91Ó°ÊÓ

The time in minutes required to serve a customer at a certainfacility has an exponential distribution for which the value of the parameterθis unknown and the prior distribution ofθis a gamma distribution for which the mean is 0.2 and the standard deviation is 1.

The mean of the gamma distribution with parameters\(\alpha \,\,{\rm{and}}\,\,\beta \)is\(\alpha /\beta \)and standard deviation is\({\alpha ^{1/2}}/\beta \)

\(\begin{aligned}{\rm{Coefficient}}\,{\rm{of}}\,{\rm{variation}} = \frac{{{\rm{standard}}\,{\rm{deviation}}}}{{{\rm{mean}}}}\\ = \frac{{{\alpha ^{1/2}}/\beta }}{{\alpha /\beta }}\\ = {\alpha ^{ - 1/2}}\end{aligned}\)

Therefore, the coefficient of variation is\({\alpha ^{ - 1/2}}\). Since the coefficient of variation of the prior gamma distribution of\(\theta \)is 2, it follows that\(\alpha = 1/4\)in the prior distribution. Furthermore, it now follows from Theorem 7.3.4 that the coefficient of variation of the posterior gamma distribution of\(\theta \)is\({\left( {\alpha + n} \right)^{ - 1/2}} = {\left( {n + 1/4} \right)^{ - 1/2}}\). This value will be less than 0.1

\(\begin{aligned}{\left( {n + \frac{1}{4}} \right)^{ - 1/2}} \le 0.1\\{\left( {n + \frac{1}{4}} \right)^{1/2}} \ge 10\\n + \frac{1}{4} \ge 100\\n \ge 99.75\end{aligned}\)

Thus, the required sample size is\(n \ge 100\)

So the required sample size is 100