\documentclass[a4paper,12pt]{article}
\begin{document}
QUESTION A random sample of size $n$ is taken without replacement
from a very large sample of components and $r$ of the sample are
found to be defective. Write down an approximate$99\%$ confidence
interval for the proportion of the population which are defective
stating clearly \underline{three} reasons who your interval is
only approximate.
If $n=400$ show that the length of the longest such interval is
about 0.13.
ANSWER
99\% CI approximately
$\frac{r}{n}\pm 2.58 \sqrt{\frac{\frac{r}{n}(1-\frac{r}{n})}{n}}$\\
The distribution is really Hypergeometric but the batch is very
large so the approximate distribution is Binomial n,p, n large
hence we can use the normal to approximate. Variance
=$p\frac{q}{n}$ but we use $\frac{r}{n}$ for p as an
approximation.
n=400, Length of interval $2\times2.58
\sqrt{\frac{\frac{r}{n}(1-\frac{r}{n})}{n}}$. (p(1-p)is maximum
when p$=\frac{1}{2}$)\\
Hence maximum length $2\times 2.58 \sqrt
{\frac{\frac{1}{2}\frac{1}{2}}{400}}=\frac{2.58}{20}\approx
0.13.$
\end{document}