91Ó°ÊÓ

A huge produced batch contains 0.1 percent faulty goods.

The proportion of defective items is 0.13

The distribution of the proportion\({\bar X_n}\)of defective items in the sample will be approximately the normal distribution, with the mean is,

\(\mu = 0.1\)

The variance is,

\(\begin{array}{c}{\sigma ^2} = \frac{{0.1 \times 0.9}}{n}\\ = \frac{{0.09}}{n}\end{array}\)

The standard deviation is,

\(\begin{array}{c}\sigma = \sqrt {\frac{{0.09}}{n}} \\ = \frac{{0.3}}{{\sqrt n }}\end{array}\)

The distribution,

\(\begin{array}{c}Z = \frac{{\sqrt n \left( {{{\bar X}_n} - \mu } \right)}}{\sigma }\\ = \frac{{\sqrt n \left( {{{\bar X}_n} - 0.1} \right)}}{{0.3}}\end{array}\)

Then the distribution of Z will be an approximately standard normal distribution.

It follows that,

\(\begin{array}{c}\Pr \left( {{{\bar X}_n} < 0.13} \right) = \Pr \left( {Z < 0.1\sqrt n } \right)\\ \approx \Phi \left( {0.1\sqrt n } \right)\end{array}\)

But,

\(\begin{array}{c}\Phi \left( {0.1\sqrt n } \right) \ge 0.99\\0.1\sqrt n \ge {\Phi ^{ - 1}}\left( {0.99} \right)\\0.1\sqrt n \ge 2.327\\\sqrt n \ge 23.27\end{array}\)

\(\begin{array}{c}n \ge 541.4929\\n \ge 541.5\end{array}\)

Therefore, the smallest possible value of n is 542