91Ó°ÊÓ

The samples X1,.., and Xn come from a uniform distribution

Let \(U = \max \left\{ {{X_1},...,{X_n}} \right\}\)

The CDF of U is

\(F\left( u \right) = \left\{ \begin{align}{l}0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,u \le 0\\{\left( {u/\theta } \right)^n}\,\,\,\,\,\,for\,0 < u < \theta \\1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,for\,u \ge \theta \end{align} \right.\)

Since, \(U \le \theta \) with probability 1, the event \(|U - \theta | \le 0.10\) is the same as the probability of this \(1 - F\left( {0.9\theta } \right) = 1 - {0.9^n}\) .

For this to be at least 0.95, we need

\(\begin{array}{0.9^n} \le 0.05\\n \ge \log \left( {0.05} \right)/\log \left( {0.9} \right)\\n \ge 28.43\end{array}\)

So, \(n \ge 29\) is needed.