that a randomly selected article is defective with probability $p$ and that the articles are produced independently of each other.

The company is interested in estimating $p$. Let $X$ denote the number of defects among a randomly selected on $n$ produced items. An expectant estimator for $p$ is $\hat{p} = \frac{X}{n}$ which is unbiased, i.e. $E[\hat{p}] = p$.

Assuming that $p=0.12$ and that $n=53$ we are supposed to use that $\hat{p}$ under these circumstances are approximately normal distributed. Using this we are suppposed to calculate $P(|\hat{p} - p| < 0.05)$

How do I solve these kinds of problems? I have usually solved problems where we could easily find the mean and variance, but now we only know one of them.

Answer 1

You know that the underlying distribution is binomial, and the variance of the binomial distribution can be calculated.

Denote the binomial distribution by $B(n, p)$, and let $X \sim B(1, p)$ be the distribution with $p(X=0) = 1-p$ and $p(X=1) = p$. Then $E(X) = E(X^2) = p$ and therefore $\operatorname{Var}(X) = E(X^2) - E(X)^2 = p (1-p)$.

Therefore, the variance of $B(n, p)$ is $np(1-p)$ and the variance of $B(n, p)/n$, which is the distribution of the fraction (rather than number) of defects, is $p(1-p)/n$.

Answer 2

Try this: First, evaluate $\sigma_{\hat p} = SD(\hat p) = \sqrt{\frac{p(1-p)}{n}}.$

Then $$P(|\hat p - p| < .05) = P(-.05 < \hat p - p <.05) = P\left(\frac{-.05}{\sigma_{\hat p}} < \frac{\hat p - p}{\sigma_{\hat p}} < \frac{.05}{\sigma_{\hat p}}\right).$$ Notice that $\frac{\hat p - p}{\sigma_{\hat p}} $ is approximately standard normal, plug in the known numerical values, and evaluate the probability. I got about 69%.