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{\bf Question}

Suppose that the proportion of defective items in a large
manufactured lot is 0.1.  What is the smallest random sample of
items that must be taken from the lot in order for the probability
to be at least 0.99 that the proportion of defective items in the
sample will be less than 0.13?



\vspace{.25in}

{\bf Answer}

Let $X_i = \left\{\begin{array} {cl} 1 &amp; {\rm if\ the\ ith\
chosen\ item\ is\ defective}\\ 0 &amp; {\rm otherwise}
\end{array} \right\}$

Now $\bar X_n=\frac{1}{n} \ds\sum_{i=1}^n X_i =\ $sample
proportion of defective items

The problem is what minimum $n$ we require so that $$P(\bar
X_n&lt;0.13) \geq 0.99$$

Here $E(X_i)=\mu=0.1$ and ${\rm
var}(X_i)=(0.1)(1-0.1)=0.09=\sigma^2$ since $X_i$ are Bernoulli
random variables.

CLT

$\ds Z = \frac{\sqrt{n}(\bar X_n-\mu)}{\sigma} \sim N(0,1)$
approximately when $n$ is large.

Therefore \begin{eqnarray*} &amp; &amp; P(\bar X_n&lt;0.13)\\ &amp; = &amp;
P\left(\frac{\sqrt{n}(\bar X_n-\mu)}{\sigma}&lt;\frac{(0.13-0.1)\sqrt
n}{\sqrt{0.09}}\right)\\ &amp; = &amp; P\left(Z&lt;\frac{0.03}{0.03}\sqrt
n\right)\ {\rm where}\ Z \sim N(0,1)\\ &amp; = &amp; P\left(Z&lt;\frac{\sqrt
n}{10}\right) \end{eqnarray*}

i.e. $\Phi\left(\frac{\sqrt n}{10}\right) \geq \Phi(2.326) = 0.99$

$\Rightarrow \frac{n}{100} \geq 2.326^2$ i.e. $n \geq 541.02$

Therefore the smallest $n$ we require is \underline{542}.



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