91Ó°ÊÓ

A random sample of n items with a mean of\(\mu \)and a standard deviation of\(\sigma \)

The relation is,

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) \ge 0.99\)

Using the Chebyshev inequality,

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \ge t} \right) \le \frac{{{\sigma ^2}}}{{n{t^2}}}\)

And

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le t} \right) \ge 1 - \frac{{{\sigma ^2}}}{{n{t^2}}}\)

So,

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \ge \frac{\sigma }{4}} \right) \le \frac{{{\sigma ^2}}}{n} \times \frac{{16}}{{{\sigma ^2}}}\\ = \frac{{16}}{n}\end{array}\)

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) \ge 1 - \left( {\frac{{{\sigma ^2}}}{n} \times \frac{{16}}{{{\sigma ^2}}}} \right)\\ = 1 - \frac{{16}}{n}\end{array}\)

We know that,

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) \ge 0.99\)

i.e.,

\(\begin{array}{c}1 - \frac{{16}}{n} \ge 0.99\\\frac{{16}}{n} \ge 1 - 0.99\\\frac{n}{{16}} \ge 100\\n \ge 1600\end{array}\)

Therefore, the smallest number is, \(n \ge 1600\)

Using the central limit theorem,

\(\begin{array}{c}Z = \frac{{{{\bar X}_n} - \mu }}{{{\raise0.7ex\hbox{$\sigma $} \!\mathord{\left/{\vphantom {\sigma {\sqrt n }}}\right.}\!\lower0.7ex\hbox{${\sqrt n }$}}}}\\ = \frac{{\sqrt n }}{\sigma }\left( {{{\bar X}_n} - \mu } \right)\end{array}\)

It will be approximately a standard normal distribution.

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) = \Pr \left( {\left| Z \right| \le \frac{{\sqrt n }}{4}} \right)\\ = 2\Phi \left( {\frac{{\sqrt n }}{4}} \right) - 1\end{array}\)

Now,

\(\begin{array}{c}2\Phi \left( {\frac{{\sqrt n }}{4}} \right) - 1 \ge 0.99\\2\Phi \left( {\frac{{\sqrt n }}{4}} \right) \ge 1 + 0.99\\\Phi \left( {\frac{{\sqrt n }}{4}} \right) \ge \frac{{1.99}}{2}\\\Phi \left( {\frac{{\sqrt n }}{4}} \right) \ge 0.995\end{array}\)

\(\begin{array}{c}\frac{{\sqrt n }}{4} \ge {\Phi ^{ - 1}}\left( {0.995} \right)\\\frac{{\sqrt n }}{4} \ge 2.567\\\sqrt n \ge 10.268\\n \ge 105.431824\end{array}\)

Therefore, the smallest possible sample size is 106