91Ó°ÊÓ

A physicist makes 25 independent measurements of the specific gravity of a particular body, i.e.,\(n = 25\)

The standard deviation for each measurement is \(\sigma \) units

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,

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \ge \frac{\sigma }{4}} \right) \le \frac{{{\sigma ^2}}}{n} \times \frac{{16}}{{{\sigma ^2}}}\\ = \frac{{16}}{{25}}\end{array}\)

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) \ge 1 - \left( {\frac{{{\sigma ^2}}}{n} \times \frac{{16}}{{{\sigma ^2}}}} \right)\\ = 1 - \frac{{16}}{{25}}\\ = \frac{9}{{25}}\\ = 0.36\end{array}\)

\(\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) = 0.36\)

Therefore, the lower bound is 0.36

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{5}{\sigma }\left( {{{\bar X}_n} - \mu } \right)\end{array}\)

Will be approximately a standard normal distribution.

\(\begin{array}{c}\Pr \left( {\left| {{{\bar X}_n} - \mu } \right| \le \frac{\sigma }{4}} \right) = \Pr \left( {\left| Z \right| \le \frac{5}{4}} \right)\\ \approx 2\Phi \left( {1.25} \right) - 1\\ = \left( {2 \times 0.8944} \right) - 1\\ = 0.7888\end{array}\)

Therefore, the approximate value for the probability for part (a) is 0.7888